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Solution Manual for Elementary Linear Algebra 8th Edition by Larson

ISBN 1305658000 9781305658004

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C H

Matrices Section2.1 Operations withMatrices 30 Section2.2 Properties ofMatrix Operations..................................................... 36 Section2.3 The Inverseofa Matrix 41 Section2.4 ElementaryMatrices 46 Section2.5 Markov Chains .............................................................................. 53 Section2.6 More Applications ofMatrixOperations....................................... 62 Review Exercises.................................................................................................. 66 Project Solutions.................................................................................................. 75

A P T E R 2

C H A P T E R 2

Matrices Section 2.1 Operations with Matrices

 +       =2+ 5 4 + 4 5 + 8 7

2. x = 13, y = 12 4. x + 2 = 2x + 6 2 y = 18 4 = x y = 9 2x = 8 x = 4 y + 2 = 11 y = 9  6 1  1 4  6 + 1 1 + 4  7 3 6. (a) A + B =  2   1  =  + ( 1) 4 +  =  1   4 +  5 2 5  9  3 5  1 10  3 + 1 5 + 10  2 15  6 1  1 4  6 1 1 4  5 5 (b) A B =  2 4  1  =  2 ( 1) 4  =  3   5  5  1  3 5  1 10  3 15 10  4 5 2 A =   6 1  2(6) 2( 1) 12 2 (c) 2 2 4 = 2(2) 2(4) =  4 8  3 5 2( 3) 2(5)    610  12 2  1 4  12 1 2 4  11 6 (d) 2 A B =  4 8  1  =  4 ( 1) 8  =  5   5  5  3  6 10  1 10  6 110 10  7 0  1 4  6 1  1 4  3 1  4 7        2  2 (e) B + 1 A = 1 5 + 1 2 4 = 1 5 +  1 2 =  0 7       2   2      3 5  1 25 A + B =  1 10  3 5  1 10  2 2  2 2   3 2 1 0 2 1 3 + 0 2 + 2 1 + 1 3 4 0 8. (a) 2 4 5 5 4 2  2 = 7  0 1 2 2 1 0 0 + 2 1 + 1 2 + 0 222 3 2 1 0 2 1 3 0 2 2 1 1  3 0 2 (b) A B = 2 4    = 2 5 4 4 5 2 =  3 0   5 5 4 2    3 0 1 2 2 1 0 0 2 1 1 2 0  2 0 2 3 2 1 2(3) 2(2) 2( 1) 6 4 2 (c) 2 A = 22 4 5 = 2(2) 2(4) 2(5) = 4 8 10  0 1 2  2(0) 2(1) 2(2)   0 2 4 (d) 2 A B = 3 2 1 0 2 1 6 4 2 0 2 1  6 2 3           22 4 5 5 4 2 = 4 8 10 5 4 2 =  1 4 8 0 1 2 2 1 0 0 2 4 2 1 0  2 1 4 0 2 1 3 2 1 0 2 1 3 1 1 3 3 1 (e) B + 1 A =   + 1   =  +  2 2 = 2 9 2 5 4 2 2 4 5 5 4 2  1 2 5  6 6  2

2   2      1 2  3  2 1 0 0 1 2 2 1 0 0 2 1 2 2 1 30

10. (a) A + B is not possible. A and B have different sizes. 14. Simplifying the right side of the equation produces

A B is not possible. A and B have different sizes.

setting corresponding entries equal to each other, you obtain four equations.

