Chapter 6 Systems of Linear Equations and Matrices

Lecture Note

6.1 Systems of Two Linear Equations in Two Variables

6.2 Larger Systems of Linear Equations

6.3 Applications of Systems of Linear Equations

6.4 Basic Matrix Operations

6.5 Matrix Products and Inverses

6.6 Applications of Matrices

Chapter 6 Systems of Linear Equations and Matrices

Lecture Note

Chapter 6.1 Systems of Two Linear Equations in Two Variables

System of Linear Equations

Two or more linear equations make up a system of linear equations. A solutionto a system of equations in two variables is an ordered pair that is a solution toeach individual equation. The solution set to a system of equations is the set ofall solutions to the system. Solving a system of equations means finding thesolution set.

Example Determine if the ordered pair is a solution to the system

3𝑥 − 𝑦 = 10

𝑥 +1

4𝑦 = 1

(a) 2,−4 (b)1

3, −9

Lecture Note


Using the Substitution MethodStep 1. Isolate one of the variables from one equation.Step 2. Substitute the expression found in step 1 into the other

equation.Step 3. Solve the resulting equation.Step 4. Substitute the value found in step 3 back into the equation in

step 1 to find the value of the remaining variable.Step 5 Check the ordered pair in each equation and write the

solution as an ordered pair in set notation.

Example 1. Solve the system

2𝑥 − 𝑦 = 13𝑥 + 2𝑦 = 4

by using the substitution method.

Lecture Note


Using the Addition (or Elimination) MethodStep 1. Write both equations in standard form: 𝐴𝑥 + 𝐵𝑦 = 𝐶.Step 2. Clear fractions or decimals (optional).Step 3. Multiply one or both equations by nonzero constants to

create opposite coefficients for one of the variables.Step 4. Add the equations from step 3 to eliminate one variable.Step 5. Solve for the remaining variable.Step 6. Substitute the known value found in step 5 into one of the

original equations to solve for the other variable.Step 7. Check the ordered pair in each equation and write the

solution as an ordered pair in set notation.


2𝑥 − 𝑦 = 13𝑥 + 2𝑦 = 4

by using the addition method.

Lecture Note

Methods of Solving a System of Two Linear Equations in Two Variables:

Substitution Method

Addition Method


Examples. Solve each system.

a 5𝑥 + 𝑦 = 43𝑥 + 2𝑦 = 1

(b) 3𝑥 − 4𝑦 = 12𝑥 + 3𝑦 = 12

Lecture Note

Dependent and Inconsistent Systems:

A system of equations that has a unique solution is called an independentsystem.

A system of equations that has infinitely many solutions is called adependent system.

A system of equations that has no solution is said to be inconsistent.

Examples. Solve the system

𝑎 −4𝑥 + 𝑦 = 28𝑥 − 2𝑦 = −4

𝑏 −3𝑥 − 2𝑦 = 4−6𝑥 + 4𝑦 = 7


Geometrical Meaning of Solving A System of Linear Equations

Geometrically, Solving a system of linear equations means finding theintersecting point of two lines.

Lecture Note

Example 6. Eight hundred people attend a basketball game, and total ticket salesare $3102. If adult tickets are $6 and student tickets are $3, how many adultsand how many students attended the game?


Lecture Note

Two systems of equations are said to be equivalent if they have the samesolutions. The basic procedure for solving a large system of equations is totransform the system into a simpler, equivalent system and then solve thissimpler system.

Chapter 6.2 Larger Systems of Linear Equations

Three operations, called elementary operations, are used to transform a systeminto an equivalent one:1. Interchange ANY TWO equations in the system: Changing the order of the

equations obviously does not affect the solutions of the equations or thesystem.

2. Multiply an equation in the system by a nonzero constant: Multiplying anequation by a nonzero constant does not change its solutions. So it does ontchange the solutions of the system.

3. Replace an equation in the system by the sum of itself and a nonzeroconstant multiple of another equation

Lecture Note


2𝑥 + 𝑦 − 𝑧 = 2 ⋯ (R1)𝑥 + 3𝑦 + 2𝑧 = 1 ⋯ (R2)𝑥 + 𝑦 + 𝑧 = 2 ⋯ (R3)


Interchange 𝑅1 and 𝑅2

⇒ −2 𝑅1 + 𝑅2, then replace 𝑅2

⇒ −1 𝑅1 + 𝑅3, then replace 𝑅3

⇒ Divide 𝑅2 by −5

⇒ 2𝑅2 + 𝑅3, then obtain 𝑧

⇒ Finally, do back substitution to obtain 𝑦 and 𝑥.

Here the first variable that appears in an equation with a nonzero coefficient iscalled the leading variable of that equation, and its nonzero coefficient is calledthe leading coefficient. This method can be used with any system.

