Introduction Linear Algebra. Session 1 Dr. Marco A Roque Sol 08/28/2018 Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Linear Algebra. Session 1
Dr. Marco A Roque Sol
08/28/2018
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
Introduction
Systems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equations
Gaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reduction
Matrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebra
DeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminants
Vector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spaces
Linear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independence
Basis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimension
Coordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basis
Linear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformations
OrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonality
Inner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and norms
The Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization process
Eigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value Decomposition
Matrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Table of contents
IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Algebra
( Medical Imaging )
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Algebra
( Medical Imaging )
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Algebra
( Medical Imaging )
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Algebra
( Medical Imaging )Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone
in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation.
It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used
by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence.
It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence,
evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction.
For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students,
it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas
where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications
ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.
Here are some concrete areas where we can find applications ofLinear Algebra.
Abstract Thinking
Chemistry
Coding theory
Coupled oscillations
Cryptography
Economics
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Classical Electromagnetism.
Geophysics.
Elimination Theory.
Game Theory.
Genetics.
Geometry.
Graph theory.
Heat distribution.
Image compression.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Linear Programming.
Markov Chains.
Networks.
Sociology
The Fibonacci numbers.
Eigenstates.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)
An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation
of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form
ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c
(5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6)
is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linear
because its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause
its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is
a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution
of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation
is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers
(x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2
suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat
ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c
(5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example,
(1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and
(−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1),
are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions.
In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,
we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write
the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as
x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5
and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as
x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2
A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).
For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1
And the graph of the lines is:
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
A Survey
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation
of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line
is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y
are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and
a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c
are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation
in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables
x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form:
a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution
of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation
is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn
such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = b
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Sistems of linear equations
General equation of a line
The General equation of a line is given by
ax0 + by0 = c
where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )
Definition
A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b
where a1, a2, · · · an and b are constants.
A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that
a1α1 + a2α2 + · · ·+ anαn = bDr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations
is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression
of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2
...am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here
x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn
are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and
aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj
are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications,
we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations
andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here,
where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study
of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra
starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
System of linear equations
System of linear equations
A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection
of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and
2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6,
in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2.
You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find
the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution
of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem
bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and
solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system
of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations
in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns
{x − y = −2
2x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System)
{x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔
{x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.1
Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {
x − y = −22x + 3y = 6
⇔ ( Equivalent System){x = y −2
2x + 3y = 6
⇔ {x = y − 2
2(y − 2) + 3y = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔
{x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔
{x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔
{x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus,
the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution
is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by
the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔ {x = y − 2
5y = 10
⇔ {x = y − 2y = 2
⇔ {x = 0y = 2
Thus, the solution is given by the point (0, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{x − y = −2
2x + 3y = 6x = 0; y = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{x − y = −2
2x + 3y = 6
x = 0; y = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{x − y = −2
2x + 3y = 6x = 0; y = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
In a similar way, we have
{2x + 3y = 22x + 3y = 6
inconsistent system (no solution)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
In a similar way, we have
{2x + 3y = 22x + 3y = 6
inconsistent system (no solution)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
In a similar way, we have
{2x + 3y = 22x + 3y = 6
inconsistent system (no solution)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
In a similar way, we have
{2x + 3y = 22x + 3y = 6
inconsistent system (no solution)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
In a similar way, we have
{2x + 3y = 22x + 3y = 6
inconsistent system
(no solution)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
In a similar way, we have
{2x + 3y = 22x + 3y = 6
inconsistent system (no solution)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{4x + 6y = 122x + 3y = 6
⇒ 2x+3y = 6 (infinitely many solutions)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{4x + 6y = 122x + 3y = 6
⇒ 2x+3y = 6 (infinitely many solutions)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{4x + 6y = 122x + 3y = 6
⇒ 2x+3y = 6 (infinitely many solutions)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{4x + 6y = 122x + 3y = 6
⇒ 2x+3y = 6
(infinitely many solutions)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
{4x + 6y = 122x + 3y = 6
⇒ 2x+3y = 6 (infinitely many solutions)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:
(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable,
solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and
eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit
from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation
used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and
return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm
reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables
(as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations),
hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after
a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops,
the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that
it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear
how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Solving systems of linear equations
Elimination Method
Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;
(2) put aside the equation used in the elimination, and return tostep (1).
