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CHAPTER Matrices

2

Section 2.1

Operations with Matrices .....................................................................30

Section 2.2

Properties of Matrix Operations ..........................................................36

Section 2.3

The Inverse of a Matrix ........................................................................41

Section 2.4

Elementary Matrices .............................................................................46

Section 2.5

Markov Chains .......................................................................................53

Section 2.6

More Applications of Matrix Operations ...........................................62

Review Exercises ...........................................................................................................66 Project Solutions............................................................................................................75

Š 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


CHAPTER Matrices Section 2.1

2

Operations with Matrices

2. x = 13, y = 12 2 y = 18 y= 9

4. x + 2 = 2 x + 6 −4 = x 2 x = −8 x = −4

y + 2 = 11 y= 9

 6 −1  1 4  6 + 1 −1 + 4  7 3       4+ 5 =  1 9 6. (a) A + B =  2 4 + −1 5 = 2 + (−1          1 10  −3 + 1 5 + 10 −35 −2 15  6 −1  1 4  6 − 1 −1 − 4  5 −5       4 − 5 =  3 −1 (b) A − B =  2 4 − −1 5 = 2 − (−1          −3 − 1 5 − 10 −5 −4  −35   1 10 

(

 6 −1

−610

 1 4 2

 

−1

5 +

 1 4

=

12 −2

2( 5)

A=

2( −3)

 12 −2  (d) 2A − B =  4 8  

(e) B +

)

2( 4)

−35

1

(

2 6 2 −1 

)

 =  2( 2)

4

(c) 2A = 2 2 

)

)

4

8

−610

 12 − 1 −2 − 4

 11 −6

5 = 4 − −1 8− 5 = 5 3 − −1 (       1 10   −6 − 1 10 − 10 −70 

 6 − 1  1 4  3 −  2 4 = − 1 5 +  1

1

5 

1 2

)

3

 

2

2

4 = 0 

7

2

7

25 

1

 1 10  2 2   2 2   1 10 −35  3 2 −1 0 2 1  3 + 0 2 + 2 −1 + 1  3 4 0         2 5 2 7 8. (a) A + B =  4 5 +  4 2 =  + 5 4 + 4 5 + 2 =  8 7         0 0  + 21 + 1 2 +   0 1 2  2 1 0  2 2 2  3 2 −1 0 2 1  3 − 0 2 − 2 −1 − 1  3 0 −2         (b) A − B = 2 4 5 −  5 4 2 =  2 − 5 4 − 4 5 − 2 =  −3 0 3         − 21 − 1 2 − 0 0 −2 02  0 1 2  2 1 0  3 2 −1  2 3 2 2 2 −1  6 4 −2 

( )

( )

(

)

2( 5) = 4 8 10 4 5  = 2 ( 2 ) 2 ( 4 ) 20  0 2 4 21 22     ( ) () ( )    3 2 −1 0 2 1 6 4 −2 0 2 1         (d) 2A − B = 22 4 4 2 = 4 8 10 −  5 4 2 5 −  5         0 1 2 2 1 0 0 2 4 2 1 0  0 2 1  3 2 − 1 0 2 1 1 −1 3        2 2 1 2 1 (e) B + 2 A =  5 4 2 + 2  4 5 =  5 4 2 +  1 2 2

(c) 2A = 22

 0 1 2

5

2 30

1 0

0

1 2

2

1 0

 6 

=  −1 

2 −3  4 8  −21 4 3 1 

 3  2  = 6 6

1

0

2

1

2 

3

 

2

2

1

9

2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.1 Operations with Matrices 14. Simplifying the right side of the equation produces

10. (a) A + B is not possible. A and B have different sizes.

w x −4 + 2 y3 + 2w   = . y x    2 + 2z −1 + 2x By setting corresponding entries equal to each other, you obtain four equations. −2 y + w = −4 w = −4 + 2 y  x= 3 + 2w x − 2w = 3   y= 2 + 2z y − 2z = 2  x  = 1 x = −1 + 2x 3 The solution to this linear system is: x = 1, y = 2, z

(b) A − B is not possible. A and B have different sizes.  3  6     (c) 2A = 2 2 =  4     −1

−2

(d) 2 A − B is not possible. A and B have different sizes. 1 (e) B + 2 A is not possible. A and B have different sizes. 12. (a) c23 = 5a23 + 2b23 = 5( 2) + 2(11) = 32 (b) c32 = 5a32 + 2b32 = 5(1) + 2( 4) = 13 16. (a)

AB =



2 − 2 4

=  (



−1 4 (b) BA = 

)(

)

(

( )

( )

3 1



( )

()

)(

(

)(

(



11

)(

)

)

(

()

(b) BA = 2

1

1

()

()

()

)( )

()

(

)( )

( )

2 1 + ( −1 1

+ 8 −3

()

)( )

 1

−3 2

  31

−1

( )

( )

)

(

( )

)

(

)( )

2 −1 + 1 −1

()

1 1 +

−3 2 + 2 3 1

( )



 3



1  1

2

20. (a)

AB =

0

 4 −2

4 −4



−1

2

1 −2

()

(

)

(

)

( )

( )

( )

+

((

)

(

()

+ −2 2

41

)

(

(

( )

)

(b)

)

( )

)()

(

()

() )

(

3 2 + 2 −1 + 1 − 2 − + + 4 − 2) 32 0 −1

()

( )

(

)

+(−2

− 4 14 2

   () ( )( ) ( ) BA is not defined because B is 3 × 2 and A is 3 × 3.

2

22. (a)  

  

()

 −1

 

= 3 2

2 1

AB = 

 1

(b) BA =  2 1 3

2

−21

( )

−1 3

23 −

()

11

(

−1 +

( )

−1 2

23

13

( )

22 

( )  =−  ( )

) (

( )

1 4

+ 2 −1 

 (   8 = 1

)

 

2 = 2 −1

(

)

− 2−1 − 3 − 2  2 6 4 

−2

−6 −4

 2

1

3

( )

12 

−4 

( )

22

( )

(

)

 4

2

+ 1 2+ 3 − 2 + 2 1 = − 4

 

21

 

−19

2 −14

− 418

)

( )

(−4 − 2 

) (

)

−1 

−2

( )

()

−22

12

 

() ()

21

 () 

−1 1

22

−2

()

 −1 2

( )

 9 0 13

) ( 

7 5

2 7 ) + 1 8 + 1 −1  = 7 −2

()

+ 2 11 7 + −3 8

)

() ()

(

( )

1 7 + 1 8 + 2 −1 

+11

(

−1 + −3 −1

= − 3 1 + 0 2 +4 1

4

) (

3 1 + 2 2 +1 1

2



−3

( )

( )

−1 2 

()( )

( ) ))

)

()

1 −1 + 1 −1 1 + 2 1

21 +12 +13

( )

( ) ((

() ))

( )

)( )

−33 2 + 1 1 +

) (

(

+ −1 1 + 7 2  6 −21 15  2 2 + ( −1 1 + 8 2  = 8 −23 19

(

( )

(

+ 1 1 + −1

1 1 +1 2 + 2 3

8 =

( ) ()

()

 

+ 7 1 1 1 + −1 1 + 7 −31 2

 3 1 + 1 2 + −1 13 1

2 −1

)( )

)( )

−17





4 −9

)

= 2 1 + ( −1 2 + 8 1

  −1 1 −3 2

 1 2  1

1

6



) ( 

4

=

−1 1 + 4 − 2

 1 1 + −1 2

1 2

 2

(

4 − 2 +1 4  7 − 4 ) ( ( )  =  2 2 + − 2 −12 − 2 + −2 4 6−12

4

 1 −17  1 AB = 2 −1 8

()

− 2 22 1 + − 2 − 2 

−1 4 + 4 2



− 2 −1

18. (a)

(

1  2 − 2 4 2 + 1 −1  = ( ) ( ( )

2 )



−2

4 2

)

1 = − 4, and w = −1.

2 4 +

1

31


 1   24. (a) AB is not defined because A is 2 × 2 and B is 3 × 2. 

 2 − 3

2 1 (b) BA = 1 3 

 

2

5

 

2 −1

( )

( )

2 2 + 1 5 = 1 2 +3 5 

 

( )

( )

22 +

(

( )

( )

2 −3 +12 1 − 3 +3 2

(

9 − 4 17 3  −1 −8

(

) )

)( ) ( )

( )

(

−1 52 − 3 + −1 2

=  

)( )

 

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


32

Chapter 2 Matrices  2 1 2  4 0 1 3    3 −1 − 2 −1 2 − 3 −1    1 − 2  − 2 1 4 3 − 2

26. (a) AB =



( )

(

)

(

)

2 4 + 1 −1 + 2 − 2 ( )

(

= 34 + 

−24

3

)

7

(

)

( )

()

20 +12 ( )

+ 21

(

)( )

(

)( )

3 0 + −1 2 + − 2 1 −20 + 12 + − 21

()

(

)

( )

2 1 + 1 −3 + 2 4 ()

(

)(

)

)

( )

( )

)( )

(

3 1 + −1 − 3 + − 2 4 − 2 1 + 1− 3 + − 2 4

) (

) ( ) ((

4

 = 17

(

)

−1 −1 + − 2 − 2 + 1 −1 + − 2 −2

( )

 

)(

)(

()

()

( )

)

( )

( )

(

)(

)

(

 )( )

3 3 + −1 −1 + − 2 3 − 2 3 + 1−1 + −2 3  ) (

) ( ) ( )

(

(

2 3 + 1 −1 + 2 3

( )

( )

()

11

 − 24

−4

 − 5

0 −13 −13

(b) BA is not defined because B is 3 × 4 and A is 3 × 3. 28. (a) AB is not defined because A is 2 × 5 and B is 2 × 2. (b) BA =  = 

 1 6  1 0 3 −2 4  ()

(

()



( ) ( ) ( ) ( ) ( ) ( ) 1 1 + 6 6 ) 1 0 + 6 13 1 3 + 6 8 1 −2 ) + 6 −17 ) 1 4 +6 20  4 1 + 2 6 4 0 + 2 13 4 3 + 2 8 4 −2 + 2 −17 4 4 + 2 20

4 2 6 13 8 −17 20

37 78 =

(

)

()

(

)

( )

( )

( )

( )

(

(

(

)

(

)

51 −104 124 

16

26 28 −4256 30. C + E is not defined because C and E have different sizes. 32. − 4A is defined and has size 3 × 4 because A has size 3 × 4.

40. In matrix form Ax = b, the system is 2 3  x1 

 5

1 4 x

10



 = 

.

 2    Use Gauss-Jordan elimination on the augmented matrix.

2 3 5  1 0 −2   

34. BE is defined. Because B has size 3 × 4 and E has size 4 × 3, the size of BE is 3 × 3.

36. 2D + C is defined and has size 4 × 2 because 2D and C

So, the solution is

4 10

0

1  x1  

3

=

−2 

.

x

have size 4 × 2.

 3 42. In matrix form Ax = b, the system is  2

38. As a system of linear equations, Ax = 0 is x1 + 2x2 + x3 + 3x4 = 0 x1 − x2

1

+ x4 = 0.

x2 − x3 + 2x4 = 0 Use Gauss-Jordan elimination on the augmented matrix for this system. 1 2 1 3 0 1 0 0 2 0     0 1 0  0 1 0 1 0  1 −1     0  −1 2 0 0 1 −1 0 0 1 Choosing x4 = t, the solution is x1 = − 2t , x2 = −t , x3 = t, and x4 = t, where t is any real number.

−49  x1 



−3 x

−13 = .

 1  2  12 Use Gauss-Jordan elimination on the augmented matrix. −49  

1 −3

−13 1 0 −23   − 35  12 0 1   3

So, the solution is  x1 

x2

=  −23. 

− 35

3


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Section 2.1 Operations with Matrices 44. In matrix form Ax = b, the system is 1 1

−3  x1 



50. The augmented matrix row reduces as follows. 1

−1

2   2 =  1.       1 −1 1  x3  2 Use Gauss-Jordan elimination on the augmented matrix. 0

−1

1 1 

x

 1 0 

− 3 −1 

−1

2 0 1

−1

1

0   3 2

1 2

x

 1 

.

46. In matrix form Ax = b, the system is

 1 −1 4  x1

 1 

So,thesolutionisx2=3

 17

 

  2 = −11.     0  −6 5  x3  40 Use Gauss-Jordan elimination on the augmented matrix.  1 −1 4 17  1 0 0 4     1 0 −5  1 3 0 −11  0     0 0  −6 540  1 2 0  4  x1  3 0

x

So, the solution is

 =

x  2

−5.

 

x

0

0 0   x1 

0

 0  0

1 0

1 1

0 0   x2   1 0  x3 

0

  = 0.

 0

0

0

1 1   x4 

0

 

x

1

0

0

1 0

1 1

0 1

5

 0 

0

0

1

So, the solution is  x1 

  x  2 

x3  =

 

0 0 1 0 0

0 0 0 0 1

−1    1  

 

x

5 

4  02.



3

−1

 

−1

1 0  0

4 

1 −2.

0

 0

0

−2

 3 −2 a    21

a

a

3a11

22 

2a12 − a22 

 1 0

3a12 − 2a22 

0 1

 =

= 1

21

a22 = 0

− 2a21

= 0

3a12 − 2a22 = 1. Solving by Gauss-Jordan elimination yields

a

a

= 2, a12 = −1, 21 = 3, and a22 = −2. 2 −1 So, you have A =  . 11

3

−2

−1 .

 x4

0 1 0 0 0 1

0 0 0 1 0 1 

1 0 

− 1 1 − 1 1 − 1 5 

1 0 0 0 0 −1 

0

a12 

2a12

 

0 0  0 0  0 0 

2 −1  a11 A=

2a11

Use Gauss-Jordan elimination on the augmented matrix. 1   0  0

3

 

− 1 1 − 1 1 −1  5 

0

 32  4 −8 54. Expanding the left side of the equation produces

 2a11 − a21 = 3a11 − 2a21 and you obtain the system

 

0

−3 5 −22  1 −3 10     4  0 9 −18  3 4        4 −8 32 −4 8 0 1 −3 10    0 1 −2    0  1 −2 So, −22 −3  5       4 4 3 b=   =   + ( −2) 4.      

1

0

1 3 2

There are an infinite number of solutions. For example,

1

 

0 1 3 2

2 −1

 3  2 48. In matrix form Ax = b, the system is 

 

−2 −3 

     0  1 52. The augmented matrix row reduces as follows.

3  

2

−1

1 0 2 4 1   0 2 3  0  

2

2

2 3

1 2

x 

3

0 0

x3 = 0, x2 = 2, x1 = −3.  1 2  1       So, b = 3 = −3−1 + 20 +      

2

0 1 0

33

 1

−1

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34

Chapter 2

Matrices  3 0 0 −7 0 0   60. AB = 0 −5 0  0 4   0 0  0 0 0

56. Expanding the left side of the matrix equation produces a b2  

1  2a + 3b a + b  3 17  =  = .

 c d 3

1

2c + 3d c + d

4

−1

2 a + 3b = a+

b = 17,

4

d = −1.

