W7 Linear Equations

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Centre for Computer Technology

ICT114ICT114Mathematics forMathematics for

ComputingComputing

Week 7Week 7

Linear System of EquationsLinear System of Equations


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March 20, 2012March 20, 2012 http://mathworld.wolfram.comCopyright Box Hill Institute

ObjectivesObjectives

Determinant of a MatrixDeterminant of a Matrix

Inverse of a MatrixInverse of a Matrix

Inverse of a Matrix using Gauss JordanInverse of a Matrix using Gauss JordanMethodMethod

Solution of Equations using GaussSolution of Equations using Gauss

Elimination MethodElimination MethodSummarySummary


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Determinants are mathematical objects

that are very useful in the analysis and

solution of System of Linear EquationsOnly square matrices (no. of rows = no. of

columns) have determinants.

The determinant of a (square) matrix is ascalar.


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For a 2 X 2 matrix A with elementsFor a 2 X 2 matrix A with elements

the determinant of the matrix isthe determinant of the matrix is


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For a 3 X 3 matrix (shown below) theFor a 3 X 3 matrix (shown below) the

determinant of the matrix isdeterminant of the matrix is


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For a k X k matrix, A (expanded by

minors) the determinant of the matrix is


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Some PropertiesSome Properties

det(I) = 1 for any identity matrix I.

det(AB) = det(A) det(B)

det(AT) = det(A) If the determinant of a matrix is 0, it is said

to be singular.

If the determinant is 1, the matrix is said tobe unimodular.


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The inverse of aThe inverse of a square matrix Asquare matrix A , is a, is a matrixmatrix

AA-1-1 such thatsuch that

wherewhere I is the identity matrixI is the identity matrix

A square matrix A has an inverse only if it isA square matrix A has an inverse only if it is

nonsingular i.e.nonsingular i.e. det IAI 0.det IAI 0.


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For a 2 X 2 matrix, A with elementsFor a 2 X 2 matrix, A with elements

the determinant of the matrix isthe determinant of the matrix is


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For a 3 X 3 matrix A,For a 3 X 3 matrix A,

the inverse is given bythe inverse is given by


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Gauss Jordan MethodGauss Jordan Method

One more method to find the Inverse of a matrixOne more method to find the Inverse of a matrixis by using the Gauss Jordan method.is by using the Gauss Jordan method.

For a n X n square matrix A, we consider theFor a n X n square matrix A, we consider the

matrixmatrix

where I is the identity matrixwhere I is the identity matrix


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A series of row operations are performed toA series of row operations are performed to

obtain the matrix in the formobtain the matrix in the form


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The Matrix B is the inverse of AThe Matrix B is the inverse of A


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Example Gauss Jordan MethodExample Gauss Jordan Method

To find the Inverse of the matrix ATo find the Inverse of the matrix A


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March 20, 2012March 20, 2012

http://mathworld.wolfram.comCopyright Box Hill Institute

matrix A, augmented by the

3x3 identity matrix. The first

pivot encircled in red

row operations required

for the first pivoting

Next pivot on "3" in the (2,2)

position below, encircled in red


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March 20, 2012March 20, 2012


the result of performing theprevious operation , so the

pivot (2,2 position) is now "1"


for the second pivoting

The result of the second

pivoting is below. We now pivot

on the element in the (3,3)

position, encircled in red below


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the result of performing the

previous operation, so the

pivot (3,3 position) is now "1"


for the third pivoting

The result of the third (and last)

pivoting is below


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March 20, 2012March 20, 2012http://mathworld.wolfram.comCopyright Box Hill Institute

Therefore the inverse of the matrix ATherefore the inverse of the matrix A-1-1 isis


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QuestionQuestion

Find the inverse of the matrixFind the inverse of the matrix

22 -1-1 00A =A = -1-1 22 -1-1

00 -1-1 22


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The inverse of the matrix is AThe inverse of the matrix is A-1-1

0.7500 0.5000 0.25000.7500 0.5000 0.25000.5000 1.0000 0.50000.5000 1.0000 0.5000

0.2500 0.5000 0.75000.2500 0.5000 0.7500


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Solution of EquationsSolution of Equations

Suppose, we have to solve the equations:Suppose, we have to solve the equations:

2 x + 3 y = 7 .. (1)2 x + 3 y = 7 .. (1)

3 x + 4 y = 10 .. (2)3 x + 4 y = 10 .. (2)

