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1 Matrices: Definitions and Basic Operations

# 1 Matrices: Definitions and Basic Operations

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16.1 Matrices: Definitions and Basic Operations

F1

E X A M P L E 3 Matrix equation

F2

The forces acting on a bolt are in equilibrium, as shown in Fig. 16.1. Analysing the

horizontal and vertical components as in Section 9.4, we find the following matrix

equation. Find forces F1 and F2.

28.1°

12.7°

437

8.0 N

c

3.5 N

Fig. 16.1

0.98F1 - 0.88F2

8.0

d = c

d

0.22F1 + 0.47F2

3.5

From the equality of matrices, we know that 0.98F1 - 0.88F2 = 8.0 and

0.22F1 + 0.47F2 = 3.5. Therefore, to find the forces F1 and F2, we must solve the system of equations

0.98F1 - 0.88F2 = 8.0

0.22F1 + 0.47F2 = 3.5

Using determinants, we have

F1 =

2 8.0

3.5

-0.88 2

0.47

2 0.98

0.22

-0.88 2

0.47

=

8.010.472 - 3.51 -0.882

= 10.5 N

0.9810.472 - 0.221 -0.882

Using determinants again, or by substituting this value into either equation, we find that

F2 = 2.6 N. These values check when substituted into the original matrix equation. ■

If two matrices have the same number of rows and the same number of columns, their

sum is defined as the matrix consisting of the sums of the corresponding elements. If

the number of rows or the number of columns of the two matrices is not equal, they

E X A M P L E 4 Adding matrices

(a)

J

8

0

1

-2

-5 9

-3

R + J

3 7

6

4 6 0

8 + 1 -32

R = J

-2 6 5

0 + 6

= J

Practice Exercise

1. For matrices A and B, find A + B.

2 -4

-9 -6

d B = c

d

A = c

-7

5

7

3

5

6

1 + 4

-2 + 1 -22

-5 + 6 9 + 0

R

3 + 6 7 + 5

5 1 9

R

-4 9 12

(b) The matrices

3

C2

4

-5 8

9 0S

-2 3

and

3

C2

4

-5 8 0

9 0 0S

-2 3 0

cannot be added since the second matrix has one more column than the first matrix.

This is true even though the extra column contains only zeros.

The product of a number and a matrix (known as scalar multiplication of a matrix)

is defined as the matrix whose elements are obtained by multiplying each element of

the given matrix by the given number. Thus, we obtain matrix kA by multiplying the

elements of matrix A by k. In this way, A + A and 2A are the same matrix.

438

CHAPTER 16 Matrices; Systems of Linear Equations

E X A M P L E 5 Scalar multiplication

For the given matrix A, find 2A:

A = c

-5 7

d

3 0

2A = J

21 -52

2132

2172

-10 14

R = c

d

2102

6 0

By combining the definitions for the addition of matrices and for the scalar multiplication of a matrix, we can define the subtraction of matrices. That is, the difference of

matrices A and B is given by A - B = A + 1 -B2. Therefore, we change the sign of

each element of B, and proceed as in addition.

E X A M P L E 6 Subtracting matrices

c

7

-9

-4

-2 6

7

d - c

d = c

3

-8 5

-9

-4

2

d + c

3

8

-6

9

d = c

-5

-1

-10

d

-2

The operations of addition, subtraction, and multiplication of a matrix by a number

are like those for real numbers. For these operations, the algebra of matrices is like the

algebra of real numbers. We see that the following laws hold for matrices:

Practice Exercise

A + B

A + 1B + C2

k1A + B2

A + O

2. For matrices A and B in Practice

Exercise 1, find A - 2B.

=

=

=

=

B + A

1A + B2 + C

kA + kB

A

(16.1)

(16.2)

(16.3)

(16.4)

(commutative law)

(associative law)

Here, we have let O represent the zero matrix. We will find in the next section that not

all laws for matrix operations are like those for real numbers.

E XE R C I SES 1 6 .1

In Exercises 1 and 2, make the given changes in the indicated

examples of this section and then perform the indicated operations.

1. In Example 4(a), interchange the second and third columns of the

second matrix and then add the matrices.