Solving by Gauss-Jordan elimination yields a = 48, b = −31, c = −7, and d = 6. 2 0

58. AA =

0

0

0 2 0 0 4 0 0     9 0 −3 0 0 −3 0 = 0    

0

0

12  0 + 0 + 0 0 + 0 + 0

0+ 0+ 0 0+

  0+ 0+ 0 −21 0 0   =  0 −20 0.    0 0 0 Similarly, −21 0 0   BA =  0 −20 0.  

and c+

+ 0+ 0

= 

3

2 c + 3d =

)

(

3 −7

You obtain two systems of linear equations (one involving a and b and the other involving c and d).

0

( −5)4 + 0 0 + 0 + 0

0+

0 + 0 0 + 0 + 0

0

0 0 0 a11 0

0 0  0 0 0  b b 0  b11 12 13   a11b11 a11b12 a11b13  a  22    a b b 0 b21   22 23  = 62. (a) AB =  0 22 b21 a22 b22 a22 b23        0 0 a33  b31 b32 b33  a33b31 a33b32 a33b33  The ith row of B has been multiplied by aii , the ith diagonal entry of A.  b11 (b) BA =

b

 21

b

b

b

b

12

22

 b31 b32

13

 a11 

a

23

 0  b33  0

0

0

22

0

a b a b

 a11b11

22

 = a11b21

 0 a33 

  33 23  

12

33 13

a b a b 22

22

  a11b31 a22 b32 a33b33 

The ith column of B has been multiplied by aii , the ith diagonal entry of A. (c) If a11 = a 22 = a33 , then AB = a11B = BA. 64. The trace is the sum of the elements on the main diagonal. 1+1+1= 3

ij

(

1+ 0+ 2+

n

68. Let AB = c , where c  

66. The trace is the sum of the elements on the main diagonal.

= ij

 a b . Then, Tr ik kj

(AB

)

n

=

k =1

Similarly, if BA = d , d ij = 

ij 

 ii

n  n

=

i=1

n  ik

c

   aik bki i = 1  k =1

n

b a kj . Then Tr BA =

k =1

(

)

 dii i=1

cos α 70. AB = 

=

n 

i=1

−sin α  cos β −sin β   cos α cos β − sin α sinβ    sin α cos α sin cos β β    sin α cos β + cos α sin β cos β − sin β  cos α −sin α  cos β cos α − sin β sin α BA =    cos β   sin α cos α  sin β cos α + cos β sin α  sin β cos α + β − sin α + β  ). ) So, you see that AB = BA =  ( (

 sin(α

 

n

ba

 ik

 k =1

)

−3 = 0

.

 ki

 = Tr( AB .  

)

cos α ( − sin β ) − sin α cos β   sin α ( − sin β ) + cos α cos β  cos β ( − sin α ) − sin β cos α   sin β ( − sin α ) + cos β cos α 

+ β ) cos(α + β )

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.1 Operations with Matrices  a11 a12

72. Let A =

a

a

a

a

21



22 

b

b12 

b

21

21

.

22 

0 1

 b  b11 12  a11 a12 

 −

22 

b b

 1 0 AB − BA =  is equivalent to

Then the matrix equation  a11 a12   b11

22

21

 b11 b12 

B=

and

35

b

b 21



22

This equation implies that

a b +a b −b a −b a 11 11 12 21 11 11 12 21 a b +a b −b a −b a

a

a

21

 1 0

=

.

22 

0 1

=a b −b a 12 21 12 21 =a b −b a

=1

22 22 21 12 22 22 21 12 21 12 = 1 which is impossible. So, the original equation has no solution.

21 12

74. Assume that A is an m × n matrix and B is a p × q matrix. Because the product AB is defined, you know that n = p. Moreover, because AB is square, you know that m = q. Therefore, B must be of order n × m, which implies that the product BA is defined.

n

cij =

 aik bkj . Then, c sj k =1

sj n

n

=

 a sk bkj , and ctj

=

k =1

, n. Let AB = c , where

= a for j = 1,

76. Let rows s and t be identical in the matrix A. So, a

 ij 

ij

 atk bkj . Because a sk

= atk for k

= 1,, n, rows s and t of AB

k =1

are the same. 70

78. (a) No, the matrices have different sizes. 80.

(b) No, the matrices have different sizes.

1.2

(c) Yes; No, BA is undefined.

35

50 25 84  = 100 70

42

60 30  120 84

1 82. (a) Multiply the matrix for 2010 by 3090 . This produces a matrix giving the information as percents of the total population.

A=

1 3090

12,306 35,240  41,830 16,095  27,799

 5698 

72,075

7830   9051   14,985

13,717

12,222 31,867

2710  

3.98   5.21

9.00

1.84 

11.40 2.53  13.54 2.93  23.33

4.85

4.44

0.88 

3.96 10.31 1.91 1 Multiply the matrix for 2013 by 3160 . This produces a matrix giving the information as percents of the total population. 12,026  15,772

B=

1 3160

(b) B− A=

27,954

 5710 

5901 

35,471 8446   41,985 9791   73,703

14,067

16,727

3104  

 3.81  ≈

12,124 32,614 6636   3.81 11.23 2.67 3.98    13.29 3.10 4.99 5.21    8.85 23.32 5.29 − 9.00 

1.81 

4.45

0.98 

1.84 

11.23

2.67 

4.99

13.29

8.85

23.32

1.81 

3.10

4.45

5.29

0.98 

3.84

10.32 2.10 2.53 − 0.18 − 0.18 0.14    13.54 − 0.25 2.93 0.17 − 0.22    23.33 4.85 = − 0.15 − 0.001 0.44 11.40

4.44 0.88 

− 0.04 

0.01

0.11 

3.84 10.32 2.10 3.96 10.31 1.91 − 0.12 0.01 0.19 (c) The 65+ age group is projected to show relative growth from 2010 to 2013 over all regions because its column in B − A contains all positive percents.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


36

Chapter 2 Matrices

84. AB =

0 1 0  1 2 3 4  1 2    0 0 0 1 5 6 7 8  5 6 −10 0   1 2 3  =  −1 −2

     

0

0

4



0 −1 0 0 5

−5

6 7 8



3

4 

7 −3

8

−4

−6 −7 −8

86. (a) True. The number of elements in a row of the first matrix must be equal to the number of elements in a column of the second matrix. See page 43 of the text. (b) True. See page 45 of the text.

Section 2.2 2.

6 8  0  +

  

2 (

4.

1

5 −11 −7  6 + 0 + −11  +  = ( ( )

−1 0

−3

−1

2 −1

 



−1 +

)

6. −1 −2 −1 +



4 11  

 9 3

 −5 −1    1 6   

 

7 5  





  

2

+

1

2

−3

−4 +

0 1  1 3  =  

10. (a + b B = ( 3 + ( −4 )

)

0

14. (a) X = 3A − 2B −6−3   = 3 0    9 −12  6 −9   = −1 0   17

−10

0

1 = 0

−1 2

( ab)O = ( 3)( −4) 0 0 = ( −12)0 0 = 0 0

2 9

−1

()

6 12

−3 + 2

−9

−3

11

−11 + 3  − 3 1 +1  =  1 2 

2  3

1  2 

1 2

31

3  3

2 

 −8−1

−1 2

12.

2 −7

2 6

 0 1 = (−1 0

))

−9+ 1

)

 19

( −1

−4 −11  13    1 + −1 = 2

2

19 4 − 14 9 =

)

2

3

0 + 6 13 +

6

=  2 + (−1

2 1

4+

1−63

+

1

1

−1 2

−2

−1 + 5

  2 4

−9

−2

)

( −9 )

3+

−3

−4 −11

3 4

6

−9



6 

−1

 −5 + 7 1

=

8. A +

)

(

−4 −11 =

6 −1

 

 1 2  B=  + 

 −5 =

)

(

5 + 14 − 2 + 64 + − 18 0 + 9 =

 

( ) (

−3 + 2 0 + −1 +

2 1

3 4 + −9 −1

0 13

8 + 5 + −7 

)

5 − 24 0 + 14 6 − 18 9 = 

Properties of Matrix Operations

 0

0 

   

0 6  4 0

−8 −2

1

−1 

−2

 0 0  (b) 2X = 2A − B −4 −2   2X =  2 0    6 −8 −4−5   2X =  0 0    10 −7 

−4 −1

5 − 2

−2

X =  

 0 3    2 0  

0 5−

0

7

2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.2 Properties of Matrix Operations 2X + 3A = B

(c)

(d)

−6−3 

 0 3   0 =  2 0   

3  9

2X + 

−4

−12

−1

 6  2X =  −1 

6  0 

−13 11

 3 X =  −1

16. c ( C B) =

 0

( −2)  



0

=

2

2

−1

−8

2 A + 4 B = −2X  0 12    8 0 = −2X   −16

−4

 −4   10 

10 

−10

−12

0

 2 −5   −5

0

 5

6

= −2X

= X

− 3 4 − 4 2  1 − 5 0    24. (a) AB C =    0 1 3 3  ( )   1 −3 −2  

2



2 

( − 2)

3 



−2  0 + 

11

1  1  −1 0

−1

3

2  − 13

−4   2   6

= 2 − 4  





2

 −31 

2



 1 3 − 2 − 4 − 6 =  =

− 1 2 

0

−22



26. AB =

−2 −100 

 2

− 3 4 − 5 0    3 3  0 1

 

−12

−1 2 − 1 0 − 2 −1  1 3   1 2 3  =   ( −2)  1 −1  −1  0 

 − 4 2 − 3 −1 =     1 − 3  3 − 2  18 0 =  



1 3  0 1  = 

0  5

− 1 0 − 2 − 1  3 −1 0 0  1 3   0 1 20. B ( C + O) =    +   −1  −1 0 0 0

0 10

BA =

 1 1  1 4 2  2 1  1 1 2 2   2  1 1  1 2 2   4 1  1 1

2

4



2

1 2 1

5

3

4 1 2

 

  

 

−1

1 

1 4

= 8  1

3

 2 8

3

=     2 1

1 8 2

 8

1

3 −2 1 4

≠ AB

3

4

 1 2 3 0 0 0   28. AC = 0 5 0 4 0     =  

1

 − 4 2   1   (b) A( BC) =  1 − 3  − 2



 0 1  −3 1 − 2 −1 =   = 



 −1

 −1 1

−12

22. B ( cA)

 



 18 = 

 6  0 1   1 3  0 1  18. C ( BC) =     −1 0  −1 0 −1 2

=



2

  

− 3 4 26 −8 6   7 −14 − 9  0 1 = 

 − 1 − 3

37

12 −6 9  

= 16 −8 12

0

  4 −23

−2 3

4 −6 3 0 0 0   5 4 4 0 0 0



= BC

−10 1 4 −2 3

But A ≠ B. 2 4  1 −2 0 0 30. AB =   1  =  = O 

2

4

 2

1

0

0

But A ≠ O and B ≠ O.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


38

Chapter 2 Matrices

 1 2  1 0  1 2 32. AT =    =   0 0  −1 01  −1 1 34. A + IA = 

 1 0  1

−1

0 1 0 −1

 +

0

1 =  

2

0

−1

2



2  1  +

 1 2  1 2  1 0 2 36. A =    =   = I2 0 1 0 −1 0 −1       1 0 I 2 2 4 2 = I2 = 2 =  So, A = ( A ) .

0

2 2  = 

0

−1

38. In general, AB ≠ BA for matrices.

4 

0

 6 −7

−2

T

40. D

= −7  

42. ( AB)

  1 2 −3 −1 

= 

 0

−2





T

 1 1



=



2 1

 T

T

−4 −2

−3 −1  1 2 B T AT =    =

T

= 

23

T

 6 −7  = −7 0  

19 0

 19 T

1

 

23 − 32

 19

19 23

 

23 − 32

1 −4 1 −2

 

−3 2  10 1 −4   = 

1

2 1 0 −2  −1 1 2 −2  −2 T 4 0 −7  2 1 −1  1 0 −1 T  4 2 4        T 44. ( AB) =  0 1 3 2 1 = 2 4 4 2 −2 7 =  0         4 0 2 0 1 3  4 2 2 −7 7 2 T T  1 0 −1 2 1 −1  1 2 0  2 0 4  4 2 4          T T B A = 2 1 −2 01 3 =  0 1 1  1 1 0 =  0 4 2        

0

46. (a)

1 3 4 0 2 −1 −2 3 −1 3 2  1 −1  1 3 0 11   10   3 4 T  = 11 21 A A = −1 4 −2  

0

 1 −1   1 T  (b) AA =  3 4 −1 

2

−84 0  252 8   8 77 2 0 11 −1    =  168 −70 −2 12    50 4  6 8 −5 4 −104

11 12

−9



 4 −3 2 0  4 2 14 6   0 −2 0 11 −1 −3  2 T (b) AA =    2 11 12 −2 12 

−9

14

6  

0

17

 29  8  30  =  86   4 −10   −5



0

0

0

0

0

1

( )

50.

1

17

A

0 

= 0 

( )

−1 0

17

1

0

0

 0

0

0



0

0

17

()

 0 

0 

(

0

−117

)

0

()

17

4  0 −1 −9 

8 −5

168 −104  −70 50  294

2

−5 14

−1 −9

 6  4 −3 

−30 −2 8 

T

AA =  2   0

−2

 −2  4 2 14  48. (a)

7 2

 2 −1 2 3 0    25 −8 4 −2 = −1



0

−7

0

0

−139

30 86 −10  126 169 −47  169 425 −28

−47 −28

0

141 

0

0 0 1 0

0

 0

98

−1 00 0

0

=

−139

0

0

−1 0   0 0 0 0 1



1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.2 1

52.

(

0

20

)

0

A

(

0

0

0

0

=  0

−1 0

 0

0

0

 0

0

0

20

)

0

()

0

20

0



−1

(

0

20

)

0 0

0 0 

 

=

 0 0 1 0 0

0

0 0

 

1 0  0 0 0 0 1

( )20 

0

39

 0 1 0 0 0

20

1

1 

 

0

Properties of Matrix Operations



1 

3

54. Because

−1

= 0

A

 3 2  

8 0 0 

0 =

0

(

0

0   0 , you have 3

0 )3

−1

0 27

A=

0

2 0 0 

0.

0

( 3) 

0

0 −1

0 3

1

56. (a) False. In general, for n × n matrices A and B it is not true that AB = BA. For example, let A = 2 Then AB =

1 1

0≠

0 0

 1 1

(b) False. Let A =

0 0

11

0

1 0

1 

0

,B

1 0

=

. 1 0 

= BA.

1 0 =  , C

,B

2 0 = . Then AB

2 =

0

0

0

0

0

= AC, but B ≠ C.

(c) True. See Theorem 2.6, part 2 on page 57.

(

)

aX + A bB

58.