2 A B is not possible. A and B have different

possible. A and B have different

Section 2.1 Operations with Matrices 31 =       2

w x  4 + 2 y 3 + 2w  3  6  y x  2 + 2z 1 + 2x (c) 2 A = 2 2  = 4   1  2     By

w = 4 + 2 y sizes. x = 3 + 2w  2 y + w = 4  x 2w = 3   (e) B + 1

sizes. y = 2 + 2z x = 1 + 2x y 2z = 2  x = 1 12. (a) c = 5a + 2b = 5(2) + 2(11) = 32 The solution to this linear system is: x = 1, y = 3 , 23 23 23 z = 1 , and w = 1. 2 (b) c32 = 5a32 + 2b32 = 5(1) + 2(4) = 13 4 AB =  2 2 4 1 = 2(4) + ( 2)(2) 2(1) + ( 2)( 2) = 4 6 16. (a)          1 4 2 2   1(4) + 4(2) 1(1) + 4( 2)   4 9 4 1  2 2 4(2) + 1( 1) 4( 2)+ 1(4)  7 4 (b) BA =     =   =   2 2  1 4  2(2) + ( 2)( 1) 2( 2) + ( 2)(4)  6 12 18. (a)  1 1 7   1 1 2   1(1) + ( 1)(2) + 7(1) 1(1) + ( 1)(1) + 7( 3) 1(2) + ( 1)(1) + 7(2)  6 21 15  AB = 2 1 82 1 1 = 2(1) + ( 1)(2) + 8(1) 2(1) + ( 1)(1) + 8( 3) 2(2) + ( 1)(1) + 8(2)= 8 23 19  3 1 1   1 3 2  3(1) + 1(2) + ( 1)(1) 3(1) + 1(1) + ( 1)( 3) 3(2) + 1(1) + ( 1)(2)   4 7 5  1 1 2   1 1 7    1(1) + 1(2) + 2(3) 1( 1) + 1( 1)1 + 2(1) 1(7) + 1(8) + 2( 1)  9 0 13  (b) BA = 2 1 1 2 1 8 =  2(1) + 1(2) + 1(3) 2( 1) + 1( 1) + 1(1) 2(7) + 1(8) + 1( 1) = 7 2 21 1 3 23 1 1 1(1) + ( 3)(2) + 2(3) 1( 1) + ( 3)( 1) + 2(1) 1(7) + ( 3)(8) + 2( 1)   1 4 19  3 2 11 2 3(1) + 2(2) + 1(1) 3(2) + 2( 1) + 1( 2)   8  2 20. (a) AB = 3 0    3(1) + 0(2) + 4(1) 3(2) + 0( 1) + 4( 2) = 1  14   4 2 4 2 1 =  ( ) + ( 2)(2) + ( 4)(1) 4(2) + ( 2)( 1) + ( 4)( 2)  4  41 2 4 1    18 (b) BA is not

3  2 and A is 3  3.  1  1(2) 1(1) 1(3) 1(2)  2 1 3 2 AB =  2 2 1 3 2 =  2(2) 2(1) 2(3) 2(2) =  4 2 6  22. (a)      2(2) 2(1) 2(3) 2(2)  4 2 6 4   2    4  1  1  1(2) 1(1) 1(3) 1(2)   2 1 3 2 BA = 2 1 3 2 2 = 2( 1) + 1(2) + 3( 2) + 2(1) =  4 (b)       

(b)

(d)

not

defined because

32 Chapter 2 Matrices  1 24. (a) AB is not defined because A is 2  2 and B is 3  2. 2 12 3 2(2)+ 1(5) 2( 3) + 1(2)   9 4  (b)  3 = (2) + 3(5) 1( 3) +3(2)    BA = 1 5 2 1  = 17 3 2 1 2(2) + ( 1)(5) 2( 3) + ( 1)(2)    1 8 

30. C + E is not defined because C and E have different sizes.

32. 4 A is defined and has size 3  4 because A has size

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

36. 2hDav+ e C sizie s 4 defin2ed and has size 4

2 because 2D and C

38. As a system of linear equations, Ax = 0 is 42. In matrix form Ax = b, the systemis

elimination on the augmented matrix. Use Gauss-Jordan elimination on the augmented matrix for this system.