2. In Example 5, find the matrix - 2A.

In Exercises 3–10, determine the value of the literal numbers in each

of the given matrix equalities.

3. c

5. c

a b

1

d = c

c d

4

x

r>4

2y

-s

-3

d

7

-2

z

d = c

- 5t

12

10

-4

4. c

x

2

d = c d

x + y

5

-9

d

5

6. 3a + bj 2c - dj 3e + f j4 = 35j a + 6 3b + c4

1j = 1- 12

C + D

5

7. C 2C - D S = C 4 S

D - 2E

6

8. c

2x - 3y

13

d = c d

x + 4y

1

x - 3 x + y

5 3

x

2

9. C x - z y + z S = c

d 10. c

d = c

4 -1

x + y

4

x + t y - t

0

d

-3

In Exercises 11–14, find the indicated sums of matrices.

11. c

12. c

2 3

-1

d + c

-5 4

5

1

3

0

-5

50

13. £ -34

-15

4.7

14. C -6.8

- 1.9

7

d

-2

9

4

d + c

-2

2

-1

0

7

d

-3

-82

- 55

82

57 § + £ 45

14 §

62

26 -67

2.1 -9.6

-4.9

4.8

7.4 S + C 3.4

0.7

5.9

5.6

-9.6

0.7

10.1

- 2.1

0.0 S

- 1.6

In Exercises 15–34, use matrices A, B, and C to find the indicated

matrices.

A = c

-1

2

4

-6

-7

d

11

B = c

7

4

9

-1

-6

d

-8

C = c

3

7

19

d

-5

15. A + B

16. A - B

17. A + C

19. 2A + B

20. 2B + A

21. A - 2B

18. B + C

22. 3A - B

23. -4A

24. -3B

25. -C - A

26. - 12A + B

16.2 Multiplication of Matrices

27. 3A

28. - 2C

31. A - C

32. - 4A - 3B 33. - 6B + A

29. A + 3B

30. B - 0.5A

two-car garage. The following matrix shows the number of

houses of each type and the type of garage.

34. A - B + 2C

In Exercise 35–38, use matrices A and B to show that the indicated laws

hold for these matrices. In Exercise 35, explain the meaning of the result.

-1

A = C 0

9

2

-3

-1

3

-1

0

7

4S

-2

-1

0

11

4

B = C5

1

-3

-1

8

36. A + O = A

37. - 1A - B 2 = B - A

38. 31A + B 2 = 3A + 3B

In Exercises 39 and 40, find the unknown quantities in the given

matrix equations.

39. An airplane is flying in a direction 21.0° north of east at 235 km>h

but is headed 14.5° north of east. The wind is from the southeast.

Find the speed of the wind vw and the speed of the plane vp relative to the wind from the given matrix equation. See Fig. 16.2.

c

Carport

1-car garage

2-car garage

0

1S

2

35. A + B = B + A

vp cos 14.5° - vw cos 45.0°

235 cos 21.0°

d = c

d

vp sin 14.5° + vw sin 45.0°

235 sin 21.0°

21.0

14.5

45.0

E

Fig. 16.2

40. Find the electric currents shown in Fig. 16.3 by solving the following matrix equation:

I1 + I2 + I3

0

C - 2I1 + 3I2 S = C 24 S

- 3I2 + 6I3

0

24 V

Type B

Type C

Type D

96

£ 62

0

75

44

35

0

24

68

0

78

42. The inventory of a drug supply company shows that the following

numbers of cases of bottles of vitamins C and B3 (niacin) are in

stock: vitamin C—25 cases of 100-mg bottles, 10 cases of 250mg bottles, and 32 cases of 500-mg bottles; vitamin B3—30 cases

of 100-mg bottles, 18 cases of 250-mg bottles, and 40 cases of

500-mg bottles. This is represented by matrix A below. After two

shipments are sent out, each of which can be represented by

matrix B below, find the matrix that represents the remaining

inventory.

A = c

vp

Type A

If the contractor builds two additional identical developments,

find the matrix showing the total number of each house-garage

type built in the three developments.