)

(

= b AB +

IB

Original equation

( Ab ) B = b ( AB + B)

aX +

Associative property; property of the identity matrix Property of scalar multiplication; distributive property

aX + bAB = bAB + bB

( −bAB ) =

aX + bAB +

( −bAB ) bAB + bB + ( −bAB) bAB + ( −bAB ) + bB

Add − bAB to both sides.

bAB + bB +

aX =

Additive inverse

aX = aX = bB

Commutative property Additive inverse

X= bB a

2

1

   0 0 1  − 1 1 3 10 0 0  10 5 − 5     = −  0 100 +  5 0 10 

 0

=

−5  0

= −5 =

0 0 10 5 − 5

5 5 5 − 5 5 − 10 10

 0   5  − 8

0 3

 2 1 −1 2 1 −1

   2 1 02 1 0 2     −1 13 − 1 1 3

   −1 13

 2 1 − 1 2 1 − 1

2

5 +

   − 4 212

 12 2 − 6

3 −11 21

   − 1 1 3

− 5 5 15  6 1 − 3

5 − 10 10 − 2

5  46 −12

3

 2 1 − 1  2 1 −1 0  2 1 − 1        0 + 5 1 0 2 − 2 1 0 2 +  1 0 2

1 0

60. f ( A) = −10 0

Divide by a.

6 10 + 4 24

1 − 1 6 1 − 3

1 0 2



0 3 5

   − 1 13 − 4

 16 3 −13

− 2 5 21

  −18

+

8 44

2

    1 02 1 0 2     −1 13 −1 1 3

 

2 12

 

−15 9 25 

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


40

Chapter 2 Matrices

(

)

(

( )  ij 

) ij 

(

 

ij

)

ij 

(

)

62. cd A = cd a  =  cd a  = c da = c da  = c dA 64. c + d A = c + d a  =  c + d a  = ca + da  = ca  + da  = c a  + d a  = cA + dA ( )  ij  ( ij   ij   ij   ij   ij  ) ( ) ij   ij

66. (a) To show that A( BC) = ( AB)C, compare the ijth entries in the matrices on both sides of this equality. Assume that A has size n × p , B has size p × r, and C has size r × m. Then the entry in the kth row and the jth column of BC is

lr=1 bkl clj . Therefore, the entry in ith row and jth column of A(BC) is p

r

k =1

l =1

=

 aik  bkl clj

 aik bkl clj . k,l

The entry in the ith row and jth column of (AB)C is l =1 d il clj , where dil is the entry of AB in the ith row and the r

lth column. So, d il = r

kp=1 aik bkl for each l =

1, , r. So, the ijth entry of ( AB)C is

p

  aik bkl clj =aik bkl cij . i =1 k =1

(

k,l

)

Because all corresponding entries of A(BC) and (AB)C are equal and both matrices are of the same size n × m , you (

)

(

)

conclude that A BC =

AB C.

(b) The entry in the ith row and jth column of ( A + B)C is ( ail + bil ) c1 j + ( ai 2 entry in the ith row and jth column of AC + BC is ( ai1c1 j + distributive law for real numbers.

+ bi 2 ) c2 j ++ ( ain

+ bin )cnj , whereas the

+ ain cnj ) + (bi1c1 j ++ bin cnj ), which are equal by the

+a b

i 2 2 j ++ ain bnj . The corresponding entry for ( cA)B is (c) The entry in the ith row and jth column of c( AB) is c ai1b1 j cb . ca b + (ca b ++ (ca b and the corresponding entry for A(cB) is a cb + a cb ++ a

(

i1 )

i2)

1j

(

in ) nj

2j

)

Because these three expressions are equal, you have shown that c AB T

(

=

T

)

i1 (

1j)

)

i2(

2 j)

in (

nj )

cA B = A cB . T

= a  + b  = a + b  = a + b  = a  + b  = A + B ( ) (  ij  ij  )  ij ij   ji  ji   ji  ji  T T T cA = ca  A (3) = c a  T = ca  = c a  = c ) ( ) (  ij  )  ij   ji  (  ji 

68. (2) A + B

(

T

T

(4) The entry in the ith row and jth column of ( AB) is a j1b1i T

T

T

and jth column of B A is b1i a j1

+ a j 2b2i +a jn bni . On the other hand, the entry in the ith row

+b a 2i

70. (a) Answers will vary. Sample answer:

j2 ++ bni a jn , which is the same. 0 1 −1 1  1 0   =  

1 0

1 0



−1 1

(b) Let A and B be symmetric. (

)

T T T If AB = BA, then AB = B A = BA = AB and AB is symmetric.

If ( AB ) = AB, then AB = T

72. Because A =

2  1  1  3

74. Because − A =

( AB )T

= BT AT = BA and AB = BA.

T

= A , the matrix is symmetric.

0 − 2 1  T 0 3 = A , the matrix is skew-symmetric.

2

1 − 3 0 

76. If AT = − A and B = −B, then ( A + B ) = A + B = − A − B = −( A + B), which implies that A + B is skewT

T

T

T

symmetric.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.3 The Inverse of a Matrix

A=

78. Let



a

11

a

a

a

a a

12

21

a

a

22

1n

13



23 2n

 

.

a

a

n2

n1

T A − A =  a11 a12

a

a  21

22

a a  n 3 nn  a a a23a2

n

 

a

a

n1

n2

0

a  21

− a12

 

a

 a11

13 1n

a

a  12

a a  n 3 nn 

a

21

22

 

=

41

a

1n

2n

a

a

an2 

a  nn

3n

32 2 n

an1 − a1n an 2 − a2 n an 3 − a3n a −a a −a 12 21 13 31 0 

a −a

0

= 

− an2  0

  − ( a12 − a21)

− an1 

a −a a 23

32

a12 − a21 a13 − a31a1n 0

an1 

31

a

23

a1n − an1 

a2 n − an2 

32

0 

 − ( a1n

− an1 ) − ( a2 n − an 2 ) − ( a3 n − an3 )

T

So, A − A is skew-symmetric.

Section 2.3 2. AB =

 

The Inverse of a Matrix

1 −1 2 1  

−12 

 2−1 1 − 1  1 0 =  =  −1 + 2 −2 + 2 0 1 2 − 1 −2 + 2  1 0 = =  

1 1

2 1  1 −1 BA =    2  1 1 −1 4. AB =

 1   2

 BA = 

−1  3



6. AB =

5 2

1

5

2 3



0 

4 −3 −1 11 −7

 3 6 −5 

1



3

=

d

−b,

− bc −c a

1 ad

where

a A= 

c d

b  2 − 2 .  =  2

2

So, the inverse is  1 1

0 1

1 1 2  

=

  6 −2

=

 1 0 0   

3 −2  3 6 −5

 1 1 2  2 −17 11

BA = 2 

0

1

−1

A

A−1

1 −1 1 0  = 

−111 −7 2 4 −3

2 −17 11  

 

0

1

5

1  5 5  1 − 2   5 5 3

−1 + 2

 1 0  5 = 

3 −

1−1

8. Use the formula

0 1 0 

2( 2 ) − ( − 2 )( 2) −

=

 

4 4 

.

1 1  4 4 

10. Use the formula

0 0 1   1 0 0

 2 2  =  2 2 

 

1

A−1 =

d

1 ad

−b 

− bc −c

,

a

0 1 0

0 0 1

where a A= 

b  1 −2  = .

c

d

2 −3

So, the inverse is

A−1 =

 −3 2

1 ( )(

)

(

)(

1 −3 − −2 2

 −3 2

= 

)

−21

.

−21

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


42

Chapter 2 Matrices

12. Using the formula

A−1

d

1

=

−b

a

ad − bc −c

,

where a b −1 1 A=   =  c  d  3 −3

)( ) ( )( )

(

you see that ad − bc = matrix has no inverse.

−1 −3 −

1 3 = 0. So, the 22. Adjoin the identity matrix to form  

14. Adjoin the identity matrix to form  1   A I =  3 

2

2 1 0

7

9 0

0 

1 0.

 1 −4 −7 0 0  Using elementary row operations, reduce the matrix as follows. 6 4  1 0 0 −13   − 1   = 0 1 0 12 −5 −3    −1

I A

0

 0 1 −5 2 16. Adjoin the identity matrix to form

 A I =

 10 

5 −7 1 0

1

0 

1 4 0 1 0.   1   3 2 −2 0 0 Using elementary row operations, reduce the matrix as follows.  1 0 0 −10 − 4 27    −1  = 01 0 2 1 −5    −5

I A

0 0 1 −13 −5 35

Therefore, the inverse is −10 −4 27  1 − 1 −5 . A = 2   

0.1 0.2 0.3 1 0 0   A I = − 0.3 0.2 0.2 0 1 0.    0.5 0.5 0.5 0 0 1 Using elementary row operations, reduce the matrix as follows. 1 0 0 0 −2 0.8     −1 = 01 0 −10 4 4.4   0  0 1 10 −2 −3.2 Therefore, the inverse is  0 −2 0.8 I

A

1 A− = −10 

−5 35 18. Adjoin the identity matrix to form  3 2 5 1 0 0

  2 2 4 0 1 0 .

−4

 A I =

4 0 0 01 Using elementary row operations, you cannot form the identity matrix on the left side. Therefore, the matrix has no inverse.

4.4 .

4

 10 −2 −3.2 24. Adjoin the identity matrix to form 1 

1 0 0   A I = 3 0 0 0 1 0   5 5 0 0 1 2 Using elementary row operations, you cannot form the identity matrix on the left side. Therefore, the matrix has no inverse.

0 0

26. Adjoin the identity matrix to form 1 

 − 13

20. Adjoin the identity matrix to form  5  1 11 1 0 0 − 6 6 3 2  A I =  0 2 0 1 0. 3  1 −1 −5 0 0 1   2 2 Using elementary row operations, you cannot form the identity matrix on the left side. Therefore, the matrix has no inverse.

0

 A I =

0

0 0 1 0 0

2

0 0 0 1 0

0

0 3 0 0 0

0

.

−2 0 0 0 1 0

0 0

0

0  

1 

Using elementary row operations, reduce the matrix as follows. 1 0 0 0 1 0 0 0  

0

−1 

1 0

0 0

A  =

I

0

1 0

0

1 2

0 0

0 0 0 1 0 0  Therefore, the inverse is

A

−1

1 0

0

0

0

0

0

= 

0

1 2

0 0 

−2 0

1 −2

0

 

 0

1

3

 

1

0

0 0

.

0 1

3

 

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.3 The Inverse of a Matrix 28. Adjoin the identity matrix to form 4 8 −7 

 A I =

2

1 0 0 0  6 0 1 0 0

−4

5

1 −7 0 0 1 0

3 6

−5 10 0 0 0 1

−7

A−1

36. A = 

−1

0 

A

0

0 1

( )

A− 2 = ( A2 )

0

9

5

8

A−2 =

38.

1

)

5

− 1 3  4

− 9

− 3

2 9

4

9

3

( − 14 )( 89 ) − ( 94 )( 53 ) =

 8 36  9 143  5

( ) 1 A−

0

0

9

1

A − 2 = ( A2 )

−1

−4 

−

= 

32 143 60

143 36

81  143  9 

−4   143 143   1 6 − 7  2  =    = 1  3

2

 47

0

2 

5

2209



1  − 31 56 −

= 

− 40

1

=

1 2209

1 − 56  40 − 31

 1 − 56   40

− 31

The results are equal.

0

0.2 

2.6  −0.8

0.1

.

0.2 

 −15 − 4  1 =  2  −1 0

2

28  



2

 

−1

1

1

(

−12

2 

44

−3   =

−5

0

.

0 5 0 0 0

Therefore, the inverse is  1 −1.5 −4  0.5 1 0 −1 A = 0 0 −0.5

1 A−

−  4

= −

3

0

2 40. A− =

− 12 − 2 − 3 5 = 24 − 15 = 9

−2

1 9

ad − bc =

2

=

2

)

(

8

−1 

1 −

ad − bc =

5 −2 )(

Using elementary row operations, reduce the matrix as follows.  1 0 0 0 1 −1.5 − 4 2.6    1 0 0 0 0.5 1 − 0.8 0 −1 I .  A  = 0 0 − 0.5 0.1 0 1 0 0

0 

−2 1 0 0 1 0 0

−123

0

0

1

2 −1

 1 3 −2 0 1 0 0  2 4 6 0 1 0  

2  = 

4 3

34. A =

)( )

(

− −2 −3 = − 4

12

2

20

0

2 )

A−1 = − 1 

Therefore the inverse is  27 −10 4 −29   5 −2 18 −1  −16 A =  −17 . 4 −2   2 −1  −7 8   30. Adjoin the identity matrix to form

 A I = 0

( )(

Using elementary row operations, reduce the matrix as follows. 1 0 0 0 27 −104 −29   0 −16 5 −2 18 0 1 0   A I = 0 0 1 −220 0 −17 4 0 0 0 1

ad − bc =

 1 −2

−3

.

0 2

32. A =

14

43



873

= 4

61

1

228 −1604  16 −112 

  23 − 42  2420 6 344 −1317 − 1 32 − 4 228 −1604  48  873    1  = − 29 48 −17 = 4  61 16 −112 

 22 The results are equal.

9

15

−1317

− 344

2420

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


44

Chapter 2 Matrices

42. (a)

( AB )−1

= B − A− 1

5

2

11

11

=3

(b)

( A T )−1 (

77  (

−1

)

=

1 A−1

(

(b)

7

)

− 2

T

=

 3

2

2

=

7

2

1 7 =

3

7

−  7

 

7

 1

2

7

− 2 

 7

3

7 −  7

1

1 T 7

7

=

7

1

1 14

3

1

7

(c)

 

4

2

1

1

and the formula for the inverse of a 2 × 2 matrix produces 1 A− =

 1 1

1

1

 =

1

1

4

4

− 1 2 

.  2

−3

1 



7

2

3

−1 . 

2

3

−1

−1

1 2

b

1

2

  0  

1

−1

1 3 2

= 3

1

3

3

− 1 

0



 =

1

 1

−1

− 1

1 

x

3

= 1.

−1

1

2

3

3



−1 

 

0

1

x

x

The solution is: 1 = 1, x2 = 0, and 3 = 1. 50. Using a graphing utility or software program, you have Ax = b  1

 2

x = A−1b = −1

 0

 1 where 1 1

 2 1 A = 1 1  2 1 4

−13 −1 1 −1

1

 1

 x1

   x2 

   2 −1 , x = x3 , and b = 

1 −1

 x4 

 3

   4 

3 .

−1

 x5  1 −2 1  5 x The solution is: x1 = 1, x2 = 2, 3 = − 1, x4 = 0, and

1

3

−1

The solution is: x1 = 1, x2 = 1, and 1 1  −1  1      −1 2 1 (b) x = A b =  3 3 − 1  2 = 0

4

1

−1 b

1

 =  4 2 −  2

2 + 2 −2

(a) x = A

1

 2 1  2 46. The coefficient matrix for each system is 2 −1 A=   

2

(a) x = A

1 3

 −1 

3

1

( 2 A) −1

1 −1

 6 5 −3  1 −4 2    −2 4−1 0

1.

1

   4 4 2 1  1 3 −6 −25 24   7 = −6 10   15 7  17 T  1 −4 2  1 0 4     T = ( A−1) = 0 1 3 = −4 1 2     4 1 1  2 3  2     1 −4 2  1 −2  2 1 −1 1 1 3 = 2 A = 2 0 1 3 =  0 2 2

( AT ) −1

A−1 =

7

14

 2

−2

Using the algorithm to invert a matrix, you find that the inverse is 1 1 

−91

1 1 1 AB )− = B − A−

=

2

A = 1

9

2

44. (a)

7

 3

 

−1 2A

(c)

1

2 − 7

11 7

1 −4

= A



1



11 =

48. The coefficient matrix for each system is 1 1 −2

1

=

5

x

 1

5

3 1

= 1.