Section 2.1 Operations with Matrices 33 2(4) + 1( 1)+ ( 2)( 2) ( ) ( ) ( )( ) ( ) ( ) ( )( ) 2(3) + 1( 1) + ( 2)(3)  4 x   =       26. (a)  2 1 2 4 0 1 3 AB =  3 1 2  1 2 3 1  2 1 2 2 1 4 3 2(4) + 1( 1) + 2( 2) 2(0) + 1(2) + 2(1) 2(1) + 1( 3) + 2(4) 2(3) + 1( 1)+ 2(3)  = 3(4)+ ( 1)( 1)+ ( 2)( 2) 3(0) + ( 1)(2) + ( 2)(1) 2 0 + 1 2 + 2 1 3(1) + ( 1)( 3) + ( 2)(4) 2 1 + 1 3 + 2 4 3(3)+ ( 1)( 1)+ ( 2)(3)    3 4 7 11 = 17 4 2    5 0 13 13  (b) BA is not defined because B is 3  4 and A is 3  3.

(a) AB is not defined because A is 2  5 and B is 2  2. (b) 1 6 1 0 3 2 4 BA =     4 2 6 13 8 17 20 =  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)  =

28.

40. In matrix form Ax = b, the system is 2 3 x   5   1=  

 4.

1 4x2 10

Gauss-Jordan

the

2 3 5 1 0 2      1 4 10 0 1 3 So, the

is  x1 =  2  2  3



Use

elimination on

augmented matrix.

solution

x1 + 2x2 + x3 + 3x4 =

 4 9 x1  13 x1 x2 + x4 =

      1 3x2  12 x2 x3 + 2x4 = 0

Gauss-Jordan

 4 9 13  1 0 23 35 1 2 1 3 0 1 0 0 2 0  1 3 12 01 3  1 1      x1  23  0 1 0 0 1 0 1 0 So, the solution is   =   0 1 1 2 0 0 0 1 1 0 Choosing x4 = t, the solution is x1 = 2t, x2 = t, x3 = t, and x4 = t

number. x2   33 5 37 78 51 104 124 16 26 28 42 56

Use

, where t is any real

In matrix form

b, the system is

The augmented matrix row reduces as follows.

Use Gauss-Jordan elimination on the augmented matrix. There are an infinite number of solutions. For example,

34 Chapter 2 Matrices 1  3  2 2 1 2 1a a  1 0 44.

50.

 1 1 3  x1  1  1 2 4 1 1 0 2 3            1 2 0 x2 =  1  1 0 2 3  0 1 3 2  1 1 1   x3  2  0 1 3 2 0 0 0 0

x =

 1 1 3 1 1 0 0 2 x3 = 0, x2 = 2, x1 = 3. 1 2 0    3 1  1 2 4  1 0 1 0 2          1 1 1 2 0 0 1 3 So, b = 3 = 3 1 + 20 + 02  2   x1 2 2  0 1  3  So, the solution is  x  = 3  2  x3  2  52. The augmented matrix row reduces as follows. 3   3 5 22 1 3 10 2     46. In matrix form Ax = b, the system is  3 4 4  0 9 18 1 1 4  x   17  4 8 32 0 4 8    1   1 3 10 1 0 4 1 3 0 x2 =  11      0 6 5   x3   40 0 1 2  0 1 2 Use Gauss-Jordan elimination on the augmented matrix. 0 1 2 0 0 0 1  1 4 17 1 0 0 4 So, 3 0     22  3  5  11  0 1 0 5       0 6 5 40 0 0 1 2 b =  4 = 4 3 + ( 2) 4  x 4  32  4  8  1      So, the solution is x2 =  5 54.

produces x      A =    11 12 48. In matrix form Ax = b, the system is 3 2 3 2 a21 a22   1 1 0 0 0  x  0  2a11 a21 2a12 a22 =  = 1 0          0 1 1 0 0  x2   3a11 2a21 3a12 2a22 0 1  0 0 1 1 0  x3 = 0 . and you obtain the system        0 0 0 1 1  x4 0  1 1 1 1 1 x5 5 2a11 2a12 a21 = 1 a22 = 0 Use Gauss-Jordan elimination on the augmented matrix. 3a11 3a1 2 2a21 2a22 = 0 = 1. 1 1 0 0 0 0 10 00 0 1  0 1 1 0 0  0 0 1 0 0 0 1  Solving by Gauss-Jordan elimination yields  0 0 1 1 0 0  0 0 1 0 0 1 a11 = 2, a12 = 1, a21 = 3, and a22 = 2.      0 0 0 1 1 0 0 0 0 1 0 1   1 1 1 1 1 5 0 0 0 0 1 1 So, the solution is x1  1     x2  1  x  =  1 .  3    x4  1      x5  1 So, you have A = 2 3  1 2