The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2 x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :x = y + 2
2x − y− z = 3x + y+ z = 6
⇔Eliminate x from the 2nd and 3rd equations:
x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2
x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :x = y + 2
2x − y− z = 3x + y+ z = 6
⇔Eliminate x from the 2nd and 3rd equations:
x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2 x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :x = y + 2
2x − y− z = 3x + y+ z = 6
⇔Eliminate x from the 2nd and 3rd equations:
x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2 x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :
x = y + 2
2x − y− z = 3x + y+ z = 6
⇔Eliminate x from the 2nd and 3rd equations:
x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2 x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :x = y + 2
2x − y− z = 3x + y+ z = 6
⇔
Eliminate x from the 2nd and 3rd equations:x = y + 2
2(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2 x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :x = y + 2
2x − y− z = 3x + y+ z = 6
⇔Eliminate x from the 2nd and 3rd equations:
x = y + 2
2(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Example 1.2 x − y = 2
2x − y− z = 3x + y+ z = 6
Solve the 1st equation for x :x = y + 2
2x − y− z = 3x + y+ z = 6
⇔Eliminate x from the 2nd and 3rd equations:
x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔
Simplify: x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way
we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that,
the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced
to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations)
of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations
in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :
x = y+ 2y = z− 12y + z = 4
⇔
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
⇔Simplify:
x = y+ 2y− z = −12y + z = 4
In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.
Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4
⇔Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:
x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify
x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,
the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process,
has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed.
Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now,
thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem
is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by
back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
Eliminate y from the 3rd equation:x = y+ 2y = z+ 1
2(z − 1) + z = 4⇔
Simplify x = y+ 2y = z+ 13z = 6
⇔
Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is,
we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z
from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation,
then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it
in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and
find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y ,
then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z
in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and
find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally,
the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:
x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution,
the point (x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point
(x , y , z) = (3, 1, 2) .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Systems of linear equations
That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .
x = y + 2y = z + 1
z = 2
x = y + 2y = 1z = 2
x = 3y = 1z = 2
Finally, the System of linear equations:x − y = 2
2x − y− z = 3x + y+ z = 6
has as a solution, the point (x , y , z) = (3, 1, 2) .Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember
that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations
is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression
ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn
are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and
aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system
is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equations
present in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system
of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations
can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have
one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution,
infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or
no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
System of linear equations
Remember that a System of linear equations is an expression ofthe form:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
Here x1, x2, · · · xn are variables and aij , bj are constants.
A solution of the system is a common solution of all equationspresent in the system.
A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method
we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown
the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand
using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again
we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate
the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other
two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :
⇔ ( Equivalent Systems)x = −y+ 2z+ 1
y − z = 3−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1
y − z = 3−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases
Example 1.3 x + y− 2z = 1
y− z = 3−x + 4y− 3z = 14
Solve the 1st equation for x :⇔ ( Equivalent Systems)
x = −y+ 2z+ 1y − z = 3
−x + 4y− 3z = 14
Eliminate x from the 3rd equation :Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔ x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify : x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔
x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify : x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔ x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify : x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔ x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify :
x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔ x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify : x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔ x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify : x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for y
x = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
⇔ x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 14
Simplify : x = −y + 2z + 1
y − z = 35y − 5z = 15
Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:
x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔
Simplify : x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now,
the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed.
The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation
isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0.
Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z ,
is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is,
it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned
an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value.
Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y
are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by
backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3
5(z + 3)− 5z = 15
⇔Simplify :
x = −y+ 2z+ 1y = z + 3
0 = 0
Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
z = t (a parameter)
y = z + 3x = −y + 2z + 1
⇔ z = t
y = t + 3x = t − 2
Thus, the system x + y− 2z = 1
y − z = 3−x + 4y− 3z = 14
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
z = t (a parameter)
y = z + 3x = −y + 2z + 1
⇔
z = t
y = t + 3x = t − 2
Thus, the system x + y− 2z = 1
y − z = 3−x + 4y− 3z = 14
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
z = t (a parameter)
y = z + 3x = −y + 2z + 1
⇔ z = t
y = t + 3x = t − 2
Thus, the system x + y− 2z = 1
y − z = 3−x + 4y− 3z = 14
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
z = t (a parameter)
y = z + 3x = −y + 2z + 1
⇔ z = t
y = t + 3x = t − 2
Thus, the system
x + y− 2z = 1
y − z = 3−x + 4y− 3z = 14
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
z = t (a parameter)
y = z + 3x = −y + 2z + 1
⇔ z = t
y = t + 3x = t − 2
Thus, the system x + y− 2z = 1
y − z = 3−x + 4y− 3z = 14
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form
( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions
is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line
in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3
passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint
P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0)
in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction
v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
has the General Solution
(x , y , z) = (t − 2, t + 3, t)
or in vector form ( vector equation of a line )
(x , y , z) = (−2, 3, 0) + t(1, 1, 1)
The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.4
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
Solve the 1st equation for x :
⇔ (Equivalent System )x = −y + 2z + 1
y − z = 3−x + 4y − 3z = 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.4
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
Solve the 1st equation for x :
⇔ (Equivalent System )x = −y + 2z + 1
y − z = 3−x + 4y − 3z = 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.4
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
Solve the 1st equation for x :
⇔ (Equivalent System )x = −y + 2z + 1
y − z = 3−x + 4y − 3z = 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.4
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
Solve the 1st equation for x :
⇔ (Equivalent System )x = −y + 2z + 1
y − z = 3−x + 4y − 3z = 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.4
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
Solve the 1st equation for x :
⇔ (Equivalent System )
x = −y + 2z + 1
y − z = 3−x + 4y − 3z = 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.4
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
Solve the 1st equation for x :
⇔ (Equivalent System )x = −y + 2z + 1
y − z = 3−x + 4y − 3z = 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify: x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:
x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify: x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify: x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify:
x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify: x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify: x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :
x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate x from the 3rd equation:x = −y + 2z + 1
y − z = 3−(−y + 2z + 1) + 4y − 3z = 1
Simplify: x = −y + 2z + 1
y − z = 35y − 5z = 2
Solve the second equation for y :x = −y + 2z + 1
y = z + 35y − 5z = 2
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:
x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify:
x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now,
the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed.
The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equation
actually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us
a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Eliminate y from the 3rd equation:x = −y + 2z + 1
y = z + 35(z + 3)− 5z = 2
Simplify: x = −y + 2z + 1
y = z + 315 = 2
Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Thus, the system x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
has the General Solution
φ = empty set
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Thus, the system
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
has the General Solution
φ = empty set
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Thus, the system x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
has the General Solution
φ = empty set
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Thus, the system x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
has the General Solution
φ = empty set
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Thus, the system x + y− 2z = 1
y− z = 3−x + 4y− 3z = 1
has the General Solution
φ = empty set
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination
is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of
the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination method
that allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only
so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations
for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of
linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply
an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation
by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add
an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation
multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar
to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange
any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Gaussian elimination
Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .
Elementary operations for systems of linear equations:
(1) to multiply an equation by a nonzero scalar;
(2) to add an equation multiplied by a scalar to another equation;
(3) to interchange any two equations.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result,
is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important,
because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures
that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations
will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not
modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set
of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying
elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations
to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system
of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equations
does not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change
the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set
of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any
elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation
can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone
by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by another
elementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.
Theorem
(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.
(ii) Any elementary operation can be undone by anotherelementary operation.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply
the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation
by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by
r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo
the operation, multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation,
multiply the ith equation by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation
by r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by
r−1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 1: Multiply the ith equation by r 6= 0
a11x1 + a12x2 + · · ·+ a1nxn = b1...
ai1x1 + ai2x2 + · · ·+ ainxn = bi...
am1x1 + am2x2 + · · ·+ amnxn = bm
⇒
a11x1 + a12x2 + · · ·+ a1nxn = b1...
(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...
am1x1 + am2x2 + · · ·+ amnxn = bm
To undo the operation, multiply the ith equation by r−1.Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add
r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times
the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation
to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...
To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo
the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation,
add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times
the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equation
to the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 2: Add r times the ith equation to the jth equation
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi
...To undo the operation, add −r times the ith equationto the jth equation .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange
the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and
jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo
the operation, apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation,
apply it once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it
once more .