The solution is: x = 1 and y = 5. (b) x = A

−1 b

 =

1 4

− 1  2 

1 4  1 

 −1 

2

 −3

 

=

−1

 −1

The solution is: x = −1 and y = −1.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.3 The Inverse of a Matrix 52. Using a graphing utility or software program, you have Ax = b

60. Using the formula for the inverse of a 2 × 2 matrix, you have

−1

2 

A−1

1

x = A−1b =  

 3 

=  sec θ

 4 −2 4 2 −5 −1

A=

   

6 −5 −6 3 3 −3

−15 2 −3 −5

  −2

b= 

,

x

5

x

4

 

 x6 

I

. −9

−11

 x3  and 

 0

115.56

1

−12

−tan θ  .

0.010 0.008 1 0 0   F I = 0.010 0.012 0.010 0 1 0.   1 0.010 0.017 0 0  0.008 Using elementary row operations, reduce the matrix as follows. 1 4.44 0 0 115.56 −100    F −1  = 0 1 0 −100 250 −100 

 0

  

1 2

3 −4 −6 1

,x = −14 −4 −6 2 4

So, F − =    1

 x2 = 2, x3 = 1, x4 = 3,

x5 = 0, and x6

=

1

x − 4

.

64. A

)

=

66. ( I − 2A)( I − 2A) = I

 0

 

−1

=

2

− 2IA − 2AI + 4A

= I − 4A + 4A

I − 4A + 4A

=

( because A =

A2

= I

Then, multiply by

1 4

1

1 8

=

4

3

16

1 2 −4   4 + 12  3 2

1

=

1

8

=

3

16

1

− 4  8 1

68. Because ABC = I , A is invertible and A− = BC. So, 1

ABC A = A and BC A = I.

So, B− = CA. 1

to obtain −4  1  8

1

 32 3  64 

T

1

)

)

2

1

− 16  1  32  .

70. Let A2 = A and suppose A is nonsingular. Then, A−1 exists, and you have the following. A−

1

(A2)=

(A −1 A) A =

So, ( I − 2 A)− = I − 2 A.

4A =  4A −1  (

( 4

2

58. First, find 4 A.

4A

( AT )−1.

3 that 4 x = −6 or x = − 2.

1

 −15

0

0.15 = 37.5 .



= I n = In and

1 T

−3 4 ad − bc = ( x)( 4) − ( −3)( 2) = 0, which implies

A=

 

( A − 1 ) T = ( A −1 A ) T

So, ( A −

 x 2 The matrix   will be singular if

250 − 100 −100 115.56

2

1

1 1 Letting A− = A, you find that x − 4 = −1. So, x = 3. 56.

T

4.44



1 − 2 − x

−100 and

115.56 4.44 −100 115.56 − 100 4.44 

w = F − d =  − 100

= 1.

250

54. The inverse of A is given by

A−1

1 4.44 −100 115.56 −100 4.44

0

−100

The solution is: x1 = −1,

0.017 

 x2 

1 3 −1 −2

3

  x1  

−tan θ  sec θ 

−tan θ sec θ  62. Adjoin the identity matrix to form

where 3 2

  −ca = ad − bc  sec θ 1  = sec2 θ − tan2 θ −tan θ 

    1 

−b

0

d

1

45

A− A 1

I

A= I

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

)

−15


46

Chapter 2 Matrices

72. (a) True. See Theorem 2.8, part 1 on page 67.

a

74. A has an inverse if ii ≠ 0 for all i = 1n and  1  0 0 0 

1 1 

, (b) False. For example, consider the matrix  0 0   which is not invertible, but 1 ⋅ 1 − 0 ⋅ 0 = 1 ≠ 0.

−2

1

 −34

 2 4

(b) A( 5 ( 2 I − A)) 1

+

−4

1 = 5

5 0

   0

1 

0

(2A−

( 1( 2 I − A)) A = 5

I. Or,

0

0

a

nn 

0 0

=

 −3  −4 A2

0 . 

22

A2 − 2A + 5I =

Similarly,

a

 1 2

(a)

1

1 A− =  0

(c) False. If A is a square matrix then the system Ax = b has a unique solution if and only if A is a nonsingular matrix.

76. A =

a  11

2 

)= 1 5

1

0 5 

5 ( 5I ) = I

( 2I − A)

=

0

0

−2

1 1 5

2

=

A −1 directly.

1

 (c) The calculation in part (b) did not depend on the entries of A. 78. Let C be the inverse of ( I − AB), that is C = ( I − AB)− . Then C( I − AB) = ( I − AB)C = I. Consider the matrix I + BCA. Claim that this matrix is the inverse of I − BA. To check this claim, 1

show that ( I + BCA)( I − BA) = ( I − BA)( I + BCA) = I. First, show ( I − BA)( I + BCA) = I − BA + BCA − BABCA = I − BA + B ( C − ABC ) A = I − BA + B (( I − AB )C ) A = I

= 1 − BA + BA = 1 Similarly, show ( I + BCA)( I − BA) = I. 80. Answers will vary. Sample answer:  A= 

1 0  or

1 0 A=  

1

 −1

 0 0 a b   d −b d −b 1 1 a b  1 ad − bc 1 82. AA− =     =    =  a − bc  c d − ca  c d  ad − bc − c ad ad − bc  0  d −b a b ad − bc 0   1 0 1 1 1 A− A =    =   =  ad − bc − c

Section 2.4

a



c d

ad − bc 

0

ad − bc

0 1

0   = ad − bc

 

1 0  0 1

Elementary Matrices

2. This matrix is not elementary, because it is not square. 4. This matrix is elementary. It can be obtained by interchanging the two rows of I2. 6. This matrix is elementary. It can be obtained by multiplying the first row of I3 by 2, and adding the result to the third row.

8. This matrix is not elementary, because two elementary row operations are required to obtain it from I4. 10. C is obtained by adding the third row of A to the first row. So, 1 0 1 E=

 0  0

1 0 . 0

1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.4 Elementary Matrices

47

12. A is obtained by adding −1 times the third row of C to the first row. So, 1

0 −1

 0  0

E=

1 0 .

0 1 14. Answers will vary. Sample answer: Matrix  1 −1 2 − 2

  0 3 −3  0 0 2

2

0 1 −1

2

2

0 0 2

2

0 1 −1

 0 0 1

So,

R1 ↔ R2

0 1 0

   1 0 0    0 0 1

(

 1 −1 2 − 2

Elementary Matrix

6

 1 −1 2 − 2 

Elementary Row Operation

1 3

)R

1 0 2

0

→ R2

 

1

0

1 0

1

0





(

)R

1

2

 3

3

 0 −1   3 −2

0 1 0 0

3



1 −1





1

 0  0

−2

2



( − 2)R1

0

( −1 R

1 

0 −11 − 4  1 3 0

0 1 1

 0

  0

0

0

−1

1

2. 

0

1

0 1

  − 2 1 0    001 

+ R2 → R2

(

0 0

 0

1 0

)2

 0

2

)2

 

0 1 0 0

1

→ R

−1 0

  0 0 1 1 0 0   0 1 0    0 11 1 

11 R + R → R 3

3

1 0 0 1

(7

0 7 

 1

  −3

+ R3 → R3

)R

 3

→ R3

0 1  1 0 0  1 0 0  1 0    So, 0 1 0 0 1 0 0 −1   1  

Elementary Matrix

0 7  1 3 0 0 1 1

0 1 0

( − 3)R1

2

−11 − 4 1 3 0 1

 1 0 0

−1

=

− 4 3 0

−1

1  0 0  2  6  1 −1 2 − 2   

0 −1

→ R3

2 0 0 2  0 0 1 0 0 1 0 0 16. Answers will vary. Sample answer: Matrix Elementary Row Operation 1

0

3

0 0 1 1 0 0

 1 0 0  1 0 0 0 1 0 0 3 − 3     1 

0

1

0

11 1 0

0

1

 0  1 0   

0

0 1 − 3 0

0

7

0  1 0 0  1   0 1 0 − 2



1



1 0 2



3 5

 1 3 0   1 1. −1 = 0    0 

0 0 1   3 − 2 − 4

0

0 1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


48

Chapter 2

Matrices

18. Matrix 1 − 6 0 − 3  0 17  8 4 1 − 6 0 − 3  0 17  0  32 1 − 6  1 0 0 17  0 32 1 − 6  1 0 0 0  0  32 1 − 6 0 1  0 0  0 0 1 − 6  1 0 0 0  0  0 1 − 6 0 1  0 0  0 0 1 − 6  1 0 0 0 

0

0 So, 1  0

 0

0

Elementary Row Operations 0 2 3 9   −1 − 3  −5 1  0 2 3 9   −1 − 3  − 5 − 7 0 2  −1 − 3 −1 − 3  − 5 − 7 0 2  −1 − 3 16 48  − 5 − 7 0 2  −1 − 3  16 48  27 89  0 2  −1 − 3 1 3  27 89  0 2  −1 − 3  1 3 0 0

0

→ R3

R4 +

(

( − 4)R2

→ R4

1 

)

1 − 3 R2 →

R2

( −17)R2

( − 32)R2

( 16 )R 1

1 0

3

8

1

1 

0

0

0

0

0

( − 27)R3

(18 )R4

→ R4

1 0  1 0 0 0  0 0 1 0  1 0 0 0 0 0

→ R4

2

0

0

0

1 0 0

0



1

− 27 1 

01  00



0 0 1 0

0

3

0

0



0 0



1 16

0 0

 0 0 0

1 

0 0 0

0 0 0 1  1

0

− 5 1

0

0 1 0 0 0 1   0

0

0

1 0 0 

 0 0



0

0 1 0 − 2

0

 

 1 

0 0 1

0  0

0 

− 27 1 0 0  0 0 1 0 0 1  

8

0 0  0 0  1 0 0 1

1 0 − 32

0 0 0  1 

0

1

 16 0 0 0 

R4 +

0

0 0  1 0 0 −17 1 0  0 0 1 0 0 0  1 0 0 0 1 0  − 32 0 1 0 0 0  1 0 0

0 

→ R3

4

0

0 

→ R2

0 1  10 0 0 1− 4 0 0 1    1 − 6 0 2 1 − 6 0 2     0 9 0 −3 3 1 −1    − 3   =   0 1 0 3 . 5 −1 1 2 0

0

0



−17 0

0

1 

0

0

0

0

1

0 1

1  R4 +

0 

−3 0

0

→ R3

0

1

0

 

8

0

0 

R3 +

 0 0 1

− 4

3

0

1 0 0   0 0 1 0

− 3

0

0 0 0  1 0 0 0 1 0  0 0 1 0 0 0

0

0 1  0 0 0 1  1 0 0 0 0 1  0 0 1  0

0

( − 2)R1

1   0 − 2   0 1

−1 1

1 

8

R3 +

Elementary Matrix

0 0 0  1

0

0

 0 1 0 1 0 0 

0

0

1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.4 Elementary Matrices

49

20. To obtain the inverse matrix, reverse the elementary row operation that produced it. So, multiply the 1 first row by 25 to obtain 

1

E−1

0

=5

.

0

1

 22. To obtain the inverse matrix, reverse the elementary row operation that produced it. So, add 3 times the second row to the third row to obtain  1 0 0  1 E− = 0 1 0 .   

0 3 1  24. To obtain the inverse matrix, reverse the elementary row operation that produced it. So, add −k times the third row to the second row to obtain 1 0 0 0   1 −k 0 −1 0 . E = 0 1  

0

0

0 1 0 0   26. Find a sequence of elementary row operations that can be used to rewrite A in reduced row-echelon form. 1 

1 E1 =  2

0 1 R → R ( 2 ) 1 1

1

 0

1

0 

1

 1 0  

1

E

0 

2 =  R2 − R1 → R2 −1 1 Use the elementary matrices to find the inverse.  1  1 1 A− = E E =  1 0 2 0 =  2 0

0 1

2 1



−1

1



−

0

1

1

1

2

28. Find a sequence of elementary row operations that can be used to rewrite A in reduced row-echelon form.  1 0 −2

1

0 1

2

 

( )R 1

→ R2

2

2

E1 =

0 1

0 

1

0 1

0 1 0

0

1

(2)

2

0

→ R

R

3

1

E3 = 0 1

1

R −

2

 00

1 0

1 1

0 1  1 0 2

0

0

0 0

0

0

2

 

E2 =

2

0 

0 1  1 0 0 R1 + 2R3 → R1  1 0

 1 0 0  1

 

2

2

0 0

1

1

Use the elementary matrices to find the inverse. 1 0

0  1 0 2  1 0

1 A − = E3 E2 E1 = 0 1

1 0

= 0 1 

0

2

− 1  0 2

1 2

0 1 0



2





0 1

0 



0 0

2  1 0 

0

1

0

 

1 0 0 1 0 0 1

0

1 0

0 = 0 

1  0 0 1 

0

1 2

0

2 − 1

2

1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


50

Chapter 2 Matrices

For Exercises 30–36, answers will vary. Sample answers are shown below.

0 30. The matrix

A

1

=   is itself an elementary matrix, so the factorization is

10

0 1

A=

 1 0

.

32. Reduce the matrix A =

 1 1 

as follows.

2

 1 Elementary Row Operation

Matrix 1 

 1 0

1

Add −2 times row one to row two.

0

E =

1  1 0

Multiply row two by −1.

1 E2 = 

Add −1 times row two to row one.

1 −1 E3 =  

0

1

0

So, one way to factor A is

= E−1 E−1 E−1

A

1 2 3

0 

0 −1

0 

−2 1

1

 −1  1 1 

Elementary Matrix

1

=  1 0  1 0  1 1. 





2 1 0

−1 0 1 2 3  5 34. Reduce the matrix A = 2 6 as follows.    1 3 4 Matrix Elementary Row Operation 1 

1 2  1 0  1  3 1 2

0 1

3  

0

4

Elementary Matrix

Add −2 times row one to row two.

E1 =

Add −1 times row one to row three.

E2 =

3 0

0 1

 

0

 0

0

0

0 0 1  1 2 0  1 0  0 1  1 0 0  1 0 

E

Add −1 times row two to row three.

3

  0  1

0 0  1 0  0 1 0 0

1

0 0

−2

0 1 1  1 2 3

 1 

   0 1 0    −1 01 

  0 1 0    0 −1 1

=

1 0 −3   E4 = 0 1 0   0 1 0 1 −2 0   E5 = 0 1 0   0 1  0 

Add −3 times row three to row one.

Add −2 times row two to row one.

0 0 1

So, one way to factor A is  1 0 0  1 0 0  1 0 0  1 0 3  1 2 0



A = E1−1 E2−1 E3−1 E4−1 E5−1 = 2 1 0 



0 0 1

  1

0 1 0



0 1 0



0 1

  0





1 1



0 1 0

0 1 0 .



