Stewart J. College Algebra: Concepts and Contexts

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614 CHAPTER 7

■

Systems of Equations and Data in Categories

Similarly, we calculate the remaining entries of the product as follows.

Entry Inner product of Value Product matrix

Thus we have

The product BA is not defined, however, because the dimensions of B and A are

The inner two numbers are not the same, so the rows and columns won’t match up

when we try to calculate the product.

■ NOW TRY EXERCISES 11 AND 13 ■

Graphing calculators are capable of performing all the matrix operations. In the

next example we use a graphing calculator to multiply two matrices.

2 * 3

and2 * 2

AB = c

- 11723

1 - 5 - 2

- 11723

1 - 5 - 2

d1- 12

2 + 0

7 =-2c

- 1 0

- 152

047

- 117 23

1 - 5

d1- 12

5 + 0

4 =-5c

- 1 0

- 1 5 2

0 4 7

- 1 17 23

d1- 12

1- 12+ 0

0 = 1c

- 1 0

- 1 52

0 47

- 1 17 23

2 + 3

7 = 23c

1 3

- 10

- 152

047

- 1 17

5 + 3

4 = 17c

1 3

- 10

- 1 5 2

0 4 7

2 * 32 * 2

example

Multiplying Matrices (Graphing Calculator)

Calculate the products AB and BA.

Solution

Using a graphing calculator, we find the product AB.

Calculator output Product AB

18 37 5

- 10 - 20 30

524- 3

A = £

30 5

22- 4

- 11 2

B = £

1 - 15

076

38- 2

[A]

[B]

[[18 37 5 ]

[-10 -20 30]

[5 24 -3]

Not equal, so product

not defined.

SECTION 7.5

■

Matrix Operations: Getting Information from Data 615

Similarly, we find the product BA.

Calculator output Product BA

■ NOW TRY EXERCISE 19 ■

- 43 19

820- 16

27 14 - 21

[B]

[A]

[[-4 3 19 ]

[8 20 -16]

[27 14 -21]]

■ Getting Information from Categorical Data

We will see in the next example how we can get information about categorical data

by multiplying matrices. (Compare to Example 5 in Section 7.4.)

example

Getting Information by Multiplying Matrices

Josh sells organic produce at an open-air market three days a week: Thursdays,

Fridays, and Saturdays. He sells oranges, broccoli, and beans. Matrix A tabulates the

number of pounds of produce that he obtains from his suppliers every week; he is al-

ways able to sell his complete supply by the end of each day. Matrix C gives the

prices per pound (in dollars) he charged on two different weeks. Find the product AC,

and interpret the entries of AC.

Produce

Oranges Broccoli Beans

Prices

Week 1 Week 2

Solution

We use a graphing calculator to find the product AC.

Calculator output Product AC

84.00 111.00

150.00 197.00

96.00 125.50

C = £

0.90 1.20

1.20 1.60

1.50 1.90

A = £

50 20 10

70 35 30

45 15 25

TTT

Thursday

Friday

Saturday

Day

Produce

Oranges

Broccoli

Beans

[A]

[C]

[[84 111 ]

[150 197 ]

[96 125.5]]

616 CHAPTER 7

■

Systems of Equations and Data in Categories

The product matrix gives us the revenue for each day of each week.

■

Week 1: A total of $84.00 of produce was sold on Thursday, $150.00 on

Friday, and $96.00 on Saturday.

■

Week 2: A total of $111.00 of produce was sold on Thursday, $197.00 on

Friday, and $125.50 on Saturday.

■ NOW TRY EXERCISE 27 ■

7.5 Exercises

Fundamentals

1. We can add or (subtract) two matrices only if they have the same _______.

2. We can multiply two matrices only if the number of

_______ in the first matrix is the

same as the number of

_______ in the second matrix.

3. If A is a matrix and B is a matrix, which of the following matrix

multiplications are possible?

(a) AB (b) BA (c) AA (d) BB

4. Fill in the missing entries in the product matrix.

Think About It

5. Which of the following operations can we perform on a matrix A of any dimension?

(a) (b) 2A (c) AA

6. If A is a matrix for which AA is defined, what must be true about the dimension of A?

7–14

■ Perform the matrix operation, or explain why the operation is not defined.

7. 8.

9. 10.

11. 12.

13. 14. £

2 - 3

§c

- 14

1 - 23

22- 1

212

634

d£

1 - 2

- 20

§£

1 - 2

- 20

2 £

110

101

011

§+ £

§3 £

4 - 1

011

110

d- c

21- 1

13- 2

- 53

d+ c

- 1 - 3

A + A

31 2

- 12 0

13- 2

§ £

- 13- 2

3 - 2 - 1

210

§= £

ⵧ

- 7

7 - 7

ⵧ

- 5 - 5

2 * 33 * 3

CONCEPTS

SKILLS

SECTION 7.5

■

Matrix Operations: Getting Information from Data 617

15–26 ■ Matrices A, B, C, D, E, F, G, and H are defined as follows. Perform the indicated

matrix operation(s), or explain why it cannot be performed.