Lecture Note


The Elimination Method for Solving Large Systems of Linear EquationsUse elementary operations to transform the given system into an equivalent oneas follows:1. Make the leading coefficient of the first equation 1 either by interchanging

equations or by multiplying the first equation by a suitable constant.2. Eliminate the leading variable of the first equation from each later equation

by replacing the later equation by the sum of itself and a suitable multiple ofthe first equation.

3. Repeat Steps 1 and 2 for the second equation: Make its leading coefficient 1and eliminate its leading variable from each later equation by replacing thelater equation by the sum of itself and a suitable multiple of the secondequation.

4. Repeat Steps 1 and 2 for the third equation, fourth equation, and so on,until it is not possible to go any further.

Then solve the resulting system by back substitution.

Lecture Note


Matrix Methods: This idea of solving a system can be done in matrix.

Matrix. A matrix is a rectangular array of elements or entries arranged in rows (horizontal) and columns (vertical):

𝐴 =

𝑎11 𝑎12𝑎21 𝑎22

⋯𝑎1𝑛𝑎2𝑛

⋮ ⋮ ⋱ ⋮𝑎𝑚1 𝑎𝑚2 ⋯ 𝑎𝑚𝑛

= [𝑎𝑖𝑗]=

column (vector)

row (vector)

diagonal element (𝑚 = 𝑛)

Size 𝑚 × 𝑛

Lecture Note


Augmented Matrix: The augmented matrix consists of two parts, coefficients ofvariables and constants of equations separated by a vertical line. For example,the augmented matrix of

2𝑥 + 𝑦 − 𝑧 = 2 ⋯ (R1)𝑥 + 3𝑦 + 2𝑧 = 1 ⋯ (R2)𝑥 + 𝑦 + 𝑧 = 2 ⋯ (R3)

is𝟐 𝟏 −𝟏 𝟐𝟏 𝟑 𝟐 𝟏𝟏 𝟏 𝟏 𝟐

Lecture Note


Row Operations: Row operations on the augmented matrix are operationscorresponding to the elementary operations and they are applied to row.

Performing any one of the following row operations on the augmented matrix ofa system of linear equations produces the augmented matrix of an equivalentsystem.1. Interchange any two rows.2. Multiply each element of a row by a nonzero constant.3. Replace a row by the sum of itself and a nonzero constant multiple of

another row of matrix.

Lecture Note


Example 2. Use matrices to solve the system.

𝑥 − 2𝑦 = 6 − 4𝑧𝑥 + 13𝑧 = 6 − 𝑦−2𝑥 + 6𝑦 − 𝑧 = −10

Row Echelon Form:A matrix, such as the last one in Example 2, is said to be in row echelon formwhen All rows consisting entirely of zeros (if any) are at the bottom. The first nonzero entry in each row is 1 (called a leading 1); and Each leading1 appears to the right of the leading 1s in any preceding rows.

Lecture Note


The Gauss-Jordan Method:The Gauss-Jordan method is a variation on the matrix elimination method usedin Example 2. It replaces the back substitution used there with additionalelimination of variables.

Example 2. Use the Gauss-Jordan method to solve the system in Example 2.

𝑥 − 2𝑦 = 6 − 4𝑧𝑥 + 13𝑧 = 6 − 𝑦−2𝑥 + 6𝑦 − 𝑧 = −10

Lecture Note


The Gauss-Jordan Method:The Gauss-Jordan method is a variation on the matrix elimination method usedin Example 2. It replaces the back substitution used there with additionalelimination of variables.

Example 2. Use the Gauss-Jordan method to solve the system in Example 2.

𝑥 − 2𝑦 = 6 − 4𝑧𝑥 + 13𝑧 = 6 − 𝑦−2𝑥 + 6𝑦 − 𝑧 = −10

Reduced Row Echelon Form:The final matrix in the Gauss-Jordan method is said to be in reduced row echelonform, meaning that it is in row echelon form and every column containing aleading 1 has zeros in all its other entries.

Lecture Note


Reduced Row Echelon Form:The solutions of the system can be read directly from the reduced row echelonmatrix. The reduced row echelon form of a matrix is unique. This is not true of amatrix but is simply in row echelon form. This is why a row echelon formproduced by hand may differ from a row echelon form provided by a calculator,but each will lead to the same solution of the system of equations.In the Gauss-Jordan method, row operations may be performed in any order, butit is best to transform the matrix systematically. Either follow the procedure inExample 3 (which first puts the system into a form in which back substitutionscan be used and then eliminates additional variables) or work column by columnfrom left to right.

Lecture Note


Dependent and Inconsistent Systems:A system has exactly one solution (an independent system), an infinite numberof solutions (a dependent system), or no solutions at all (an inconsistent system).