0 0 1

  0

0 1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.4 Elementary Matrices

51

36. Find a sequence of elementary row operations that can be used to rewrite A in reduced row-echelon form. 1 

0 0

1

0

1 0

1

0

0 −1 0 0

2

1  0

0

0

1 

0 0 1 0

2  1

0

0 −1

1 

−2

 1 0 0 0 0 

0  1 

 1

00

00

1 

00

0 

10 

0

−2

− R3 → R3

00

1

( 12 )R

R1 −

4

0

0 0

1

0 0

0

10 0 1

0

0

0 0

−2  5

0 0

0 

1 0 0 −1

0

0 0

1

0

 

−2

0

0 1

0

0

0 0 00

1

0

 

1 

0 

10 0 1

−1

0

0

00

1

1 

00

0 

0

10 0 1

0

00

1 

0

E6 =  

E7 =  

0

0 

0

0 0 10

1

0

E5 = 

R3 + 2R4 → R3

1

1

0

0

R2 − R4 → R2

0  0

→ R1

1 0 0 1

0

0 0 1 0

0 0 0 0

= 

1

0 1

0

0

4

0

0

E

0

0 1

0

→ R4

00

=

0  1 

1

0 

4

−2

 0 1 0 0

(− 5)

R

00 10 0 1

3

1

0

0 

2

1

 0

1 

2 

0 

E

 

0 0 10

2

00 10

0 0 1

R4 − R1 → R4

1

0

5 

2 

0 0 0 1

1

2

2  1 

0 0 1 0

0 −1 0 0 −

0

1 4 0 E1 =  0  0  1   0 E2 =  0  −1

1

−2 

1

R → R

1

1

 (4) 1

2

0

2

So, one way to factor A is A = E1−1 E2− 1 E3− 1 E4−1 E5−1 E6−1 E7−1

4 0 0  0 1 0

= 

0

0

0 0 0 

0  1 0 0 0  1 0 0   1 0 0 0 0 0 1 0 1  0 1  0 1 0



0

1  1 0 0 

0



0

0 1 0 0  00 1 0



00 0

1 0 0 0 − 5 0 0 0  2 

0  1 0 0  0 0

1  

0

2

0 0 0  1 0 0 0 

1 0 0 0 1 0 1 0 1 0 0



−1 0 0

1  1



0 0



1 0 0 0 1 0 0 0 1 0 1

 0 0 0



1 0 0 0 

 . −2

1 

38. (a) EA has the same rows as A except the two rows that are interchanged in E will be interchanged in EA. (b) Multiplying a matrix on the left by E interchanges the same two rows that are interchanged from In in E. 2

So, multiplying E by itself interchanges the rows twice and E = In.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


52

Chapter 2

Matrices

1 1 1  1 0 0 −  1 0 0 −  1 a 0 −

40. A

−1



= 0 1 

0 

1

c

0 0 1

b



0 0



  1 0 

= 0 1 0   0 1 

0 0



0

0

1

Ek , where each Ei is an

elementary matrix, then A is invertible (because every elementary matrix is) and A− = Ek− 1

1

1

(c) True. See equivalent conditions (2) and (3) of Theorem 2.15. 44. Matrix −2  −6

Elementary Matrix

1 = A

4

−2 

1 = U

0

1

1 E1 = 

3 1

0

10

2  0

0 12 

2  0

0

3

0 0 1 0

1



− b

   0 

−a

0

 

1 + ab 0 .  1 0  c

48. Matrix

Elementary Matrix

 2 0  1 −2  6 2   0 0

0 −1 1

0 1

6 2 1 0

0 0 0 2 0 0 0 

3

0 0 0 

 1 0 0 0 

 1 1 0 0

E1 =

0 0 1 0

−10

0 0 0 1 

 1 0 0 0   0 1 0 0

E2 =

0 1 

−1

2 0 0 0  0

 = A

−1

0

0 0

−10

0 

0 −1

1  = LU

=

0 1

0 0 0

  1  

−3 01 0

 0 0 0 1

 1 0 0 0  

−10

0

E3 = 

= U 0 

−1

0

1 0 0

−2

1 0

0 0 0 1 

1 1 1 E3 E2 E1 A = U  A = E1− E2− E3− U

 1 0 0 0 2 0 0 0   −10 −1 1 0 0 0 1  0 0   3 2 1 3     0 0 0 1 0 0 0 −1 

 1 0 0   E1 =  0 1 0

0 0  −31 = U  0 7

 0 1 Elementary Matrix

0 0  −3 1 = A  12 3 0 0  −31

1 0



0 2 1 0

0 

 1 0−2 1 E1 A = U  A = E1− U =  

2 

0 

0 1

−3

46. Matrix

0−b1 0 0

−a

2 0 0 0 

E2− E1− . 1

 1 0 0  1  

 1  0  c

42. (a) False. It is impossible to obtain the zero matrix by applying any elementary row operation to the identity matrix. (b) True. If A = E1 E 2

 0

1  

0

 1 0 0 0  y1

0 1 0

4

1

0



= LU

0 0 

E2 E1 A = U  A = E1− E2− U 0 0 2 0 0 1    = 0 −3 1 1 0 0    −4 1 0 0 7 5 = LU 1

0

−5 01

E2 =

=

Ly = b: 

1





 

4

4

−1 1 0 0  y2  − 4 =      y 15   3 2 1 0  3   −1  0 0 0 1  y 

y1 = 4, − y1 + y 2 = −4  y2 = 0, 3y1 + 2 y 2 + y3 = 15  y3 = 3, and y4 2 0 0 0  x1   4 





4

0 1 −10   :

= −1.

 x2   0 x =   Ux = y   3 0 3 0 0   3    0 0 0 −1  x  −1 x4 = 1, x3 = 1, x 2 − x3 = 0  x2 = 1, and x1 = 2. So, the solution to the system Ax = b is: x1 = 2,

x

2

= x3 = x4 = 1.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.5 Markov Chains 53

2

50. A =

1 0

Because A

2



1 0

=

 A, A is not idempotent.



54. Assume A is idempotent. Then

(A2 )T = AT (A

A

)=A

0

 1

(

)

= A AB B = ( AA)( BB) = AB

0 . 

( ) A BA B

So, ( AB ) = AB, and AB is idempotent. 2

58. If A is row-equivalent to B, then A = E kE 2 E1 B, where E1,

2

A =A

T

= ( AB)( AB) 2

0 1

0 1 0 0 1 0  1 0  2 52. A =  1 0 0  1 0 0 = 0 1     0 0  0 1 0 0 1  0 2 Because A  A, A is not idempotent.

T

56. ( AB)

  A.  =  0 1 0 1  1 0

, Ek are elementary matrices.

So, B = E1− E2− 1

1

Ek− A, 1

T

which shows that B is row equivalent to A. T

which means that A is idempotent. T

Now assume A is idempotent. Then T

T

A A = A

(AT AT )T

=

T

( AT )T

AA = A which means that A is idempotent.

60. (a) When an elementary row operation is performed on a matrix A, perform the same operation on I to obtain the matrix E. (b) Keep track of the row operations used to reduce A to an upper triangular matrix U. If A row reduces to U using only the row operation of adding a multiple of one row to another row below it, then the inverse of the product of the elementary matrices is the matrix L, and A = LU. (c) For the system Ax = b, find an LU factorization of A. Then solve the system Ly = b for y and Ux = y for x.

Section 2.5

Markov Chains

2. The matrix is not stochastic because every entry of a stochastic matrix satisfies the inequality 0  aij  1. 4. The matrix is not stochastic because the sum of entries in a column of a stochastic matrix is 1. 6. The matrix is stochastic because each entry is between 0 and 1, and each column adds up to 1.

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54

Chapter 2 Matrices 8. 60%

70% 40%

Gas (G)

Liquid (L)

30%

50% Solid (S)

50% The matrix of transition probabilities is shown. From

G

L 0

0 .60

P=

S 0  G  

0.70 0.50 L To     0 0.50 S 0.30 The initial state matrix represents the amounts of the physical states is shown. 0.40

(

)

0.20 10,000  X

0

2000

= 0.60 10,000  = 6000 

(

 

)

(

)

 

0.20 10,000  2000 To represent the amount of each physical state after the catalyst is added, multiply P by X 0 to obtain 0 0.60 0  2000 1200       PX0 = 0.40 0.70 = 6000 . 0.50 6000        0 0.30 0.50 2000 2800 So, after the catalyst is added there are 1200 molecules in a gas state, 6000 molecules in a liquid state, and 2800 molecules in a solid state. 0 .6 0.2

10. X1 = PX0 =

 0.2

0.2

X 2 = PX1 =

0.7

0.1

0 .6 0.2 

0.2

0 .2

0.7

0.1

0.6 0.2 = PX

X 3

= 2

 

0.2

 0.2

0.7

0.1

1 3

4  15 

 3

3 

0

 0.1

1 = 

0.9  1 

3 4

0 15    0.1 

 

0.9

1

3

3

  

 5

  75  49

 150    0.9  67 

 150 

  0.3

= 

5 

12. Form the matrix representing the given transition probabilities. A represents infected mice and B noninfected.

0.26

 2 

From

0.4

A 0.2 P=  0.8

17

 75  0.226 = 49  =  0.326   150  

67

0.446

 150 

0  17  0.1

1

=

 151   750  239  750  

12

 25

=

 

The state matrix representing the current population is

 

0.3186

B 0.1 A   To 0.9 B

 0.48

X0

= 

0.3 A 

.

0.7 B (a) The state matrix for next week is 0.2 0.1  0.3 0.13 X1 = PX 0 =   =  0.8 0.9 0.7 (

)

. 

0 .87

So, next week 0.13 1000 = 130 mice will be infected. 0.2 (b) X 2 = PX1 = 

0.1 0.13  0.113  =  

0.8 0.9 0.87

0.887

0.2 0.1 0.113  0.1113 X3 = PX 2 =    =   0.8 0.9 0.887 (

)

0.8887

In 3 weeks, 0.1113 1000  111 mice will be infected.

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Section 2.5

Markov Chains

55

14. Form the matrix representing the given transition probabilities. Let S represent those who swim and B represent those who play basketball. From

S B 0 .30 0.40 S  P=    To

0 .70

0.60 B

The state matrix representing the students is 0.4 S

X = 0

.

0.6 B

(a) The state matrix for tomorrow is 0 .30 0.400 .4 0.36 X1 = PX 0 =    =  . 0 .70 0.600 .6 0.64 So, tomorrow 0.36 ( 250 ) = 90 students will swim and 0.64 ( 250 ) = 160 students will play basketball. (b) The state matrix for two days from now is 0.37 0.360.4 0.364 2 . X 2 = P X 0 =   =   0.63 0.64 0.6 0.636      So, two days from now 0.364 ( 250 ) = 91 students will swim and 0.636 ( 250 ) = 159 students will play basketball. (c) The state matrix for four days from now is 0.363637 0.3636370.4 0.36364 X 4 . 4 = P X 0 =   =   0.6363630.6  0.636363 0.63636

So, four days from now, 0.36364 ( 250 )  91 students will swim and 0.63636 ( 250 )  159 students will play basketball.

0.2511

16. Form the matrix representing the given transition probabilities. Let A represent users of Brand A, B users of Brand B, and N users of neither brands.

(b)

0.325 

From

A B N 0.75 0.15 0.10 A    P = 0.20 0.75 0.15 B To    0.10 0.75 N 0.05 The state matrix representing the current product usage is 2  A 11

3

X 0 = 11  B 5 11 N

(a) The state matrix for next month is 0.75 0.15 0.10  2    3  11

1 X1 =

P X0

 

X 2 = P X0  0.3330   2

=

0.20

0.05

 

 

0.15  11 

 5 

=  

0.309 .

0.10 0.75  11

0.372

0.75

In 2 months, the distribution of users will be 0.2511 ⋅ 110,000 = 27,625 for Brand A, 0.3330 ⋅ 110,000 = 36,625 for Brand B, and 0.325 ⋅ 110,000 = 35,750 for neither. 0.3139 (c)

X18 = P

18

X0  0.3801 

  0.2151 

In 18 months, the distribution of users will be 0.3139 ⋅ 110,000  34,530 for Brand A, 0.3801 ⋅ 110,000  41,808 for Brand B, and 0.2151 ⋅ 110,000  23,662 for neither.

So, next month the distribution of users will be

0.2227 ⋅ 110,000 = 24,500 for Brand A, 0.309 ⋅ 110,000 = 34,000 for Brand B, and 0.372 ⋅ 110,000 = 41,500 for neither.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


56

Chapter 2 Matrices

18. The stochastic matrix

22. The stochastic matrix  2 7

0 0.3

P= 

10

P = 5

3

 1 0.7

10

3

 5

2

is regular because P has only positive entries. 0 0.3 x1  PX = X      1 0.7 x2  x

 x1  = 

x

 x1   x1  PX = X   5 10    =   x x

3

x1 − 0.3x2 = 0 1

+

x1 = 1 − 

3

10

7 10

x2 = 0

x1 +

x2 = 1

6 The solution of the system is x2 = 13 and

13 = 13 .

x1 = 1 − 136 = 137. 7

10

So, X = 13.

13

6

 

13

20. The stochastic matrix 0.2 0

24. The stochastic matrix

0.8

2

1

is not regular because every power of P has a zero in the second column. 0.2 0 x1   x1  PX = X     =   x  2 0.8 1 x2   0.2x1 0.8x + x2

= x1 = x2

9

1 4

3

2

1

1

P=

1

4

1 3 1 3 1

9 4 3  1 is regular because P has only positive entries. 2 1 9 4 1 1  2 4 1

1

 

PX = X

3

x

= 0

0.8x1

3

9

So, X =  .

1

 

1

+

4

1

+

1

4x

= 0

x1 + x2 = 1 The solution of the system is x1 = 0 and x2 = 1. 0

 

3

9 4 2x 1 9 1

 x  

x 1   =

  x

1

Because x1 + x2 = 1, the system of linear equations is as follows.

3  1 1 3   x2 

1

− 0.8x1

x 2 = x2

10

Because x1 + x2 = 1, the system of linear equations is as follows. − 3 x1 + 7 x2 = 0 10

13 and

3

So, X = 13.

P= 

10 3

5

The solution to the system is x2 = 10

+

x

5 1

3 x 5 1

=1

2

5

Because x1 + x2 = 1, the system of linear equations is as follows. − x1 + 0.3x2 = 0

x

3

3

 5 10  2   2  2x 7 x 1 + 2 = x1

0.3x2 = x1 + 0.7x2 = x2

1

x

7

2

2

1

is regular because P has only positive entries.

2

1

1 +

3

x2 + x2 + x2 +

4

 x2 

x 

1x 3 3 1x 3 3 1x 3

3 

= x1 = x2

3 = x3

Because x1 + x2 + x3 = 1, the system of linear equations is as follows. − 7 x1 + 9

1 3 4 9

x − 1

x1 +

1 4 1 2 1 4

x2 + x2 + x2 +

1x 3 3 1x 3 3 2 x 3 3

=

0

=

0

=

0

x1 + x2 + x3 = 1 The solution of the system is x3 = 0.33, x2 = 0.4, and x1 = 1 − 0.4 − 0.33 = 0.27.