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Operation 3: interchange the ith and jth equations.
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...
⇒
...aj1x1 + aj2x2 + · · ·+ ajnxn = bj
...ai1x1 + ai2x2 + · · ·+ ainxn = bi
...
To undo the operation, apply it once more .Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5
x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add
−2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times
the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to
the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:
x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add
−1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times
the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to
the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:
x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Example 1.5 x − y = 2
2x − y − z = 3x + y + z = 6
Add −2 times the 1st equation to the 2nd equation:x − y = 2
y − z = −1x + y + z = 6
R2 := R2 − 2 ∗ R1
Add −1 times the 1st equation to the 3rd equation:x − y = 2
y − z = −12y + z = 4
R3 := (−1) ∗ R1 + R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add
−2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times
the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to
the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:
x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note:
At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point,
the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process
is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and
wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve
the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by
back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution.
However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However,
we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue
with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply
the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:
x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add −2 times the 2nd equation to the 3rd equation:x − y = 2
y − z = −13z = 6
R3 := (−2) ∗ R2 + R3
Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations
Multiply the 3rd equation by 1/3:x − y = 2
y − z = −1z = 2
R3 := (1/3) ∗ R3
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add
the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to
the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:
x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add
the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to
the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:
x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus,
the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system
x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has
as the solution set, the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set,
the point (3, 1, 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Gaussian Elimination. Gauss-Jordan Reduction
Add the 3rd equation to the 2nd equation:x − y = 2
y = 1z = 2
R2 := R2 + R3
Add the 2nd equation to the 1st equation:x = 3
y = 1z = 2
R1 := R2 + R1
Thus, the system x − y = 2
2x − y − z = 3x + y + z = 6
has as the solution set, the point (3, 1, 2)Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrix
n × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector
1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Matrices
Definition. A matrix is a rectangular array of numbers.
The dimension of a matrix is given by
dimensions = (number of rows) X ( number of columns)
Thus we have
n × n : Square matrixn × 1 : Column vector1× n : Row vector
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)
(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)
(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)
(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Examples of matrices are:
2 7−1 03 3
(3× 2)
(2 7 0−1 1 5
)(2× 3)
358
(3× 1)
(2 4 9
)(1× 3)
(−2 01 5
)(2× 2)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:
a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find
the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and
column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector
of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1
a21x1 + a22x2 + · · ·+ a2nxn = b2...
am1x1 + am2x2 + · · ·+ amnxn = bm
We can find the coefficient matrix and column vector of theright-hand sides:
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
b1
b2...
bm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also
associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system
we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,
remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination,
the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations
splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and
(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
and also associated to the linear system we have the Augmentedmatrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:
(A) Elimination and(B) Back substitution.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts
can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done
by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number
of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Both parts can be done by applying a finite number of elementaryoperations:
(1) to multiply a row by a nonzero scalar;
(2) to add the ith row multiplied by some r ∈ R to the jth row;
(3) to interchange two rows.
Notation
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Augmented matrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
=
v1v2...
vm
where vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Augmented matrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
=
v1v2...
vm
where vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Augmented matrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
=
v1v2...
vm
where vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Augmented matrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
=
v1v2...
vm
where
vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Augmented matrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
=
v1v2...
vm
where vi = (ai1 ai1 ai1 · · · ai1, bi )
is a row vector.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Augmented matrix:
a11 a12 · · · a1n b1
a21 a22 · · · a2n b2...
am1 am2 · · · amn bm
=
v1v2...
vm
where vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Elementary row operations on Matrices
Operation 1: To multiply the ith row by r 6= 0
v1...
vi...
vm
⇒
v1...
rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Elementary row operations on Matrices
Operation 1: To multiply the ith row by r 6= 0
v1...
vi...
vm
⇒
v1...
rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Elementary row operations on Matrices
Operation 1:
To multiply the ith row by r 6= 0
v1...
vi...
vm
⇒
v1...
rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Elementary row operations on Matrices
Operation 1: To multiply the ith row by r 6= 0
v1...
vi...
vm
⇒
v1...
rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Elementary row operations on Matrices
Operation 1: To multiply the ith row by r 6= 0
v1...
vi...
vm
⇒
v1...
rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Elementary row operations on Matrices
Operation 1: To multiply the ith row by r 6= 0
v1...