By (1) * 3 .. 6 x + 9 y = 21By (1) * 3 .. 6 x + 9 y = 21

and (2) * 2 .. 6 x + 8 y = 20and (2) * 2 .. 6 x + 8 y = 20

By subtraction, y= 1By subtraction, y= 1

On substitution value of y in (1), x=2On substitution value of y in (1), x=2


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Solution of EquationsSolution of EquationsAnother method to solve the equations:Another method to solve the equations:

2 x + 3 y = 7 .. (1)2 x + 3 y = 7 .. (1)

3 x + 4 y = 10 .. (2)3 x + 4 y = 10 .. (2)

Divide (1) by 2 .. 1 x + 1.5 y = 3.5 . (3)Divide (1) by 2 .. 1 x + 1.5 y = 3.5 . (3)Multiply (3) by 3 and subtract from (2)Multiply (3) by 3 and subtract from (2)

0 x - 0.5 y = - 0.5 . (4)0 x - 0.5 y = - 0.5 . (4)

Divide (4) by 0.5 0 x + 1 y = 1 . (6)Divide (4) by 0.5 0 x + 1 y = 1 . (6)

Multiply ((6) by 1.5 and subtract from (3)Multiply ((6) by 1.5 and subtract from (3)

1 x + 0 y = 2 . (5)1 x + 0 y = 2 . (5)

Hence, x=2 and y =1Hence, x=2 and y =1


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March 20, 2012March 20, 2012



Consider the system of equationsConsider the system of equations

aa1111xx11+a+a1212xx22+a+a1313xx33 = b= b11

aa2121xx11+a+a2222xx22+a+a2323xx33 = b= b22

aa3131xx11+a+a3232xx22+a+a3333xx33 = b= b33

The above can be represented in the formThe above can be represented in the form

of a matrix Ax = bof a matrix Ax = b


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March 20, 2012March 20, 2012



A x = bA x = b

aa1111 aa1212 aa1313 xx11 bb11aa2121 aa2222 aa2323 xx22 = b= b22

aa3131 aa3232 aa3333 xx33 bb33

The solution of the above system ofThe solution of the above system ofequations isequations is x = Ax = A-1-1bb


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Gauss Elimination MethodGauss Elimination Method

For a system of equations with k variables theFor a system of equations with k variables the

matrix will be of the formmatrix will be of the form


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The Augmented Matrix is of the formThe Augmented Matrix is of the form


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A series of elementary row operations are performed onA series of elementary row operations are performed on

the augmented matrix to obtain a matrix in the formthe augmented matrix to obtain a matrix in the form

Solve the equation of the kSolve the equation of the kthth row for xrow for xkk, then substitute, then substitute

back into the equation of the (k-1)back into the equation of the (k-1)stst row to obtain arow to obtain a

solution for xsolution for xk-1k-1, etc., etc.


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Solution of EquationsSolution of Equations

Find the solution to the following set ofFind the solution to the following set of

equationsequations

x + y + 2z = 8x + y + 2z = 8

-1x - 2y + 3z = 1-1x - 2y + 3z = 1

3x - 7y + 4z = 103x - 7y + 4z = 10


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Therefore the solution to the set ofTherefore the solution to the set of

equations isequations is x = 3, y = 1, z = 2x = 3, y = 1, z = 2


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QuestionQuestion

Find the solution to the system ofFind the solution to the system of

equations using gauss elimination methodequations using gauss elimination method

-3x + 2y - 1z = -1-3x + 2y - 1z = -1

6x 6y + 7z = -76x 6y + 7z = -7

3x 4y + 4z = -63x 4y + 4z = -6


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The solution of the system of equations isThe solution of the system of equations is

x = 2, y = 2, z = -1x = 2, y = 2, z = -1


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SummarySummary

For the above matrix,For the above matrix,perform a series of rowperform a series of rowoperations to obtain theoperations to obtain thematrix in the formmatrix in the form

The matrix B is theThe matrix B is theinverse of the matrixinverse of the matrix


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SummarySummary

A x = bA x = b

aa1111 aa1212 aa1313 xx11 bb11

aa2121 aa2222 aa2323 xx22 = b= b22

aa3131 aa3232 aa3333 xx33 bb33

The solution of the above system ofThe solution of the above system of

equations isequations is x = Ax = A-1-1bb


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ReferencesReferences

Gerald W. Recktenwald, NumericalGerald W. Recktenwald, Numerical

Methods with MATLAB, ImplementationMethods with MATLAB, Implementation

and Application, Prentice Halland Application, Prentice HallSolving Linear Systems of Equations,

Gerald Recktenwald, Portland State

University, Mechanical Engineering

Department, [email protected]

http://mathworld.wolfram.comhttp://mathworld.wolfram.com

W7 Linear Equations

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