0.27

 0.33 

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.5

5

1

1

3

5

0

P=

0.1 0 0.3 

1

0.2

 1 3 0 6 5  is regular because P2 has only positive entries. 1  1 2 1 x   x  5  1  1 PX = X   1 1 x 3 5 0 2  =  x2   

3

x 

0

3

 3

5

2

1

x

3 1 1 6

1

x2 + x3 = 0

5 4

− x2

x1 +

5 3

5 22 , x2

x1 = 1 −

5

=

22

6   11 

So, X = 

 

5

22

5

 

 22 

5 17

5 22

5 5 17 ( 22

= 6. 11

 0.54 

≈ 0.227 .     0.227

PX = X 

 x1 

 

1 0.3 x2 

=  x2    x  0.4 x

0.7

0  3  3  0.1x1 + 0.3x3 = x1 0.7 x1 + x2 + 0.3x3 = x2

0.2

0.2 x1 + 0.4x3 = x3 Because x1 + x2 + x3 = 1, the system of linear equations is as follows. − 0.9 x

0.3 x

+

1

=

0

3

+ 0.3 x3 = 0 − 0.6 x3 = 0

0.7 x1 0.2 x1

x1 = 1 − 1 − 0 = 0.

0 

x1 + x2 + x3 = 1 The solution of the system is x3 =

0.1 0 0.3 x1

x1 + x2 + x3 = 1 The solution of the system is x3 = 0, x2 = 1 − 0 = 1, and

= 0

x2 − x3 = 0

5

0.4

  is not regular because every power of P has two zeros in the second column.

Because x1 + x2 + x3 = 1, the system of linear equations is as follows.

0.3

0

x 

 6 5  1 2 x1 + 1 x2 + x3 = x1 5 1 1x 3 x1 + = x2 2 5 3 1 = x3 6 x1 + x2

− 1 x1 +

1

P = 0.7

1

57

28. The stochastic matrix

26. The stochastic matrix  1 1 2

Markov Chains

So, X = 1.

  0 

5

) = 22, and

30. The stochastic matrix 1 0 0 0   0 1 0 0 P=   1 0 0 0 0 0 0 1  is not regular because every power of P has three zeros in the first column. 1 0 0 0 x1  x1  

PX = X



0 0 1 0 x2  

 

0

   x3 =  x3      x  4 0 0 0 1 x4 x2

100

x1 = x1 x3 = x2 x2 = x3 x4 = x4 Because x1 + x2 + x3 + x4 = 1, the system of linear equations is as follows. 0= 0 − x2 + x3 x2 − x3

= 0 = 0 0= 0

x1 + x2 + x3 + x4 = 1 Let x3 = s and x4 = t. The solution of the system is x x4 = t, x3 = s, x2 = s, and 1 = 1 − 2s − t, where 0 ≤ s ≤ 1, 0 ≤ t ≤ 1, and 2 s + t ≤ 1.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


58

Chapter 2

Matrices x   1 

32. Exercise 3: To find X , let X =

x

2

x

Let x2 = r , x3 = s, and x4 = t, where r, s, and t are

. Then use the

real numbers between 0 and 1.

The solution of the system is

3 

x1 = 1 − r − s − t , x2 = r , x3 = s, and x4 = t, where r, s, and t are real numbers such that 0 ≤ r ≤ 1, 0 ≤ s ≤ 1, 0 ≤ t ≤ 1, and r + s + t  1.

matrix equation PX = X to obtain 0.16 0.25  x1 

0 .3 

0 .3

0.25  x  =

0.6

 or

2

  3 

0.5  x

0.16

0.3

 x1  x  2

So, the steady state matrix is 1 − r − s − t    r . X = s  

 3  x

0. 3x1 + 0.16x2 + 0.25x3 = x1

 

0. 3x1 +

0.6x2 + 0.25x3 = x2 0. 3x1 + 0.16x2 + 0.5x3 = x3 Use these equations and the fact that x1 + x2 + x3 = 1 to write the system of linear equations shown.

 

t

x x 1  X =

Exercise 6: To find X , let

− 0.6x1 + 0.16x2 + 0.25x3 = 0

+ 0.5x3 = 0

1x 6

1x

4

matrix equation PX = X to obtain   x3    0 0 0 1 x  x

2

0

 4

= x1 = x2 . = x3 x4 = x4

Use these equations and the fact that x1 + x 2 + x3 + x4 = 1 to write the system of linear equations shown. x1 + x 2 + x3 + x4 = 1

  x

 2

x3  x  4

 x  5  4  1

4

2 1x + 4 x = x 1 + x2 + 3 4 1 9

15

4

1 1x + 4 x = x 1 + x2 + 3 4 2 3

15

4

2 1 x + 4 x4 = x3 1 + x2 + 3

1 +

9 2x

4

1x + 2 + 3

15 1x

9

4

5

− 1 x1 + 2

or

 x1 

.

4 = x4

Use these equations and the fact that x1 + x 2 + x3 + x4 = 1 to write the system of equations shown.

 x1 

  x2  =x   3 x  4

4   15 x2  = 4   x3  15

4

6

 x  3 

4

15   x1

1

6 9

6

 x

3

2 9 2

1x

Exercise 5: To find X , let X =  2 . Then use the

x

1 4 1

3

1x

x

x2

1

2

 x1 

x1

1

1

13

4

4

0 0 1

1

9

1

13 

2

6   or

13

0 1 0 0

1

2

 6

 

. Then use the

4 

X =  6 .



matrix equation PX = X to obtain

So, the steady state matrix is 3  

2

x3 

x

x1 + x2 + x3 = 1 The solution of the system is 3 6 4 x1 = 13 , x2 = 13 , and x3 = 13 .

 1 0 0 0 x1 

 

0.3x1 − 0.3x2 + 0.25x3 = 0 0. 3 x1 + 0.16 x2

1

6 x1

1

6 x1

1

+

x 6 1

2 9 2 3 2 9 2 9

x2 + x2 + x2 − x2 +

1x 3 4 1 4 x3 3 4 x3 1 x 4 3

+ + + −

4x 4 15 4 15 x4 4 15 x4 4 x 5 4

= 0 = 0 = 0 = 0

x1 + x2 + x3 + x4 = 1 The solution of the system is

x

1

=

24 ,

x2 =

18

73

x

,

3

=

73

16, 73

and x4 =

15

.

73

So, the steady state matrix is  24   73

18

 73    73   15 

X =  16 

0.3288   0.2466

≈ 

0.2192

0.2055

.  

 73

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Section 2.5

Markov Chains

59

34. Form the matrix representing the given transition probabilities. Let A represent those who received an “A” and let N represent those who did not. From

A 0.70 P= 

0.30

N 0.10 A   To 0.90 N 

To find the steady state matrix, solve the equation PX =

 x1 

X , where X =

x

 and use the fact that x1

2

+ x2

= 1

to write a system of equations. − 0.3x1 + 0.1x2 = 0

0.70x1 + 0.10x2 = x1 0.30x1 + 0.90x2 = x2  x1 +

0.3x1 − 0.1x2 = 0

x2 =1

x1 +

x2 = 1  1

1 4

The solution of the system is x1 =

and x2

= 3 . So, the steady state matrix is X

=

4

4.

 3

1 This indicates that eventually

4

4 

of the students will receive assignment grades of “A” and

3 4

of the students will not.

36. Form the matrix representing transition probabilities. Let A represent Theatre A, let B represent Theatre B, and let N represent neither theatre. From

P=

A 0.10  0.05

0.85

B N 0.06 0.03 A   0.08 0.04 B To   B 0.86 0.97 

To find the steady state matrix, solve the equation PX

= X where X =

x   1  x

 2  and use the fact that x1 + x2 + x3 = 1

x 3

 

to write a system of equations. 0.10x1 + 0.06x2 + 0.03x3 = x1 0.05x1 + 0.08x2 + 0.04x3 = x2  0.85x1 + 0.86x2 + 0.97x3 = x3 x1 +

x2 +

x3 =1

− 0.90x1 + 0.06x2 + 0.03x3 = 0 0.05x1 − 0.92x2 + 0.04x3 = 0 0.85x1 + 0.86x2 − 0.03x3 = 0 x1 +

x2 +

x3 = 1

  5 The solution of the system is x1 = 1194 , x2 = 1195 , and x3 = 110 . So, the steady state matrix is X =  119 119 . This indicates    110  119  4 5  3.4% of the people will attend Theatre A,  4.2% of the people will attend Theatre B, and that eventually 4

119

119

119

110 119  92.4% of the people will attend neither theatre on any given night. 38. The matrix is not absorbing; The first state S1 is absorbing, however the corresponding Markov chain is not absorbing because there is no way to move from S2

40. The matrix is absorbing; The fourth state S4 is absorbing and it is possible to move from any of the states to S4 in one transition.

or S3 to S1.

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60

Chapter 2 Matrices 46. Let Sn be the state that Player 1 has n chips. From

42. Use the matrix equation PX = X , or  0.1 0 0 x1  

 x1 

   0.2 1 0  x2  =  x2 

    0.7 0 1  x3   x3 

S0 S1 1 0.7 

0

along with the equation x1 + x2 + x3 = 1 to write the linear system − 0.9x1

= 0.

= t, where t is a real number such that 0  t  1.  0    So, the steady state matrix is X = 1 − t, where    t  0  t  1.





4

+ 0.2 x3 + 0.1x4 = 0

0.1x1

+ 0.5 x3 + 0.6 x4 = 0 − 0.8 x3 + 0.2 x4 = 0 .

0.2 x1

+ 0.1x3 − 0.9 x4 = 0

x1 + x2 + x3 + x4 = 1 The solution of this system is x1 = 0, x2 = 1, x3 = 0, 0   = 0. So, the steady state matrix is

  2  To and X 0 =  1

49

0 0  

0 0.3 0 0S3  0 0 0.3 1 S   4 

58 

 0 n

P X 0 → PX 0 =

   0 .

 0 

9

  So, the probability that Player 1 reaches S4 and wins the

n

along with the equation x1 + x2 + x3 + x4 = 1 to write the linear system

and x4

0 0.7 0

0

S

compute X n = P X0 , where X 0 is the initial state matrix. (b) To find the steady state matrix of a Markov chain,

 4

− 0.3 x1

 

n

 0.1 1 0.5 0.6 x2  x2   0 0 0.2  x3 =  x3 0.2      0.2 0 0.1 0.1 x   x 

0

48. (a) To find the nth state matrix of a Markov chain,

 x1  

0

0 S1

9 tournament is 58  0.155.

44. Use the matrix equation PX = X or 0 0.2 0.1 x1

0.3

3

0

S3 S4 0 0S0 

0 0.7

So,

= 0

x1 + x2 + x3 = 1 The solution of this system is x1 = 0, x2 = 1 − t, and

0.7

0 0 

= 0

0.2x1 0.7x1

x

P=

S2 0

X =

1 

.

0

 

0  

determine the limit of P X 0 , as n → ∞, where X 0 is the initial state matrix. 2

3

(c) The regular Markov chain is PX 0 , P X 0 , P X 0 , , where P is a regular stochastic matrix and X 0 is the initial state matrix. (d) An absorbing Markov chain is a Markov chain with at least one absorbing state and it is possible for a member of the population to move from any nonabsorbing state to an absorbing state in a finite number of transitions. (e) An absorbing Markov chain is concerned with having an entry of 1 and the rest 0 in a column, whereas a regular Markov chain is concerned with the repeated multiplication of the regular stochastic matrix.

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Section 2.5 50. (a) When the chain reaches S1 or S4 , it is certain in the next step to transition to an adjacent state, S2 or S3, (b)

0.6 0 0 0.7

0

0 

(c)

1 6

1

0

0

P

30

5

6

0

0

1 6

5

P

0

12

6

7

0

30

5

0

0

5 12

12

≈ 0

5

6

7

0 7

0

1

 12

(d)

 X = 

5 24

part (a) t = 0 and in part (b), t

0 7 

12  1  6 

t

,

 

=

6

P =

 5  6 

6. 11

b 

a

1 − a 1 − b

be a 2 × 2 stochastic matrix, and consider the system of equations PX = X .  a b  x1   x1    =  x 1 1 − b x2   − a  2

  

You have

ax1 +

( − a) 1

 12 

 24  Half the sum entries in the corresponding columns of n n 1 P and P + approach the corresponding entries in

1

x

+ 1(− b

bx2 = x1 )

2

x

2

= x

or (

)

1

( − a)

1

a

−1x

1

X .

x

+ bx 2 = 0 − 2 = 0. bx

Letting x1 = b and x2 = 1 − a, you have the 2 × 1 state matrix X satisfying PX = X  b  X = 

.

1

 − a

56. Let P be a regular stochastic matrix and X 0 be the initial state matrix. = n n lim P X 0 lim P ( x1 + x2 + + xk ) n→∞

. In

11

54. Let

5 7

t

where t is any real number such that 0  t 

 

5 12

5 6

results where the columns alternate. 

6

0 12  12 Other high even or odd powers of P give similar 

0

the steady state matrix is X = 11

0

0

6

 11   5 − 5 t

5

0

Both matrices X satisfy P X = X . The steady state matrix depends on the initial state matrix. In general,  6 − t

1

1 6

61

52. (a) Yes, it is possible. (b) Yes, it is possible.

respectively, so S1 and S4 reflect to S2 or S3. 0 0.4 0 0   1 0 0 0.3   P=

Markov Chains

n→ ∞

n

n

= lim P ⋅ x1 + lim P ⋅ x2 ++ lim P n→ ∞

n→∞

= Px1 + Px2 +

+ Pxk

= P( x1 + x2 +

+ xk )

n

⋅ xk

n→∞

= PX 0 = X , where X is a unique steady state matrix.

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62

Chapter 2 Matrices

Section 2.6

More Applications of Matrix Operations

2. Divide the message into groups of four and form the uncoded matrices. H E L 

P _ I

S _ C O M I N G _ _

8 5 12 16 0 9 19 0 3 15 13 9 14 7 0 0 Multiplying each uncoded row matrix on the right by A yields the coded row matrices 3 − 1 −1 − 2    −1 1 1 1  − 1 −1 1 2    3 1 − 2 − 4 8 5 12 16 A = 8 5 12 16

3 15 13 9

19 0 A = −28

0 9

= 15 33 −23 − 43

−10

28 47 

A = −7 20 7 2 

= −35 49 −7 −7 .

14 7 0 0 A

So, the coded message is 15, 33, − 23, − 43, − 28, −10, 28, 47, −7, 20, 7, 2, −35, 49, − 7, − 7. 4. Find

85

A−1

84 117 A−

1

A−1

1 125 A− 

 O, R

0 14  _, N

=

= 15

 O, T

20

A−1 45 A−1

60 80 30

 T, O

5 0  E, _

= 15 18

 42 56

to find the associated uncoded row matrix.

0 2  _, B =

−1

90

and multiply each coded row matrix on the right by A

−2

6 8 A− = 1 10 15 A−

,

3

1

3

 3 −2 −4 3  = 20 15

120 

−4

=

= 0 20 

 _, T

= 15 0  O, _ 

19 26 A = 2 5  B, E So, the message is TO_BE_OR_NOT_TO_BE. −1

 11 6. Find A−1 =  4 

112 −140

 

1 83 A− =

72

−76 61 A−

1

=

95

−118 71 A−

1

=

1

1 21 38 A− −1 

=

=

35

−23 36 A

=

42

1 −48 32 A−

=

6

 11 

112 −140 83

−25 13 A−

20 

19 

−1

−8

−3 , and multiply each coded row matrix on the right by A−1 to find the associated uncoded row matrix.