235 km/h

vw

439

25 10 32

d

30 18 40

B = c

10 5 6

d

12 4 8

43. One serving of brand K of breakfast cereal provides the given percentages of the given vitamins and minerals: vitamin A, 15%;

vitamin C, 25%; calcium, 10%; iron, 25%. One serving of brand G

provides: vitamin A, 10%; vitamin C, 10%; calcium, 10%; iron,

45%. One serving of tomato juice provides: vitamin A, 15%; vitamin C, 30%; calcium, 3%; iron, 3%. One serving of orangepineapple juice provides vitamin A, 0%; vitamin C, 100%; calcium,

2%; iron, 2%. Set up a two-row, four-column matrix B to represent

the data for the cereals and a similar matrix J for the juices.

44. Referring to Exercise 43, find the matrix B + J and explain the

meaning of its elements.

2⍀

3⍀

6⍀

I1

I2

I3

Fig. 16.3

In Exercises 41–44, perform the indicated matrix operations.

1. c

-7

0

- 10

d

8

2. c

20

- 21

8

d

-1

41. The contractor of a housing development constructs four different

types of houses, with either a carport, a one-car garage, or a

16.2 Multiplication of Matrices

.VMUJQMJDBUJPOPG.BUSJDFT t *EFOUJUZ

.BUSJY t *OWFSTFPGB.BUSJY

The definition for the multiplication of matrices does not have an intuitive basis.

However, through the solution of a system of linear equations we can, at least in part,

show why multiplication is defined as it is. Consider Example 1.

E X A M P L E 1 Reasoning for the definition of multiplication

If we solve the system of equations

2x + y = 1

7x + 3y = 5

we get x = 2, y = -3. Checking this solution in each of the equations, we get

2122 + 11 -32 = 1

7122 + 31 -32 = 5

440

CHAPTER 16 Matrices; Systems of Linear Equations

2 1

Let us represent the coefficients of the equations by the matrix c

d and the solu7 3

2

tions by the matrix c

d . If we now indicate the multiplications of these matrices and

-3

perform it as shown

c

2 1

2

2122 + 11 -32

1

dc

d = c

d = c d

7 3 -3

7122 + 31 -32

5

we note that we obtain a matrix that properly represents the right-side values of the

equations. (Note the products and sums in the resulting matrix.)

Following reasons along the lines indicated in Example 1, we now define the multiplication of matrices. If the number of columns in a first matrix equals the number of

rows in a second matrix, the product of these matrices is formed as follows: The element in a specified row and a specified column of the product matrix is the sum of the

products formed by multiplying each element in the specified row of the first matrix by

the corresponding element in the specific column of the second matrix. The product

matrix will have the same number of rows as the first matrix and the same number of

columns as the second matrix. We summarize this as follows.

Multiplication of Matrices

r If A is an n * k matrix, and B is a k * m matrix, then their product AB is an

n * m matrix.

r The ijth element of AB is formed by multiplying each element of row i in A

by the corresponding element of column j in B and then adding these

products.

r The product BA requires that n = m, so just because you can form the product AB does not mean that you can form the product BA. Clearly, AB ≠ BA

in general, so that matrix multiplication is not commutative.

E X A M P L E 2 Multiplying matrices

Find the product AB and the product BA, where

2 1

A = C -3 0 S

1 2

B = c

-1 6 5

3 0 1

-2

d

-4

Matrix A is 3 * 2, and matrix B is 2 * 4, so the product AB can be formed, resulting in

a 3 * 4 matrix. The calculations are shown below, with the elements used to form the

element AB11 and the element AB32 outlined in colour.

2 1

-1 6 5

C -3 0 S J

3 0 1

1 2

21 -12 + 1132

-2

R = C -31 -12 + 0132

-4

11 -12 + 2132

1

= C3

5

Practice Exercise

1. Find the product AB.

-1

4

5

A = C 8 -2 S B = c

-7

0 12

-3

d

10

12

-18

6

11

-15

7

2162 + 1102

-3162 + 0102

1162 + 2102

2152 + 1112

-3152 + 0112

1152 + 2112

21 -22 + 11 -42

-31 -22 + 01 -42 S

11 -22 + 21 -42

-8

6S

-10

In trying to form the product BA, we see that B has four columns and A has three

rows. Since these numbers are not the same, the product cannot be formed.