vi...
vm
⇒
v1...
rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 2: To add the ith row multiplied by r to the jth row
v1...
vi...
vj...
vm
⇒
v1...
vi...
vj + rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 2:
To add the ith row multiplied by r to the jth row
v1...
vi...
vj...
vm
⇒
v1...
vi...
vj + rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 2: To add the ith row multiplied by r to the jth row
v1...
vi...
vj...
vm
⇒
v1...
vi...
vj + rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 2: To add the ith row multiplied by r to the jth row
v1...
vi...
vj...
vm
⇒
v1...
vi...
vj + rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 2: To add the ith row multiplied by r to the jth row
v1...
vi...
vj...
vm
⇒
v1...
vi...
vj + rvi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 3: To interchange the ith row with the jth row
v1...
vi...
vj...
vm
⇒
v1...
vj...
vi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 3:
To interchange the ith row with the jth row
v1...
vi...
vj...
vm
⇒
v1...
vj...
vi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 3: To interchange the ith row with the jth row
v1...
vi...
vj...
vm
⇒
v1...
vj...
vi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 3: To interchange the ith row with the jth row
v1...
vi...
vj...
vm
⇒
v1...
vj...
vi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Operation 3: To interchange the ith row with the jth row
v1...
vi...
vj...
vm
⇒
v1...
vj...
vi...
vm
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry
of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is
the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of
the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination
is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert
the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix
into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries
shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right
as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from
the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row to
the last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each
leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry
is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to
1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
Definition. Leading entry of a matrix is the first nonzero entry ina row.
The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form
1) Leading entries shift to the right as we go from the first row tothe last one;
2) Each leading entry is equal to 1.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Thus, we have this example in the echelon form
1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Thus,
we have this example in the echelon form
1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Thus, we have this example
in the echelon form
1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Thus, we have this example in the echelon form
1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Thus, we have this example in the echelon form
1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Echelon FormationDr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries
are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed
(all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All
the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below,
the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary)
staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line
are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form
General augmented matrix in row echelon form
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
1) Leading entries are boxed (all equal to 1);
2) All the entries below, the (imaginary) staircase line are zero;
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step
of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the
(imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase
has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height
1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case
of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form
that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
3) Each step of the (imaginary) staircase has height 1;
4) Each circle marks a column without a leading entry thatcorresponds to a free variable.
Strict triangular form
Is a particular case of row echelon form that can occur for systemsof n variables :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� ∗ ∗ ∗
� ∗ ∗� ∗
1) No zero rows.
2) No free variables.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� ∗ ∗ ∗
� ∗ ∗� ∗
1) No zero rows.
2) No free variables.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� ∗ ∗ ∗
� ∗ ∗� ∗
1) No zero rows.
2) No free variables.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� ∗ ∗ ∗
� ∗ ∗� ∗
1) No zero rows.
2) No free variables.Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
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Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system
of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations
is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent
if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry
in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column
of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix
inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Consistency check
The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.
Augmented matrix of an inconsistent system
� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗
� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗
� � ∗ ∗� ∗
�
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal
of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction
is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert
theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix
into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries
below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line
are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero;
2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entry
is 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1,
the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries
in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column
are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero;
3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circle
corresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to
a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :
1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗
1 � ∗ ∗1 ∗ ∗
1 � ∗1 ∗
1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6
From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example,
we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):
x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.6From a previous example, we have
x − y = 1
2x − y− z = 3x + y+ z = 6
⇒
1 −1 0 12 −1 −1 31 1 1 6
Row echelon form (also strict triangular):x − y = 1
y − z = 1x + y+ z = 1
⇒
1 −1 0 1
0 1 −1 1
0 0 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :
x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Reduced row echelon form :x = 3
y = 1z = 2
⇒
1 0 0 3
0 1 0 1
0 0 1 2
Example 1.7
x + y− 2z = 1
y− z = 3−x + 4y− 3z = 6
⇒
1 1 −2 10 1 −1 3−1 4 −3 1
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Row echelon form:
x + y− 2z = 1
y− z = 30 = 1
⇒
1 1 −2 1
0 1 −1 3
0 0 0 1
Reduced row echelon form:x + − z = 0
y− z = 00 = 1
⇒
1 0 −1 0
0 1 −1 0
0 0 0 1
Inconsistent system
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down
the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix
of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix
to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix
to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row
echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down
the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding
to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
How to solve a system of linear equations
1) Order the variables.