15. (a) (b) 16. (a) (b)

17. (a) 5A (b) 18. (a) (b)

19. (a) AD (b) DA 20. (a) DH (b) HD

21. (a) AH (b) HA 22. (a) BC (b) BF

23. (a) GF (b) GE 24. (a) (b)

25. (a) (b) 26. (a) (DA)B (b) D(AB)

27. Fast-Food Sales A small fast-food chain with restaurants in Santa Monica, Long

Beach, and Anaheim sells only hamburgers, hot dogs, and milk shakes. On a certain

day, sales were distributed according to matrix A. The price of each item (in dollars) is

given by matrix C.

(a) Calculate the product matrix CA.

(b) What is the total profit from the Long Beach restaurant?

Number of items sold

Santa Long

Monica Beach Anaheim

Item

Hamburger Hot dog Milk shake

28. Car-Manufacturing Profits A specialty car manufacturer has plants in Auburn,

Biloxi, and Chattanooga. Three models are produced, with daily production given in

matrix A. Because of a wage increase, February profits are lower than January profits.

The profit (in dollars) per car is tabulated by model in matrix B.

(a) Calculate the product AB.

(b) Assuming that all cars produced were sold, what was the daily profit in January

from the Biloxi plant?

C = 30.90 0.80 1.104

TTT

A = £

4000 1000 3500

400 300 200

700 500 9000

TTT

2H + D3B + 2CC - 5A

2C - 6BC - BB + FB + C

G = £

5 - 310

6 1 0

- 5 2 2

 H = c

3 1

2 - 1

D = 3734 E = £

§ F = £

100

010

001

A = c

2 - 5

0 7

d B = c

1 - 13

d C = c

2 -

0 2 - 3

CONTEXTS

Hamburgers

Hot dogs

Milk shakes

Item

Price

618 CHAPTER 7

■

Systems of Equations and Data in Categories

Cars produced each day

Model K Model R Model W

January February

29. Canning Tomato Products Jaeger Foods produces tomato sauce and tomato paste,

canned in small, medium, large, and giant sized cans. The matrix A gives the size (in

ounces) of each container. The matrix B tabulates one day’s production of tomato sauce

and tomato paste.

(a) Calculate the product AB.

(b) Interpret the entries in the product matrix AB.

Size

Small Medium Large Giant

Cans of sauce Cans of paste

30. Produce Sales A farmer’s three children, Amy, Beth, and Chad, run three roadside

produce stands during the summer months. One weekend they all sell watermelons,

yellow squash, and tomatoes. Matrices A and B tabulate the number of pounds of each

product sold by each sibling on Saturday and Sunday. Matrix C gives the price per

pound (in dollars) for each type of produce that they sell. Perform each of the following

matrix operations, and interpret the entries in each result.

(a) AC (b) BC (c) (d)

Saturday

Melons Squash Tomatoes

Sunday

Melons Squash Tomatoes

B = £

100 60 30

35 20 20

60 25 30

TTT

A = £

120 50 60

40 25 30

60 30 20

TTT

1A + B 2CA + B

B = ≥

2000 2500

3000 1500

2500 1000

1000 500

A = 36 10 14 284

TTTT

B = £

1000 500

2000 1200

1500 1000

A = £

12 10 0

4 4 20

8 9 12

TTT

Price per pound

C = £

0.10

0.50

1.00

Auburn

Biloxi

Chattanoogad

City

Model K

Model R

Model Wd

Small

Medium

Large

Giant

Amy

Beth

Chad

Amy

Beth

Chadd

Melons

Squash

Tomatoes

Ounces

SECTION 7.6

■

Matrix Equations: Solving a Linear System 619

7.6 Matrix Equations: Solving a Linear System

■

The Inverse of a Matrix

■

Matrix Equations

■

Modeling with Matrix Equations

IN THIS SECTION… we express a system of equations as a single matrix equation and

then solve the system by solving the matrix equation.

In the preceding section we saw that when the dimensions are appropriate, matrices

can be added, subtracted, and multiplied. In this section we investigate “division” of

matrices. With this operation we can solve equations that involve matrices.

The identity matrix is the matrix for which each main diagonal

entry is a 1 and for which all other entries are 0.

n * nI

■ The Inverse of a Matrix

First, we define identity matrices, which play the same role for matrix multipli-

cation that the number 1 does for ordinary multiplication of numbers; that is,

for all numbers a. In the following definition the term main di-

agonal refers to the entries of a square matrix whose row and column numbers

are the same. These entries stretch diagonally down the matrix, from top left to

bottom right.

Identity Matrix

a = a

1 = a

Thus, the , , and identity matrices are

Identity matrices behave like the number 1 in the sense that

whenever these products are defined.

The following matrix products show how multiplying a matrix by an identity

matrix of the appropriate dimension leaves the matrix unchanged.

- 17

1213

- 207

§£

100

010

001

§= £

- 17

1213

- 207

3 5 6

- 127

d= c

3 5 6

- 127

= AandB

= B

= c

dI

= £

100

010

001

§I

= ≥

1000

0100

0010

0001

4 * 43 * 32 * 2

620 CHAPTER 7

■

Systems of Equations and Data in Categories

If A and B are matrices, and if , then we say that B is the

inverse of A, and we write . The concept of the inverse of a matrix is analo-

gous to that of the reciprocal of a real number.