16.2 Multiplication of Matrices

441

E X A M P L E 3 Multiplying matrices

■ Note that, if the second matrix had been to

the left of the first, then the product would have

had four rows and four columns.

c

-1 9

2 0

3

-7

6

-2 1

dD

1 3

3

The product of two matrices below may be formed because the first matrix has four

columns and the second matrix has four rows. The matrix is formed as shown.

-2

0

-1162 + 9112 + 3132 + 1 -22 132

T = c

-5

2162 + 0112 + 1 -72 132 + 1132

9

-6 + 9 + 9 - 6

12 + 0 - 21 + 3

6 -31

= c

d

-6

40

= c

2 + 0 - 15 - 18

d

-4 + 0 + 35 + 9

-11 -22 + 9102 + 31 -52 + 1 -22 192

d

21 -22 + 0102 + 1 -72 1 -52 + 1192

IDENTITY MATRIX

There are two special matrices of particular importance in the multiplication of matrices. The first of these is the identity matrix I, which is a square matrix with 1’s for

elements of the principal diagonal with all other elements zero. (The principal diagonal

starts with the element a11.) It has the property that if it is multiplied by another square

matrix with the same number of rows and columns, then the second matrix equals the

product matrix.

E X A M P L E 4 Identity matrix

Show that AI = IA = A for the matrix

A = c

2

4

-3

d

1

Since A is 2 * 2, we choose I to be 2 * 2. Therefore, for this case,

I = c

1 0

d

0 1

elements of principal diagonal are 1’s

Forming the indicated products, we have results as follows:

2 -3 1 0

dc

d

4

1 0 1

2112 + 1 -32 102 2102 + 1 -32 112

2

= c

d = c

4112 + 1102

4102 + 1112

4

1 0 2 -3

IA = c

dc

d

0 1 4

1

1122 + 0142 11 -32 + 0112

2 -3

= c

d = c

d

0122 + 1142 01 -32 + 1112

4

1

AI = c

Therefore, we see that AI = IA = A.

-3

d

1

INVERSE OF A MATRIX

For a given square matrix A, its inverse A-1 is the other important special matrix. The

matrix A and its inverse A-1 have the property that

AA-1 = A-1A = I

(16.5)

442

CHAPTER 16 Matrices; Systems of Linear Equations

If the product of two square matrices equals the identity matrix, the matrices are called

inverses of each other. Under certain conditions, the inverse of a given square matrix

may not exist. In the next section, we develop the procedure for finding the inverse of a

square matrix, and the section that follows shows how the inverse is used in the solution of systems of equations. At this point, we simply show that the product of certain

matrices equals the identity matrix and that therefore these matrices are inverses of

each other.

E X A M P L E 5 Inverse matrix

For the given matrices A and B, show that AB = BA = I, and therefore that B = A-1:

A = c

Practice Exercise

2. Show that AB = BA = I.

5 -7

3

A = c

d B = c

-2

3

2

1

-2

-3

d

7

B = c

7 3

d

2 1

Forming the products AB and BA, we have the following:

7

d

5

1 -3 7

dc

-2

7 2

7 3

1

BA = c

dc

2 1 -2

AB = c

3

7 - 6

3 - 3

1 0

d = c

d = c

d

1

-14 + 14 -6 + 7

0 1

-3

7 - 6 -21 + 21

1 0

d = c

d = c

d

7

2 - 2

-6 + 7

0 1

Since AB = I and BA = I, B = A-1 and A = B-1.

E X A M P L E 6 Matrix multiplication—application

A company makes three types of automobile parts. In one day, it produces 40 of type

X, 50 of type Y, and 80 of type Z. Required for production are 4 units of material and 1

worker-hour for type X, 5 units of material and 2 worker-hours for type Y, and 3 units

of material and 2 worker-hours for type Z. By representing the number of each type

produced as matrix A and the material and time requirements as matrix B, we have

units of worker-hours

material

type X type Y

A = 3 40

50

type Z

804

number of each type

produced

4 1

B = £5 2§

3 2

type X

type Y

type Z

material and time

required for each

The product AB gives the total number of units of material and the total number of

worker-hours needed for the day’s production in a one-row, two-column matrix:

4 1

AB = 3 40 50 80 4 £ 5 2 §

3 2

total

units of

material

total

worker-hours

= 3 160 + 250 + 240 40 + 100 + 1604 = 3 650 3004

Therefore, 650 units of material and 300 worker-hours are required.

y

E X A M P L E 7 Multiplication of matrices—application to robotic arms

2m

45∘

45∘

Fig. 16.4

x

Consider the two-dimensional robotic arm in Fig. 16.4, consisting of two links, each one

2.0 m long. The first link has been rotated 45° with respect to the x-axis, and the second

link has been rotated 45° with respect to the first link. We can find the coordinates of the

end of the arm by applying transformation matrices to the origin of our coordinate system.

x

We start by representing a point (x, y) as the vector (a 3 * 1 matrix) £ y § . In par0

1

ticular, the origin is represented by the vector £ 0 § . If the coordinate system is rotated

1

443

16.2 Multiplication of Matrices

by an angle u and then translated by m units in the x-direction and n units in the y-direction, the new coordinates of the origin are found by the matrix multiplication

■ See the chapter introduction.

-sin u

cos u

0

cos u

£ sin u

0

m 0

n § £0§

1 1

For our example, u = 45°, m = 2 (the length of a link), and n = 0, so the transformation matrix becomes

cos 45°

£ sin 45°

0

-sin 45°

cos 45°

0

1

2

12

1

0 § = £ 12

1

0

1

12

-

1

12

0

2

1

Since the process is repeated twice (once for each link), the location of the end of the

arm is given by

£

Some Properties of Matrix Algebra

1

12

1

12

0

1

- 12

1

12

0

1

2 12

1

0 § £ 12

1

0

1

- 12

1

12

0

2 0

0

0§ £0§ = £1

1 1

0

-1

0

0

2 + 22 0

2 + 22

22 § £ 0 § = £ 22 §

1

1

1

which represents the point 12 + 22,222.

This method can be generalized to robotic arms in three dimensions by including

more variables. Moreover, transformation matrices are also useful for computer graph■

We have seen that matrix multiplication is not commutative (see Example 2); that is,

AB ≠ BA in general. This differs from the multiplication of real numbers. Another

difference is that it is possible that AB = O, even though neither A nor B is O.

Moreover, the only number that does not have a multiplicative inverse is 0, yet there

are many nonzero matrices that do not have an inverse. There are also some similarities, for example, AI = A, where I and the number 1 are equivalent. Also, the distributive property A1B + C2 = AB + AC holds for matrix multiplication.

E XE RC IS ES 1 6 .2

In Exercises 1 and 2, make the given changes in the indicated examples

of this section and then perform the indicated multiplications.

1. In Example 2, interchange columns 1 and 2 in matrix A and then

do the multiplication.

2. In Example 5, in A change - 2 to 2 and -3 to 3, in B change 2 to

- 2 and 3 to - 3, and then do the multiplications.

In Exercises 3–14, perform the indicated multiplications.

3. 34

- 24 c

2

5. c

0

-3

7

-8

7. £

7

2

-6

3

4

-1 0

d

2 6

90

1

d £ - 25 §

-3

50

1

-8 § c 4

2

4

5

-3

d

5

4. 3 - 12

0

6. c

4

12

8. £ 43

36

6

4

- 23 4 £ 13

0

3

-1 2

d £1

11 2

6

-47

25

-18 § c d

66

-22

8

- 12 §

9

-1

1

-1

3

9. D

10

-5

11. c

13. c

14. c

2

5

7

5 2

Tc

-1 5

12

-3 3

dc

-1 7

-9.2

-3.8

1 2

-2 4

0

-5

1

d

-3

-1

d

8

6.5

2.3 0.5

d £ 4.9

-2.4 9.2

-1.8

1

-1

-6 6 1

d E 0U

0 1 2

5

2

10. 35 44 c

12. c

-5.2

1.7 §

6.9

-7

5

4

-5

-4

d

5

8 -90

dc

0

10

100

d

40

444

CHAPTER 16 Matrices; Systems of Linear Equations

In Exercises 15–18, find, if possible, AB and BA. If it is not possible,

explain why.