2𝑥 + 4𝑦 = 43𝑥 + 6𝑦 = 82𝑥 + 𝑦 = 7


2𝑥 − 3𝑦 + 4𝑧 = 6𝑥 − 2𝑦 + 𝑧 = 9

𝑦 + 2𝑧 = −12

Lecture Note

Example 1. A rent-a-truck company plans to spend $5 million on 200 newvehicles. Each van will cost $20,000, each small truck $25,000, and each largetruck $35,000. Past experience shows that the company needs twice as manyvans as small trucks. How many of each kind of vehicle can the company buy?

Chapter 6.3 Applications of Systems of Linear Equations

Lecture Note

Example 2. An investor plans to put a total of $100,000 in a money marketaccount, a bond fund, an international stock fund, and a domestic stock fund.She wants 60% of her investment to be conservative (money market and bonds).She wants the amount in international stocks to be one-fourth of the amount indomestic stocks. Finally, she needs an annual return of $4000. Assuming she getsannual return of 2.5% on the money market account, 3.5% on the bond fund, 5%on the international stock fund, and 6% on the domestic stock fund, how muchshould she put in each investment?


Lecture Note

Example 3. An animal feed is to be made from corn, soybeans, and cottonseed.Determine how many units of each ingredient are needed to make a feed thatsupplies 1800 units of fiber, 2800 units of fat, and 2200 units of protein, giventhat 1 unit of each ingredient provides the number of units shown in the tablebelow. The table states, for example, that a unit of corn provides 10 units offiber, 30 units of fat, and 20 units of protein.


Corn Soybeans Cottonseed Totals

Units of Fiber 10 20 30 1800

Units of Fat 30 20 40 2800

Units of Protein 20 40 25 2200

Lecture Note

Example 4. The table shows Census Bureau projections for the population of theUnited States (in millions).

(a) Use the given data to construct a quadratic function that gives the U.S.population in millions) in year 𝑥 from 2000.

(b) Use this model to estimate the U.S. population in the year 2030.


Year 2020 2040 2050

Population 334 380 400

Lecture Note

Example 5. A specialty wholesaler sells espresso machines. The EZ modelweights 10 pounds and comes in a 10-cubic-foot box. The compact modelweights 20 pounds and comes in an 8-cubic-foot box. The commercial modelweights 60 pounds and comes in a 28-cubic-foot box. Each delivery van has 248cubic feet of space and can hold a maximum of 440 pounds. In order for a van tobe fully loaded, how many of each model should it carry?


Lecture Note

Matrices are also important in the fields of management, natural sciences,engineering, and social science as a way to recognize data.

Chapter 6.4 Basic Matrix Operations

Example 1 The EZ Life Company manufactures sofas and armchairs in threemodels: A, B, and C. The company has regional warehouse in New York, Chicago,and San Francisco. In its August shipment, the company sends 10 mode A sofas,12 model B sofas, 5 model C sofas, 15 model A chairs, 20 model B chairs, and 8model C chairs to each warehouse. Construct a matrix to organize these data

Lecture Note

Matrix. A matrix is a rectangular array of elements or entries arranged in rows (horizontal) and columns (vertical):

𝐴 =

𝑎11 𝑎12𝑎21 𝑎22

⋯𝑎1𝑛𝑎2𝑛

⋮ ⋮ ⋱ ⋮𝑎𝑚1 𝑎𝑚2 ⋯ 𝑎𝑚𝑛

= [𝑎𝑖𝑗]=

column (vector)

row (vector)

diagonal element (𝑚 = 𝑛)

Size 𝑚 × 𝑛


Lecture Note

Matrix Addition:

The sum of two 𝑚 × 𝑛 matrices 𝑋 and 𝑌 is the 𝑚× 𝑛 matrix 𝑋 + 𝑌 in whicheach element is the sum of the corresponding elements of 𝑋 and 𝑌.


Lecture Note

Matrix Subtraction:

The difference of two 𝑚 × 𝑛 matrices 𝑋 and 𝑌 is the 𝑚 × 𝑛 matrix 𝑋 − 𝑌 inwhich each element is the difference of the corresponding elements of 𝑋 and 𝑌.


Lecture Note

Scalar Multiplication:

The product of a scalar 𝑘 and a 𝑚× 𝑛 matrix 𝑋 is the 𝑚× 𝑛 matrix 𝑘𝑋 in whicheach element is 𝑘 times the corresponding element of 𝑋.


Negative of a Matrix:

The negative (or additive inverse) of a 𝑚× 𝑛 matrix 𝐴 is the 𝑚 × 𝑛 matrix(−1)𝐴 which is obtained by multiplying each element of 𝐴 by −1. It is denotedby −𝐴.

Zero Matrix:

A matrix consisting only of zeros is called a zero matrix (or additive identity) andis denoted 𝑂.