Inverse of a Matrix

B = A

-1

AB = BA = I

n * n

Let A be a square matrix. If there exists an matrix with the

property that

then we say that is the inverse of A.A

-1

= A

-1

A = I

-1

n * nn * n

example

Verifying That a Matrix Is an Inverse

Verify that matrix B is the inverse of matrix A.

Solution

We perform the matrix multiplications to show that and .

■ NOW TRY EXERCISE 3 ■

The inverse of a matrix can be found by using a process that involves elemen-

tary row operations. This process is programmed into graphing calculators. On

TI-83 calculators, matrices are stored in memory by using names such as [A], [B],

[C], and so on. To find the inverse of [A], we key in

[A]

ENTER

⫺1

3 - 1

- 5 2

d= c

2 + 1- 1253

1 + 1- 123

1- 522 + 2

5 1- 5 21 + 2

d= c

3 - 1

- 5 2

d= c

3 + 11- 52 21- 12+ 1

3 + 31- 52 51- 12+ 3

d= c

BA = IAB = I

A = c

2 3

5 1

dandB = c

3 - 1

- 5 2

example

Finding the Inverse of a Matrix

Find the inverse of matrix A.

Solution

A graphing calculator gives the following output for the inverse of matrix A.

A = £

1 - 2 - 4

2 - 3 - 6

- 3615

SECTION 7.6

■

Matrix Equations: Solving a Linear System 621

[A] Frac

[[-3 2 0 ]

[-4 1 -2/3]

[1 0 1/3 ]]

Calculator output Inverse of matrix A

We have used the Frac command to display the output in fraction form rather than

in decimal form.

■ NOW TRY EXERCISES 7 AND 13 ■

䉴

-1

= £

- 32 0

- 41

■ Matrix Equations

A system of linear equations can be written as a single matrix equation. For exam-

ple, the system

is equivalent to the matrix equation

If we let

then this matrix equation can be written as

Matrix A is called the coefficient matrix.

We solve this matrix equation by multiplying each side by the inverse of A (pro-

vided that this inverse exists).

Given equation

Multiply on left by

Associative Property

Property of inverses

Property of the identity matrix

X = A

-1

X = A

-1

A2X = A

-1

1AX 2= A

-1

AX = B

A = £

1 - 2 - 4

2 - 3 - 6

- 3 6 15

§X = £

§B = £

1 - 2 - 4

2 - 3 - 6

- 3 6 15

§£

§= £

x - 2y - 4z = 7

2x - 3y - 6z = 5

- 3x + 6y + 15z = 0

Solving the matrix equation

is very similar to solving the simple

real-number equation

which we do by multiplying each

side by the reciprocal (or inverse)

of 3:

x = 4

13x 2=

112 2

3x = 12

AX = B

A X B

622 CHAPTER 7

■

Systems of Equations and Data in Categories

We have proved that the matrix equation can be solved by the following

method.

Solving a Matrix Equation

AX = B

If A is a square matrix that has an inverse , and if X is a variable

matrix and B a known matrix, both with n rows and one column, then the

solution of the matrix equation

is given by

X = A

-1

AX = B

-1

n * n

example

Solving a System as a Matrix Equation

Solve the system of equations.

Solution

The system is equivalent to the matrix equation described on page 621.

Using a calculator, we find that

So because , we have

§= £

- 32 0

- 41

§£

§= £

- 11

- 23

X = A

-1

= £

- 32 0

- 41-

AX = B

x - 2y - 4z = 7

2x - 3y - 6z = 5

- 3x + 6y + 15z = 0

So the solution of the system is .

■ NOW TRY EXERCISES 19 AND 23 ■

1- 11, - 23, 72

X = A

-1

example

Solving a System Using a Matrix Inverse

A system of equations is given.

(a) Write the system of equations as a matrix equation.

(b) Solve the system by solving the matrix equation.

2x - 5y = 15

3x - 6y = 36

SECTION 7.6

■

Matrix Equations: Solving a Linear System 623

Solution

(a) We write the system as a matrix equation of the form .

2 - 5

3 - 6

d= c

AX = B

(b) Using a graphing calculator to calculate the inverse, we get

So because , we have

d= c

- 2

- 1

d= c

X = A

-1

= c

2 - 5

3 - 6

-1

= c

- 2

- 1

So the solution of the system is (30, 9).

■ NOW TRY EXERCISE 19 ■

Not every square matrix has an inverse. A matrix that has no inverse is called

singular.

X = A

-1

example

A Matrix That Does Not Have an Inverse

Find the inverse of the matrix.

Solution

If we try to calculate the inverse of this matrix on the TI-83 calculator, we get the er-

ror message shown below. That means that this matrix has no inverse.

Calculator output Inverse of matrix A

The matrix A does

not have an inverse.

■ NOW TRY EXERCISE 15 ■

2 - 3 - 7

1 2 7

1 1 4

ERR:SINGULAR MAT

1:Quit

2:Goto

A X = B