15. A = 31

16. A = c

17. A = c

-3

1

-2

B = £ 4

5

2 0

d

-4 5

- 10

42

-2

18. A = £ 3

0

-1

B = £ 5§

7

- 3 84

1

-1

2

7

-1

B = 34

1

21. A = £ 2

1

3

0

-2

0

-6 §

1

5

-2

3

24. A = c

5

1

25. A = £ 2

-1

1

26. A = £ 3

-2

-2

d

1

-4

d

-7

-2

-5

3

-1

-4

3

7

B = c

5

3

-5

3

-4

4

B = £3

1

8

B = £ 4

-1

39. Show that A2 - I = 1A + I 2 1A - I 2 for A = c

28

d

64

2 0

-3 1 §

1 3

-1

-2

-1

-5

-2

1

1

-1 §

-1

-4

-1 §

1

In Exercises 27–30, determine by matrix multiplication whether or not

A is the proper matrix of solution values.

1

27. 3x - 2y = - 1

A = c d

2

4x + y = 6

28. 4x + y = - 5

3x + 4y = 6

A = c

29. 3x + y + 2z = 1

x - 3y + 4z = - 3

2x + 2y + z = 1

30. 2x - y + z = 7

x - 3y + 2z = 6

3x + y - z = 8

2

2 § , show that A2 - 4A - 5I = O.

1

j 0

d , where j = 1- 1, show that J 2 = - I, J 3 = - J,

0 j

and J 4 = I. Explain the similarity with j 2, j 3, and j 4.

1 2

d

2 5

-4

d

-2

2

1

2

38. For J = c

- 1 54

-1

22. A = £ 4

2

B = c

a b

c d

d and B = c

d , show that AB = BA.

b a

d c

37. Using two rows and columns, show that 1 - I 2 2 = I.

In Exercises 23–26, determine whether or not B = A-1.

23. A = c

34. Show that A3B3 = AB

1

36. For matrix A = £ 2

2

-15

20. A = c

-5

-5

4

33. Show that B3 = B

35. For matrices A = c

In Exercises 19–22, show that AI = IA = A.

1 8

19. A = c

d

-2 2

32. Show that C 2 = O.

In Exercises 35–46, perform the indicated matrix multiplications.

6

B = £ - 15 §

12

25 40

d

-5 0

31. Show that A2 = A.

-2

d

3

-1

A = £ 2§

1

3

A = £ -2 §

-1

In Exercises 31–34, perform the indicated matrix multiplications,

using the following matrices. For matrix A, A2 = A * A.

2 -3 -5

1 - 2 -6

1 -3 - 4

4

5 § B = £ -3

2

9 § C = £ -1

3

A = £ -1

1 -3 -4

2

0 -3

1 -3 -4

2 4

d.

3 5

40. In the study of polarized light, the matrix product

1 0

1 0

1

c

d c

d c d occurs 1j = 1- 12 . Find this product.

0 -j 1 -j 1

41. In studying the motion of electrons, one of the Pauli spin matrices

0 -j

d , where j = 1- 1. Show that s2y = I.

used is sy = c

j 0

42. In analysing the motion of a robotic mechanism, the following

matrix multiplication is used. Perform the multiplication and

evaluate each element of the result. (See Example 7.)

-sin 60°

cos 60°

0

cos 60°

£ sin 60°

0

0 2

0§ £4§

1 1

43. In an ammeter, nearly all the electric current flows through a

shunt, and the remaining known fraction of current is measured

by the metre. See Fig. 16.5. From the given matrix equation, find

voltage V2 and current i2 in terms of V1, i1, and resistance R,

whichever may be applicable.

c

V2

d = £

i2

1

1

R

0

1

§c

V1

d

i1

V2

i1

i2

Shunt

R

V1

Fig. 16.5

44. In the theory related to the reproduction of colour photography,

the equation

X

1.0 0.1 0

x

£ Y § = £ 0.5 1.0 0.1 § £ y §

Z

0.3 0.4 1.0

z

is found. The X, Y, and Z represent the red, green, and blue densities of the reproductions, respectively, and the x, y, and z represent the red, green, and blue densities, respectively, of the subject.