1

−8

2

4

−8

0

0

7 18 

5

1 20  E,

−3

−1

6

=

_,

G, R A, T

0 23

5 

5

11

5  E,

K, E

14

4

0  N,

D,

 8 1 22  H, A, V

_, A

1

1  E,

5

2 −8

_, W, E _

The message is HAVE_A_GREAT_WEEKEND_.

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Section 2.6

8. Let A−

1

a =

b

c

d

10. You have 

c

d

S

0

19

= 0

A− = 1

= 21

−1 −2

−1

Multiply each coded row matrix on the right by A yield the uncoded row matrices. 

 

to

= 8

−18A−1

=

35

−30A−1

 81

−60A−1

= 5 = 21

42

−28A−1

= 14

75

−55A

= 20

This corresponds to the message

−1

14  H,

20  

= 0

22 −21 A−1

= 1

−10A−1

= 5

R

0  N,

_

15  

T, O

2  _, B

E, T

18  U,

−1

N

J, O

0 18  _, R

−2A−1

15

2

15 

−30A

5 18 , 19 0 , 0 19 , 21 5 .

18

0 15 , 18 4 ,

38

 

− 2  − 3 .

= 10

1 , 14 3 ,  5 12 , 

38 x − 30 z = 14.

8

−35A−1

1 A− =  1 1.

3

45

d = −2. So,

1  1

Solving this system, you find a = 1, b = 1, c = −1, and

z

(b) Decoding, you have:

+ 16d = 5.

37b

= 8 14 .

+ 16c

 10 15 and

Solving these two systems gives w = y = 1, x = −2, and z = −3. So,

d

− 19b− 19d = 19 37a



38 w − 30 y =

5.

− 19c

So, 45 w − 35 y = 10 and 45 x − 35 z = 15

E

This produces a system of 4 equations. −19a

38 −30  w x

y

U   37 16 a b = 21 

  45 −35  w x =

63

 y z

_



c

and find that

a b −19   =

 −19

More Applications of Matrix Operations

19  A, 

S

0  E, _

The message is JOHN_RETURN_TO_BASE_.

CANCEL_ORDERS_SUE.

12. Use the given information to find D. User A 0.30

B 0.20 A

D= 

 Supplier

0.40

0.40 B The equation X = DX + E may be rewritten in the

form ( I − D) X = E, that is  0.7 

− 0.2

− 0.4

0.6

X

10,000 . = 

20,000

Solve this system by using Gauss-Jordan elimination to obtain x≈ 

29,412 .

 52,941

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64

Chapter 2 Matrices

14. From the given matrix D, form the linear system X = DX + E, which can be written as ( I − D) X = E,

y

that is  0.8 − 0.4 −0.4 5000     0.8 −0.2 X = 2000. −0.4      0.8 0 −0.2 8000  21,875

18. (a) The line that best fits the given points is shown in the graph. (5, 2)

4 3

(4, 2) (3, 1)

2 1

(1, 0) (4, 1) x (2, 0) (3, 0) 5 6

14,250

−2

 

(b) Using the matrices 1 

16. (a) The line that best fits the given points is shown in the graph. y 4

(−1, 1)

(3, 2)

X =

2

(1, 1)

−1

1 2 3

(−3, 0)

 0

1 

3

1

 1  and Y =  , 4  1

1

4

 5

−1

1

1

T

you have X X =

 2

   2

 2

6

you have  0  

T

X X =

1

 2   0

4 

,

A= T

X Y=

0

4 

6

,

0

6

20

116

( X T X ) ( X TY ) = −1

T

X Y

37

 

3 1

4.

2

1 2

3

1

4

4

T

3 x 10

(c) Solving Y = XA + E for E, you have  −0.1    0.3 E= Y− XA =  .    0.1   T So, the sum of the squares error is E E = 0.2. −0.3

,

(c) Solving Y = XA + E for E, you have 1  3 1 1 1 E = Y − XA =  4 − 4 − 4 4 − 4

10

So, the least squares regression line is y =

 8 =  , and

28

So, the least squares regression line is y =

and

20   1 04  1 −1 T = A = ( X X ) X TY =  4  3 . 1   

 8

28

 1 ,  and Y =    

3 

X = 

1 

 0

3

1

(b) Using the matrices 1 −3   

2

1

1

−2

1

1

x

 0  

1 

3

(6, 2)

+ 1.

and the sum of the squares error is E E = 1.5. 20. Using the matrices 1 1 0     X = 1 3 and Y = 3,     1

6

5

you have

1 1 1 1 1   T 1 3 = X X = 1 3 5   

(

T

A= X X



)

1

0 1 1 1   T 3 X Y = 1 3 5   6

3 9 , 9 35 

5

 9 , and

= 39 

  −1

(

T

)

X Y =

3

−

2.

2

3

So, the least squares regression line is y =

3 x 2

3 . 2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

x−

3 . 4

1

4

T


Section 2.6 22. Using matrices 1 −4   1 

1

1

Y

−1    0 =  ,    5

1  

T

4 

0, 

0

40 −1

A=

( X T X ) ( X TY ) =

 8

T

1 4

32

   8  2  1   =  . 

40 

32

1

10

 −5

0.8

1

−1  and 1 

X =  

 4  

1

1

Y=

T

X X =

 A=

0, 

0

(XTX)

20 −1

10

T

X Y = 1

( X TY ) =  4  0

5

 1

0

A=

7 ,

−13 1

   

 20 So, the least squares regression line is y = −0.65 x + 1.75.

(XTX)

3

0

 = 

0 , and

−70

−27  

−27

0 

5 −70

1  1890.

= 7  7  4  =  13 . −   20

10

 205 ( X TY ) = 2961 

and

,  205

6 

 −5

27

−1

 5 27 = 

8

 −4

you have 4

 1

 0

3

0 

 1    1 1 1 1 1   4 5 8 10 

T

X Y= 

 2  , 1  

1 

1 4

1 1 1 1 1 X X = 4 5 8 0

24. Using matrices −3 

− 4  

T

So, the least squares regression line is y = 0.8 x + 2.

1 

8 

you have

, and

0

0

1 

X Y =

1 

you have X X =

 6   4  3   5 and Y =  0 ,

0 

X = 1

4

4

65

26. Using matrices

−2  and 2 

X = 

More Applications of Matrix Operations

296

 −350 

So, the least squares regression line is

−13

y = −

175

148 x + 148

945

.

28. Using matrices 1 

0.72  

9 

1

10

X = 1

11

0.92

=

and Y

 

1.17 ,

1 

12 

1.34  

13

1.60

1

you have T

X X =  

55

5

55 

and X TY

615

=  5.75 . 

65.43

−1.248 =   0.218  So, the least squares regression line is y = 0.218x − 1.248. A=

( X T X ) −1 X TY

30. (a) To encode a message, convert the message to numbers and partition it into uncoded row matrices of size 1 × n. Then multiply on the right by an invertible n × n matrix A to obtain coded row matrices. To decode a message, multiply the coded row matrices on the right by A− and convert the numbers back to letters. (b) A Leontief input-output model uses an n × n matrix to represent the input needs of an economic system, and an n × 1 matrix to represent any external demands on the system. 1

(c) The coefficients of the least squares regression line are given by A =

( X T X )−1 X T Y .

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66

Chapter 2 Matrices

Review Exercises for Chapter 2  2. −25

 1 2  −4

+

6

 4.



0

 1 4 

 1 5 6

2

2 6. 

7 1  −2 −4 56 8  54 4        8 1 2 = −10 8 +  8 16 = −2 24         

−12

= 

0

( )

 16 +

−2 8

( )

(

)

( )

−4 32 ( )

4

=

 5 

5 −2 + 

6

 0 4  5 = , the −4 

24

x

= 

−

3

 

 x1 

1 

−3

2

− 3 , x

=

3 −2  the system can be written as

x ,  2 

   x3 

 10 

and b =  

− 2

Ax = b

 

1  x1   

− 3 − 3  x2  3 x  3 −2

2

4

  

 10 

= 22.

− 2

5   =  2 .   − 6 

 3 2 12. AT =   −1 0 

 T A A= 

−3 = 

−3

1 −2 −3

−24 6

 6

 −3

 1

9

 

T

−2 −3 −2 = 14  

AA = 1 

22,

16. From the formula A− = 1

d

1

 ad − bc −c

−b 

,

a

you see that ad − bc = 4( 2) − ( −1)( −8) = 0, and so the

18. Begin by adjoining the identity matrix to the given matrix. 1 1 1 1 0 0    A I = 0 1 1 0 1 0   0  0 1 0 0 1  This matrix reduces to 1 0 0 1 −1 0   −1  = 01 0 0 1 −1.    0  0 1 0 0 1 So, the inverse matrix is 1 −1 0   1 A− = 0 1 −1.   I A

3 2  3 −1  13 −3   =  −1 0 20

 3 −1  AA =   T

matrix has no inverse.

Using Gaussian elimination, the solution of the system is x

 1   T A A = −2 1 −2   

.

4

3

   AT = −2  

 −3

10. Letting A =

−4 16

 −3 

7

2

2

−4

8

 1 14.

 5

 

16

  2 − 4  Using Gaussian elimination, the solution of the system is  6  7 x =  23.

32

=

 26 −2

4  3 9  =  

Ax = b 2 −1  x1

( )

+ 50 

2( 6) − 4( 4) 2( −2) − 4( 0) 2(8) − 4( 0)

−4 4 0 0 1  4 2 −2  +

0 −3 1  0 4 24 12  x1  2 −1 8. Letting A =  , x = , and b x 3 2    2 system can be written as 

 8 32

5 4 1 −2 + 5 01 8

−31

3 2 10  =

2 0 −1 0

6

6  4

0

0

1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Review Exercises for Chapter 2 20.

A

x

3 2  x1 

A−1

−3

A 1

0 

 1

   = 1 4 x   2 Because

26.

b 

 4

−2

= 1  10  −1

 =

3

 2  5

−1 5

1

3

− 3 − 4  z  4 − 7 Using Gauss-Jordan elimination, you find that  5 1 1

, solve the

10

 10

equation Ax = b as follows. x= A

−1

=

b

2

 

22.

 x

A

3  10

10

 1   −3



 1 −1 1  x2  = −1      21 1   2    x3 Using Gauss-Jordan elimination, you find that  2 1 3 − 5 5 5 . A−1 =  5 − 5 5 

3

1

3

1

− 2

24.

b = 

A

3 4 y 

5

=

4 1

11 − 3 

1

4

− 3  11

1

− 5   2

=

2

 5

4

− 

2 4 − 2A=   0 1

 −2 11 

−   −1 =  1   1

2

6

−9

18 17

9

3

− 7

17

 

 − 18 

2 4  =  0 1 , you can use the formula for

1

=

1

 1 − 4

1  1 − 4

 = 

0 2

.

2

 will be nonsingular if

that x ≠ 8.

 11

2

11

18

32. Because the given matrix represents 6 times the second 1

row, the inverse will be 6 times the second row. 1 0 0

2

9

 1 4 ad − bc = ( 2)( 4) − (1)( x) ≠ 0, which implies

11, solve the

3

4

9

18

6

9

8

0 2 2 − 0 1  −1 4 1  1 −4 So, A =  =  . 2 4 0 1    0   2  2 x 

1

18

  the inverse of a 2 × 2 matrix to obtain

−   =  11 

11 11

3

1

−  17

28. Because ( 2A)−1

 − 1

 11

equation Ax = b as follows. 1 x = A− b =

30. The matrix 

1

−1

= 

− 1



− 2

Because A

2

1

 5 A x b 2 −1  x   5     =  

5 

− 5

5 3

9

 5 5 5 Solve the equation Ax = b as follows.  2 3 1  1 −5 5 5  0  1 1     −1 3 x=

−2

x = A−1 b =

1

9

6

− 1

 18 9 6 Solve the equation Ax = b as follows. 1 1   0  5  − 23 

 0

9 4

 17

1   −1

=

b

 1 1 2  x1 

−5

1 −

 18

= − 8

A−1

1

5

x b 2  x   0     1  y  =  −1    

3 2

67

1

0

6 0

11

19

For Exercises 34 and 36, answers will vary. Sample answers are shown below. 34. Begin by finding a sequence of elementary row operations to write A in reduced row-echelon form.

Matrix  1 − 4 

Elementary Row Operation Interchange the rows.

−3 0

1

1

0

0

1

Add 3 times row 1 to row 2.

2

= E−1 E−1 E−1 1 2 3

= 0 1  

1



1 0  1 −4 

0 −3 1 0

1 0

= 3

1 

1

Then, you can factor A as follows. A

E

Add 4 times row 2 to row 1.

1 0

13

 1 − 4 

Elementary Matrix E 0 1 1 = 

E

4 3

=

1

0 1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


68

Chapter 2 Matrices

36. Begin by finding a sequence of elementary row operations to write A in reduced row-echelon form.

Matrix

Elementary Row Operation

Elementary Matrix

1 3

 1 0 2

  0 2 0    103 

1

Multiply row one by 3.

E1 = 0  

 1 0 2 

0

2

Add −1 times row one to row three.

0

 0 2 0    0 0 1  

0 1 0

1

0 1  0 −2

1

E3 = 0 1

1 1

 0

Add −2 times row three to row one.

 1 0 0

1

0  1 0 0

  0  −1

 1 0 0

0 0

E2 =

0

 0 1

Multiply row two by 2.

0

0 0 1

E4 = 0 2

0

0 0 1  So, you can factor A as follows.  3 0 0  1 0 0  1 0 2  1 0 0

1 1 1 1 A = E1− E 2− E3− E4− =

 

a

38. Letting A =



0 1 0



0 2 0







1 0 1 0 0 1 0 0 1



b ,

c

you have

 d 2 a b a b  a + bc ab + bd  2 A =   =  2  c d c d ac + dccb + d       So, many answers are possible. 0 0, 0 1, etc.    

0

0 1 0





0 0 1



0 1 0



0 0

.

= 

0 0

0

0  0  40. There are many possible answers. 0 A= 

0

1 , B

0

 1 0 0 1 1 0 0 =   AB =    =

0 0

 1 0 0 But, BA =  

0 0 0

0

1 0  = 0

0

00 0

0

0 = 0 0

1  ≠ 0. 0

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Review Exercises for Chapter 2 42. Because ( A− + B− 1

)

1

)( A−1 + B−1) = I , if ( A

− 1

)

+ B−1 −1 exists, it is sufficient to show that

69

( A−1 + B−1)( A( A + B)−1

B = I for equality of the second factors in each equation.