2) Write down the augmented matrix of the system.
3) Convert the matrix to row echelon form
4) Check for consistency.
5) Convert the matrix to reduced row echelon form.
6) Write down the system corresponding to the reduced rowechelon form.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system
so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables
are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the left
while everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else
is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters
to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and
write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution
in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7
{x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case
the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are
x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1,
x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2,
x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3,
x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4.
and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
7) Determine leading and free variables.
8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.
9) Assign parameters to the free variables and write down thegeneral solution in parametric form.
Example 1.7 {x2 + 2x3 + 3x4 = 6
x1 + 2x2 + 3x3 + 4x4 = 10
In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)
To get it into row echelon form, we exchange the two rows:(1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it
into row echelon form, we exchange the two rows:(1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form,
we exchange the two rows:(1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange
the two rows:(1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:
(1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)
Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check
is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed.
To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert
into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform,
add −2 times the 2nd row to the 1st row:(1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times
the 2nd row to the 1st row:(1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row
to the 1st row:(1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:
(1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)
The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables
are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and
x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence
x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and
x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4
are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
(0 1 2 3 61 2 3 4 10
)To get it into row echelon form, we exchange the two rows:(
1 2 3 4 100 1 2 3 6
)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(
1 0 −1 −2 −2
0 1 2 3 6
)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:
{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒
{x1 = x3 + 2x4 − 2
x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given by
x1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form
( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6
⇒{
x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6
and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s
(t, s ∈ R)
In vector form ( vector equation of a plane in the space ),
(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8
y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system
is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous
(all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).
Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore
it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent
( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )
The augmented matrix is: 0 1 3 01 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is:
0 1 3 01 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Example 1.8 y + 3z = 0
x + y − 2z = 0x + 2y + az = 0
a ∈ R
The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0
1 1 2 01 2 a 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since
the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row
cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve
as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one,
we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange it
with the 2nd row: 0 1 3 01 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange it
with the 2nd row: 0 1 3 01 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row:
0 1 3 01 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now
we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination.
First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First
subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row
fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row:
1 1 −2 00 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0
1 1 −2 01 2 a 0
⇒
1 1 −2 00 1 3 01 2 a 0
Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0
0 1 3 01 2 a 0
⇒
1 1 −2 00 1 3 00 1 a + 2 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row
is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row.
Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row
fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row:
1 1 −2 00 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point
row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits
into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.
Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 00 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1
1 1 −2 00 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0
0 1 3 00 1 a + 2 0
⇒
1 1 −2 00 1 3 00 0 a− 1 0
At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix
is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted
into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form.
We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards
reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form.
Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row:
1 1 −2 00 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 00 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times
the 3rd row to the 1st row: 1 1 −2 00 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row
to the 1st row: 1 1 −2 00 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row:
1 1 −2 00 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0
0 1 3 00 0 1 0
⇒
1 1 −2 00 1 0 00 0 1 0
Add 2 times the 3rd row to the 1st row: 1 1 −2 0
0 1 0 00 0 1 0
⇒
1 1 0 00 1 0 00 0 1 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,
subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row
from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row from the 1st row:
1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0
is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Finally,subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0
⇒
1 0 0 00 1 0 00 0 1 0
Thus x = y = z = 0 is the only solution
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform:
1 1 −2 00 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get
reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form,
subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row
from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row:
1 1 −2 00 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0
0 1 3 00 0 a− 1 0
⇒
1 1 −2 00 1 3 00 0 0 0
To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0
0 1 3 00 0 0 0
⇒
1 0 −5 00 1 3 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then,
z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and
the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by
{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒
{x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:
x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1
then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then
(x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).
if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1
then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then
(x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1)
t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1
Introduction
Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction
Row echelon form. Gauss-Jordan Reduction
Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0
⇒{
x = 5z = 5ty = −3z = −3t
Thus, the System of linear equations:x + 3z = 0
x + y − 2z = 0x + 2y + az = 0
has as a solution:
if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.
Dr. Marco A Roque Sol Linear Algebra. Session 1