Give the equations relating X, Y, and Z and x, y, and z.

45. The path of an earth satellite can be written as

3x y4 c

7.10

-1 x

d c d = 35.13 * 108 4

1 7.23 y

where distances are in kilometres. What type of curve is represented? (See Section 14.1.)

16.3 Finding the Inverse of a Matrix

46. Using Kirchhoff’s laws on the circuit shown in Fig. 16.6, the following matrix equation is found. By matrix multiplication, find

the resulting system of equations.

£

R1 + R2

- R2

0

- R2

R2 + R3 + R4

- R4

R1

V1

0

I1

V1

- R4 § £ I2 § = £ 0 §

R4 + R5

I3

- V2

R3

R2

I1

445

43

- 33

1. £ 54 -44 §

- 84 120

2. AB = BA = c

10

d

01

R5

R4

I2

V2

I3

Fig. 16.6

16.3 Finding the Inverse of a Matrix

Inverse of a 2 : 2.BUSJY t (BVTTo+PSEBO

.FUIPE t *OWFSTFPOB\$BMDVMBUPS

In this section we discuss two methods for finding the inverse of a square matrix. The

first method is straightforward, but it can only be used for 2 * 2 matrices.

The Inverse of a 2 : 2 Matrix

1. Evaluate the determinant of the matrix. If the determinant is zero, the inverse

does not exist.

2. Interchange the elements on the principal diagonal.

3. Change the signs of the off-diagonal elements.

4. Divide each resulting element by the determinant.

This method is illustrated in the following example.

E X A M P L E 1 Inverse of a 2 : 2 matrix—method 1

2 -3

d.

4 -7

First, we find the determinant of the original matrix, which means we evaluate

Find the inverse of the matrix A = c

22

4

-3 2

= -14 - 1 -122 = -2

-7

Since the determinant is not zero, the inverse exists. Next, we interchange the elements

on the principal diagonal and change the signs of the off-diagonal elements. This gives

us the matrix

c

-7 3

d

-4 2

signs changed

elements interchanged

We now divide each element of the second matrix by -2 (the determinant). This gives

Practice Exercise

1. Find the inverse:

3 -8

A = c

d

1 -2

■ See Exercise 37.

A-1

1 -7

=

c

-2 -4

-7

3

-2

d =

2

-4

-2

3

7 3

-2

Ơ = Ê2 2Đ

2

2 -1

-2

inverse

Checking by multiplication gives

AA-1 = c

2

4

-3 72

dc

-7 2

- 32

7 - 6

d = c

-1

14 - 14

-3 + 3

1 0

d = c

d = I

-6 + 7

0 1

Since AA-1 = I, the matrix A-1 is the proper inverse matrix.

446

CHAPTER 16 Matrices; Systems of Linear Equations

■ The Gauss–Jordan method is named for the

German mathematician Karl Gauss (1777–1855)

and the German geodesist Wilhelm Jordan

(1842–1899).

The second method, called the Gauss–Jordan method, is applicable for square

matrices of any size.

5IF*OWFSTFPGB.BUSJYCZUIF(BVTTo+PSEBO.FUIPE

1. Set up the given matrix and the identity matrix of the same size side by side.

2. Transform the given matrix into the identity matrix by performing any of the

following allowable row operations:

a. Any two rows may be interchanged.

b. Every element in any row may be multiplied by any number other than

zero.

c. Any row may be replaced by a row whose elements are the sum of a

nonzero multiple of itself and a nonzero multiple of another row.

Work one column at a time, transforming the columns in order from left to

right.

3. At every step, perform the same row operations on the identity matrix. The

resulting matrix will be the required inverse.

Note that these are row operations, not column operations, and that they are the operations used in solving a system of equations by addition and subtraction.

E X A M P L E 2 2 : 2*OWFSTF(BVTTo+PSEBONFUIPE

Practice Exercise

2. Find the inverse using the Gauss–Jordan

3 -8

method: A = c

d

1 -2

Find the inverse of the matrix

A = c

2

4

-3

d

-7

this is the same matrix as in Example 1

First, we set up the given matrix with the identity matrix side by side.

c

2

4

-3 1 0

`

d

-7 0 1

The vertical line simply shows the separation of the two matrices.