Lecture Note

Properties of Sum and Difference Operations:

If 𝐴, 𝐵, and 𝐶 are 𝑚 × 𝑛 matrices and 𝑂 is the 𝑚 × 𝑛 zero matrix, then thefollowing properties are satisfied:Commutative Property of Addition of Matrices 𝐴 + 𝐵 = 𝐵 + 𝐴Associative Property of Addition of Matrices 𝐴 + 𝐵 + 𝐶 = 𝐴 + 𝐵 + 𝐶Identity Property of Addition of Matrices 𝐴 + 𝑂 = 𝐴 = 𝑂 + 𝐴Inverse Properties of Additions of Matrices 𝐴 + −𝐴 = 𝑂 = −𝐴 + 𝐴

𝐴 + −𝐵 = 𝐴 − 𝐵


Lecture Note

Chapter 6.5 Matrix Products and Inverse

Matrix Multiplication:

Let 𝐴 be an 𝑚 × 𝑛 matrix and let 𝐵 be an 𝑛 × 𝑘 matrix. The product matrix 𝐴𝐵 isthe 𝑚 × 𝑘 matrix whose entry is the 𝑖th row and 𝑗th column is

the sum of the products of the 𝒊th row entries of 𝑨 and the 𝒋th column entries of 𝑩.

Lecture Note


Example The EZ Life Company manufactures sofas and armchairs in threemodels: A, B, and C. Suppose sofas and chairs of the same model are often soldas sets, with matrix 𝑊 showing the number of each model set in eachwarehouse

A B CNew YorkChicagoSan Francisco

10 7 35 9 64 8 2

= 𝑊

If The selling price of a model A set is $800, of a model B set is $1000, and of amodel C set is $1200, we would like to find the total value of the sets in eachcity.

Lecture Note


Properties Matrix Multiplication:

If 𝐴, 𝐵, and 𝐶 are any matrices such that all the indicated sums and productsexist, then the following properties are satisfied:Associative Property of Multiplication of Matrices 𝐴 𝐵𝐶 = 𝐴𝐵 𝐶Distributive Property 𝐴 𝐵 + 𝐶 = 𝐴𝐵 + 𝐴𝐶

𝐵 + 𝐶 𝐴 = 𝐵𝐴 + 𝐶𝐴

No Commutative Property in Matrix Multiplication:

If 𝐴 and 𝐵 are matrices such that the products 𝐴𝐵 and 𝐵𝐴 exists,𝐴𝐵 may not equal 𝐵𝐴

Lecture Note


Example 4 A contractor builds three kinds of houses, models A, B, and C, with a choice oftwo styles, Spanish or contemporary. Matrix 𝑃 shows the number of each kind of houseplanned for a new 100-home subdivision:

Spanish ContemporaryModel 𝐴Model 𝐵Model 𝐶

0 3010 2020 20

= 𝑊

The amounts for each of the exterior materials used depend primarily on the style of thehouse. These amounts are shown in matrix 𝑄 (concrete is measured in cubic yards,lumber in units of 1000 board feet, brick in thousands, and shingles in units of 100 squarefeet):

Concrete Lumber Brick ShinglesSpanishContemporary

10 2 0 250 1 20 2

= 𝑄

Matrix 𝑅 gives the cost for each kind of material:Cost per UnitConcrete

LumberBrickShingles

201806025

= 𝑅

(a) What is the total cost for each model of house?(b) How much of each of the four kinds of material must be ordered?(c) What is the total cost for material?(d) Suppose the contractor builds the same number of homes in five subdivisions. What

is the total amount of each material needed for each model in this case?

Lecture Note


Identity Matrices:

The identity matrix, denoted by 𝐼, is a square matrix whose entries are 1s on themain diagonal from upper left to lower right, and all other entries equal to 0.Let 𝐼 be the 𝑛 × 𝑛 identity matrix. For any𝑛 × 𝑛matrix𝐴,

𝐴𝐼 = 𝐼𝐴 = 𝐴

Inverse Matrix:

The inverse matrix of an 𝑛 × 𝑛matrix 𝐴, denoted by 𝐴−1, is the 𝑛 × 𝑛matrix suchthat

𝐴𝐴−1 = 𝐴−1𝐴 = 𝐼.Not all matrices have their inverses, but if a matrix has an inverse, then it is uniqueand it can be found by using the row operations.

Lecture Note


Inverse Matrix:

To obtain an inverse matrix 𝐴−1 for any 𝑛 × 𝑛matrix 𝐴 for which 𝐴−1 exists, followthese steps:1. Form the augmented matrix 𝐴 𝐼 ], where 𝐼 is the 𝑛 × 𝑛 identity matrix.2. Perform row operations on 𝐴 𝐼 ] to get a matrix of the form 𝐼 𝐵 ].3. Matrix𝐵 is𝐴−1.

Chapter 6 Systems of Linear Equations and Matrices

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