Expanding the left side of the equation

56. Expanding the left side of the matrix equation produces

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

The ith column of B has been multiplied by aii , the ith diagonal entry of A

64. The trace is the sum of the elements on the main diagonal. 1 + 1 + 1 = 3

68. Let AB = c , where c =  n a b Then, Tr( AB) =

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

+ ( 3) = 0

Similarly, if BA =

b a . Then Tr(BA)

Section 2.1 Operations with Matrices 35  0 .      = 3 7 0 0 0 0 0 0 0 0

AB = 3 0 0  7 0 0 a b2 1 2a + 3b a + b 3 17 = = . 60.   0 4  0         5 0  0 c d 3 1 2c + 3d c + d 4 1 0 0000 12

 ( ) + + + + + +  0 + 0 + 0 0 + ( 5)4 + 0 0 + 0 + 0   2a + 3b = 3  0 + 0 + 0 0 + 0 + 0 0 + 0 + 0    a + b = 17, and 2c + 3d = 4 c + d = 1.  21 0 0 =  0 20   0 00 Similarly, Solving by Gauss-Jordan elimination

a = 48,  21 0 0 b = 31, c = 7, and d = 6. BA = 0  20 0 58. 2 0 02 0 0 AA = 0 3 0 0 3 0 = 0 9 0 4 0 0  0 0 0 0 0 00 0 0 000 a11 0 0 b11 b12 b13  a11b11 a11b12 a11b13   a   b b  =  a b a b  62. (a) AB = 0 22 0 b21 22 23 22 21 22 22 a2b 2 23  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. BA =  b11 b12 b13  a1 1 0 0  a11b11a22b12a33b13  21 22 23 22 11 21 22 22 33 23 (b) b b b   0 a 0 = a b a b a b   b31 b32 b33    0 0 a33    a11b31 a22b32 a33b33  

yields

1 + 0 + 2



c =  n  n   ij  ij ik kj k =1 ii i 1  i =1  k =1 aikbki  





=  n

 n d =  n  n b a  = Tr

AB)  ij  ij ik kj k 1 ii i 1  i =1  k =1 ik ki   cos  sin   cos  sin    cos  cos  sin  sin cos  ( sin  ) sin  cos         sin  cos  sin  cos   sin  cos  + cos  sin  sin  ( sin  ) + cos  cos   BA = cos  sin   cos  sin   cos  cos  sin  sin  cos  ( sin  ) sin  cos   sin  cos  sin  cos   sin  cos  + cos  sin  sin  ( sin  ) + cos  cos   cos( +  ) sin( + ) So, you see that AB = BA =  . sin( +  ) cos( +  ) 70. AB =

, d

(

1 which is impossible. So, the original equation has no solution.

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.

76. Let rows s and t be identical in the matrix A So, asj = aij for j = 1, , n Let

where

ij =  aikbkj Then, csj =  askbkj , and ctj =  atkbkj Because ask = atk for k = 1, , n, rows s and t of AB k =1 arethesame. k =1 k =1

78. (a) No, the matrices have differentsizes.

(b) No, the matrices have different sizes.

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

the matrix for 2013 by 1 This produces a matrix giving the information as percents of the total population.