We wish to transform the left matrix into the identity matrix. Therefore, the first

requirement is a 1 for element a11. Therefore, we divide all elements of the first row

by 2. This gives the following setup:

LEARNING TIP

As in Example 2, (1) always work one

column at a time, from left to right,

and (2) never undo the work in a previously completed column.

c

1

4

- 32 12 0

`

d

-7 0 1

Next, we want to have a zero for element a21. Therefore, we subtract 4 times each element of row 1 from the corresponding element in row 2, replacing the elements of

row 2. This gives us the following setup:

c

1

4 - 4112

1

- 32

2

`

-7 - 41 - 32 2 0 - 41 12 2

0

d

1 - 4102

or

c

1

0

1

- 32

0

` 2

d

-1 -2 1

Next, we want to have 1, not -1, for element a22. Therefore, we multiply each element of row 2 by -1. This gives

c

1

0

- 32 12

`

1 2

0

d

-1

Finally, we want zero for element a12. Therefore, we add 32 times each element of row 2

to the corresponding elements of row 1, replacing row 1. This gives

c

1 + 32 102

0

- 32 + 32 112

`

1

1

2

+ 32 122

2

0 + 32 1 -12

d

-1

or

c

1 0 72

`

0 1 2

- 32

d

-1

447

16.3 Finding the Inverse of a Matrix

At this point, we have transformed the given matrix into the identity matrix, and the

identity matrix into the inverse. Therefore, the matrix to the right of the vertical bar in

the last setup is the required inverse. Thus,

- 32

d

-1

7

A-1 = c 2

2

As was done in Example 2, it is generally best to make the element on the principal

diagonal for the column 1 first and then make all other elements in the column 0. Let us

consider two more examples.

E X A M P L E 3 2 : 2*OWFSTF(BVTTo+PSEBONFUIPE

Find the inverse of the matrix c

original setup

c

-3 6 2 1 0

d

4 5 0 1

-3 6

d.

4 5

c

row 1 divided by - 3

c

1

4

-2

`

5

- 4 times row 1

Therefore, A-1 = c

5

- 39

4

39

2

13

1 d,

13

-2 2 - 13

4

13

3

0

d

1

row 2 divided by 13

- 13

0

d

0 1

1

0

c

1

0

-2

`

1

- 13

4

39

c

5

1 0 2 - 39

4

0 1

39

A-1

I

2

13

1 d

13

0

1 d

13

2 times row 2

which can be checked by multiplication.

E X A M P L E 4 3 : 3*OWFSTF(BVTTo+PSEBONFUIPE

1

Find the inverse of the matrix £ 3

-2

original setup

1

£ 3

-2

2

5

-1

1

£ 0

-2

2

-1

-1

-1 1 0 0

-1 3 0 1 0 §

-2 0 0 1

1 2

£0 1

0 3

-1 1

-2 3 3

-4 2

-1

1 0 0

3

2 -3 1 0 §

-2

0 0 1

1 0

£0 1

0 3

3 -5

-2 3 3

-4

2

1 0

£0 1

0 0

3 -5

-2 3 3

2 -7

- 3 times row 1

2 times row 1

1

£0

0

2

-1

3

-1

1 0 0

2 3 -3 1 0 §

-4

2 0 1

row 2 multiplied by - 1

-1

-1 § .

-2

2

5

-1

- 2 times row 2

0 0

-1 0 §

0 1

- 3 times row 2

row 3 divided by 2

1 0

£0 1

0 0

3 -5

-2 3 3

1 - 72

2 0

-1 0 §

3

2

2 times row 3

1

2

2 0

-1 0 §

0 1

1 0 3 -5 2 0

£ 0 1 0 3 -4 2 0 §

0 0 1 - 72 32 12

2 0

-1 0 §

3 1

1 0 0 11

2

£ 0 1 0 3 -4

0 0 1 - 72

- 3 times row 3

I

- 52

2

3

2

A-1

- 32

1

2

Therefore, the required inverse matrix is

11

2

£ -4

- 72

which may be checked by multiplication.

- 52

2

3

2

- 32

1

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