(A−1

(

+ B −1 ) A( A + B )−1 B

) = A ( A( A + B )− 1 B ) + B ( A( A + B )−1 B) −1

−1

= A − A ( A + B ) − B + B − A ( A + B )− B 1

1

1

1

= I ( A + B ) − B + B − A ( A + B )− B 1

(I+ = (B =

1

1

(( A + B )−1 B) 1 B + B A)( ( A + B )− B)

B −1A)

−1

−1

1 1 = B − ( B + A)( A + B )− B

= B−

1

(A+

B )( A + B )− B 1

= B − IB 1

1 = B− B

= I Therefore, ( A− 1 + B −1)−1 = A( A + B )−1 B. 44. Answers will vary. Sample answer: Matrix

Elementary Matrix

− 3

1  = A  12 0 − 3 1  = U 4  0  EA = U

 1

0 

E= 

1

− 4

 1 A= E− U= 

1 0− 3 1   = LU

− 4 46. Matrix

1

 0 4

Elementary Matrix

1 

1 1  2 2 = A   1  2 3  1 1 1   1

0 1 1

 

1

  0  1  E2 =  

2 3 1 1 

1  0 1 1

E1 =

 0 1 2

1 

1 1 

0 1 1

0

 1 

= U

0 1

E3 E 2 E1 A = U 

E3 =

1  

0

0 0  −1 1 0

 0 1 0 0 

0 1 0

−1 01

0 0  0 1 0

−1 1 1 

0 0  1 

1 1 

A = E1− 1 E 2− 1 E3−1U = 1 1 0 0 1 1



1

1

 0

1

= LU

0 1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


70

Chapter 2 Matrices 48. Matrix 2 

Elementary Matrix

1

1

0

−1 

3

1

−1

0

0 −2

0

2

= A

1 1 −2 2 1 1 −1   1 −1 0 3  = U −20 0 0 

0 0 0 −1

 0 1 0 0

 0 0 1 0

−1 0 0 0

 1 0 0 0 2 1 1 −1   0 1 0 0 0 3

EA = U 

 1 0 0 0

  E=  



A= 

 = LU −20

0 0 1 0 0 0

 1 0 0 1 0 0 0 −1

 1 0 0 0  y1  

0 1 0 0 y2   

Ly = b :  

0

0

1 0



 7

 y 

3

1 0 0 1 y 

4

2 1 1 −1  x1 

1 −1 x2 

0 3

Ux = y : 

0 0 −20 x3  

0 00 −1 x   4

100 90

70 30

50. 1.1

 

2

 1  4

−3

−1

 

 x=

−1

2

 1 110 99 77

=

−3 y=   2  

 8  7

   =  

 7 

−3 =  

 1 −1

−1 56. The matrix is not stochastic because the sum of entries in columns 1 and 2 do not add up to 1.

33 

 44 22 66 66  40 20 60 60 52. (a) In matrix B, grading system 1 counts each midterm as 25% of the grade and the final exam as 50% of the grade.

58. This matrix is stochastic because each entry is between 0 and 1, and each column adds up to 1. 0.307 60. X1 = PX0 =  

Grading system 2 counts each midterm as 20% of the grade and the final exam as 60% of the grade.

78 

82 80  88 85  0.25 93

84

 (b) AB = 

90 

92

 0.25 86 90 

88

78

74

96

95

 0.50

  0.20 

85.4

91.25

91

80

80 

 85.5

0.20 =  0.60

80

98

 88.5 88.8

78 96.75

3

−1 0

2 1  1 0  + 2  −1 0

 4 3  6 3 2 0

=

0 1

 −  + − 3 − 2  − 3 0  0  2  0 0

=

0

0

0.693

2

0.38246 = PX1 =   0.61754

0.3659 X3 = PX 2 =  

0.6341

 4  9 

78.4

97

(c) Two students received an “A” in each grading system.

 2 1  54. f ( A) =   − 3

X

62. X 1 = PX 0 =  5

≈

 27 

  27   37   81  22 

X 2 = PX1 = 

81 

 81   103   243

 

0.370

  0.4568   

 0.4 

0.185

10

22

 

≈ 0.2716

0.2716

 59 

0.4239  

1

3

X 3 = PX 2 = 

243 

≈ 0.2428

0.3 

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Review Exercises for Chapter 2

71

64. Begin by forming the matrix of transition probabilities. From Region 1 0.85 

2 3 0.15 0.101   P = 0.10 0.80 0.10 2 To Region    0.05 0.803 0.05 (a) The population in each region after 1 year is given by 0.85

0.15 0.10  1

 PX =



0.80

0.10

0.10

 =

1 3

0.3

0.05 0.80  1 

0.05

3

0.3

.

3      110,000 Region 1     Region 2. So, 300,000  0.3  = 100,000      0.3   90,000  Region 3 (b) The population in each region after 3 years is given by 0.665375 0.322375 0.2435  13  0.4104     1   3 P X =  0.219 0.562 0.3  0.219   =  0 .36

 

 0.4104  0.3 

So, 300,000

 

0

4 7

 

7

n

is not regular because P has a zero in the first column for all powers.  x1  X , begin by letting X =

To find

x

.

2

Then use the

matrix equation PX = X to obtain 1  

4 3

0

 x 1    x 

7  7

 x1 .

= 

1

.

 76,875  Region 3

P= 

3

0.25625

 3 123,125 Region 1   = 100,000 Region 2.  

0.25625 66. The stochastic matrix 1 

3

0.115625 0.5375  1

0.115625

x  2

Use these matrices and the fact that x1 + x2 = 1 to write the system of linear equations shown. 4 7 x2 = 0

68. The stochastic matrix  0 0 0.2   P = 0.5 0.9 0   0.1 0.8 0.5 2 is regular because P has only positive entries. To find X , let X =  x1 x2 x3 . Then use the matrix T

equation PX = X to obtain.  0 0 0.2 x1   x1       0.5 0.9 0  x2  =  x2 .      0.5 0.1 0.8  x3   x3  Use these matrices and the fact that x1 + x2 + x3 = 1 to write the system of linear equations shown. − x1

+ 0.2x3 = 0

0.5x1 − 0.1x2

4

= 0

0.5x1 + 0.1x2 − 0.2x3 = 0

− 7 x2 = 0 x1 + x2 = 1 The solution of the system is x1 = 1and x2 = 0 1 So, the steady state matrix is

X

.

= 

0

 

x1 +

x2 +

x3 = 1

1 5 The solution of the system is x1 = 11 , x2 = 11 , and 5 x3 = 11 . 1

  11

So, the steady state matrix is X =  5 .

11  5 11

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


72

Chapter 2 Matrices

70. Form the matrix representing the given probabilities. Let C represent the classified documents, D represent the declassified documents, and S represent the shredded documents. From

C

D

S

0 .70 0.20 0 C

P=

0.10

0.75

0.20

0.05

0

D To

1 S 

 x1 

Solve the equation PX = X , where X =  x2  and use the fact that x1 + x2 + x3 = 1 to write a system of equations. x  3  0.70x1 + 0.20x2

= x1

0.10x1 + 0.75x2

= x2 

− 0.3x1 +

0.20x1 + 0.05x2 + x3 = x3 x2 + x3 =1

x1 +

0.2x2

= 0

0.1x1 − 0.25x2

= 0

0.2x1 + 0.05x2

= 0

0

x1 +

x2 + x3 = 0

 0.  

So, the steady state matrix is X =

 1

 

This indicates that eventually all of the documents will be shredded. 72. The matrix 1 P=

0  0

0

0.38

0.30

0

 

0.70 0.62 is absorbing. The first state S1 is absorbing and it is possible to move from S2 to S1 in two transitions and to move from S3 to S1 in one transition. 74. (a) False. See Exercise 65, page 61. (b) False. Let A =

1

0

0 0  0

1  0 . 0

Then A + B =

and

−1

B=

0. 

 0 −1

A + B is a singular matrix, while both A and B are nonsingular matrices. 76. (a) True. See Section 2.5, Example 4(b). (b) False. See Section 2.5, Example 7(a). 78. The uncoded row matrices are 

2

5 1

B E A M _ M E _ UP _ S C O T

13 0 13

5

0 21

16

0 19

3 15

20

T Y _ 20

25

Multiplying each 1 × 3 matrix on the right by A yields the coded row matrices.

17

 

6 20

0

 

0 13

 

−32

−16 −43

 

−6 −3

7

 

11 −2 −3

115

0

 45 155

So, the coded message is 17, 6, 20, 0, 0, 13, − 32, −16, − 43, − 6, − 3, 7, 11, − 2, − 3, 115, 45, 155.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Review Exercises for Chapter 2 80. Find A− to be

73

1

−3

− 4

A−1

=

1

1

and the coded row matrices are 

 

 

 

 

11 52 , −8 −9 , −13 −39 ,

 

 



 

20 , −2 7 , 9 41 , 25 100 .

5 20 , 12 56 , 5

Multiplying each coded row matrix on the right by A− yields the uncoded row matrices. S HO W _ M E _ T H E _M ON E  Y _ 1

19

8

15 23

0 13

5

0

20

8

5

0

13 15

14

5

25 0

So, the message is SHOW_ME_THE_MONEY_. 82. Find A− to be 4 1

 13

8

 13

A −1 =

5

 13

2 13 9

1 13  2

4

− 2

− 13 −

13

,

13 

13

and multiply each coded row matrix on the right by A− to find the associated uncoded row matrix.  4 2 1  1

66 27 − 31 A−1 



5

61

46

46

−14

−9 A−

1

=

1 −73 A− = 

0

23 

S,

_, W

0 

=

31

12 A 

0 

5

18 

S,

E, R

5

19 

I,

E, S

 43

1 − 9 − 20 A− =  0 19

68 23

O, R

4

18  W, 

L, D, _

14  _, I, N 

5  _, S, 

5 14  V, E,

1 − 34 A− = 22

1 14  Y, A, N

I, N, _

19

32

12

1 33 −67 A− = 9 1 19 − 56 A− = 0 9

 47

−53 A−1 =

 

19

−4

−1

13

E, E

 66

13

25

K,

32

2

4

=

5 

9A 21 −49 A−1 

13

2

5

94

5

13

9

− −

11

=  9 14 = 23 15

−1

13

8

 13

37

13 

13

 13

= 66 27 − 31

E N

The message is YANKEES_WIN_WORLD_SERIES_IN_SEVEN.

84. Solve the equation X = DX + E for X to obtain

(I − D) X = E, which corresponds to solving the

86. Using the matrices 1 2

augmented matrix.

  

0.9 − 0.3 − 0.2 3000  0 0.8 − 0.3 3500

10,000

X =

 10,000 .

15,000 

1 3

1

   3

X = 1

= 2,

4 and Y



1 5

− 0.4 − 0.1 0.9 8500 The solution to this system is

4

   1 6

   4

you have

5 XTX =  20 A=

(X X) T

−1

20, 90

14  X TY =   , and 63

   1.8 −0.4 14  0 X Y =     =  . T

−0.4

0.1 63

0.7

So, the least squares regression line is y = 0.7 x.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


74

Chapter 2 Matrices

88. Using the matrices 1 −2  

 4   −1  2 0 and Y =  1, you have

1

X = 1 

1

1

−2

1

2

 −3

5

T

0

X X = 

0

,

 X Y = 

2

T

10

(X X) T

A=

, and

1

−1

T

X Y= 

5

0

−18

 So, the least squares regression line is y = −1.8x + 0.4, or y = −

  

2.93

8 1 9

1

90. (a) Using the matrices X = 

1

10

10 9x

+

2. 5

3.00 3.01

3 .39

8

1 9 10

 = 11

12

1 13

 6 63  

63

1

3.21

1

 1 1 1 1 1 11  9 10 11 12 13 1 

8

.

T

 and Y =  , you have 11 3 .10

   1 12    1 13

X X = 



 −18 −1.8

5

1

2  0.4

0

1

679

and 2.93

1 T X Y=  

8

   3.00 1

1

9

1

1

1 3.01 

10 11 12

13 3.10

18.64 

 

=

. 

197.23

 3.21

   3.39

Now, using ( X X )−1 to find the coefficient matrix A, you have T

A=

(X

−1

T

T

X ) X Y =

 97  15

− 3

 −3

2

5

5  18.64



197.23



35

 

2.2007 . ≈  0.0863 

So, the least squares regression line is y = 0.0863x + 2.2007. (b) Using a graphing utility, the regression line is y = 0.0863x + 2.2007. (c)

Year

2008

2009

2009

2010

2011 2012

2013

Actual

2.93

2.93

3.00

3.01

3.10

3.21

3.39

Estimated

2.89

2.89

2.98

3.06

3.15

3.24

3.32

The estimated values are close to the actual values.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Project Solutions for Chapter 2

75

Project Solutions for Chapter 2 1

Exploring Matrix Multiplication 1. Test 1 seems to be the more difficult test. The averages were: Test 1 average = 75

3.

0

0 1

0 0 index 2;  0 0

2. Anna, David, Chris, Bruce

  represents scores on the first test.

0 M   represents scores on the second test.

  4. Answers will vary. Sample answer:

0 1 0 M represents Chris’s scores.

1

M

 4 

1

represents each students’ average.

1 1 1 M represents the sum of scores on each test; 

1 1 1 1 M represents the average on each test.

8.

1 

1 11M

1

represents the overall points total for   1 all students on all tests.

7. 1

1 1 1 1M 

8

1

0 0

0

0 1 1 0 0 1

0 0 1 1  1 0 0  index 3;

0

0

0 0

0 0 0 0 

0

0 0

0

1 1 1

5. 0

0 1 1 0 0 1

0

0 0 0

1

 0

0 0 0

0

0  1 1

1 

 k

6. No. If A is nilpotent and invertible, then A = O for k 1 some k and A − ≠ O. So,

= 80.25

O, which is impossible. 7. If A is nilpotent, then ( A

)

k T

=

( AT )k

( AT ) k −1 = ( A k −1)T

= O. But T

 O, which shows that A is nilpotent with the same index. 8. Let A be nilpotent of index k. Then

(I − A)( Ak −1 + Ak − 2 +

+ A2 + A + I ) = I − Ak = I ,



1

  1.1 9. M  

which shows that

( Ak −1

k 2 + A − + +

A2 + A + I )

is the inverse of I − A.

1.0 2

0 0

1 1 k 1 k 1 k 1 A− A = I  O = A− A = ( A− A) A − = IA − ≠

1

6. 1

1



2

 0

0 

  represents the sum of the test scores for each  

student, and 1

1 index 3

 index 4

 0

5. Answers will vary. Sample answer: 1

M

0

1

0

0 0 0

 0

0

 

0 0 0 M represents Anna’s scores.

1 1

0 0 0 0   0 1 1 1

3. Answers will vary. Sample answer:  1

1

0

0

0 0 0 1    0 0 0 4.  0 0  index 2; 

Test 2 average = 85.5

M

Nilpotent Matrices 2

3

1. A ≠ 0 and A = 0, so the index is 3. 2. (a) Nilpotent of index 2 (b) Not nilpotent (c) Nilpotent of index 2 (d) Not nilpotent (e) Nilpotent of index 2 (f) Nilpotent of index 3

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution Manual for Elementary Algebra Graphs and Authentic Applications 2nd Edition by Lehmann  

Link full download: https://bit.ly/2SBLYob Language: English ISBN-10: 0321868277 ISBN-13: 978-0321868275 ISBN-13: 9780321868275 elementar...

Solution Manual for Elementary Algebra Graphs and Authentic Applications 2nd Edition by Lehmann  

Link full download: https://bit.ly/2SBLYob Language: English ISBN-10: 0321868277 ISBN-13: 978-0321868275 ISBN-13: 9780321868275 elementar...

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