36 Chapter 2 Matrices 3090  3160      72. Let A = a11 a12 and B = b11 b     12 a21 a22  b21 b22  1 0 Then the matrix equation AB BA =   is equivalent to 0 1 a11 a12 b11 b12     b11 b12 a11 a12     1 0 =  a21 a22  b21 b22 b21 b22 a21 a22 0 1 This equation implies that a11b11 + a12b21 b11a11 b12a21 = a12b21 b12a21 = 1 a21b12 + a22b22 b21a12 b22a22 = a21b12 b21a12 =



=  cij ,

n n n c

70 50 25 84 60 30 80. 1.2  =   35 100 70 42 120 84

12,306 35,240 7830  3.98 11.40 2.53  16,095 41,830 9051 5.21     13.54 2.93 A = 1 27,799 72,075 14,985  9.00 23.33 4.85 3090      5698 13,717 2710  1.84 4.44 0.88  12,222 31,867 5901  3.96 10.31 1.91

12,026 35,471 8446  3.81 11.23 2.67     15,772 41,985 9791 4.99 13.29 3.10 B = 1 27,954 73,703 16,727  8.85 23.32 5.29 3160      5710 14,067 3104  1.81 4.45 0.98     12,124 32,614 6636  3.84 10.32 2.10 (b) 3.81 11.23 2.67 3.98 11.40 2.53      0.18 0.22 0.18 0.14 0.25 0.17 4.99 13.29 3.10 5.21 13.54 2.93    B A = 8.85 23.32 5.29 9.00 23.33 4.85 = 0.15 0.001 0.44     1.81 4.45 0.98 1.84 4.44 0.88     3.84 10.32 2.10 3.96 10.31 1.91    0.04 0.01 0.11    0.12 0.01 0.19

Multiply



(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.

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 Properties of Matrix Operations

Section 2.1 Operations with Matrices 37 3 3 1 1 1 3 3 3  0 0  0 1 0 1 2   84. AB = 0 0 1 5 6  1 0  0 0 1 2   3 4  1 2   3 4   0 1 0 0 5 6 78  5 6 7 8

2.  6 8 +  0 5 +  11 7 = 6 + 0 + ( 11) 8 + 5 + ( 7) =  5 6            1 0  3 1  2 1  1 + ( 3) + 2 0 + ( 1) + ( 1)  2 2 4. 1(5 2 4 0 + 14 6 18 9) = 1 5 + 14 2 + 6 4 + ( 18) 0 + 9 = 119 4 14 9 = 19 2 7 9  2 2 2  2 2  6.   4 11 1 5 1    7 5     4 11  1   5 + 7 1 + 5  1 2 1 + 6  3 4 +  9 1 =  2 1 + 6 3 + ( 9) 4 + ( 1)  9 3  0 13  6 1  9 3 0 + 6 13 + ( 1)      4 11  2 4  4 11  1 2   1      3 3 =  2 1 + 6  6 3 =  2 1 +  1 1       2  9 3  6 12  9 3  1 2  4 + 1 11 + 2   11 31 = 2 + ( ) 1 +  =   1 2  0 1 1 3 8. A + B =   +   =   3 4  1 2 2 6  2  2  9 + 1 3 + 2  8 1  0 1  0 1 0 1 10. (a + b)B = (3 + ( 4))  = ( 1)  =    1 2  1 2 1 0 0 0 0 0 0 12. (ab)O = (3)( 4)  = ( 12)  =   0 0 0 0 0 0 2 14. (a) X = 3A 2B (b) 2 X = 2 A B  6 3  0 6  4 2  0 3 =  0  4 0     2X 0  3    =  2   2 0  9 12  8 2  6 9  6 8  4 1  4 5   2X   =  1 0 =  0 0 17 10 10  2  7 5  2 X =  0 0  5 7  3 4    1 2 3 4  7 8 =  5 6 7 8

38 Chapter 2 Matrices  2

Section 2.2 Properties of Matrix Operations 37   7   10            =     =   = 2 2 4 2 = 8 2   0 (c) 2 X + 3A = B (d) 2 A + 4B = 2 X 2 X +  6 3  0 3  4 2  0 12  3 0 =  2 0  2   8 0 = 2 X      0+    9 12  4 1  6 8  16 4  6 6  4 10 2X =  1   10 0 = 2X      13 11  10 12  3 3  2 5 X  1  =  2 0 5 = X  0 16. 13 11  2 2  c C B = 2  0 1 1 3  5 6 AB C =   4 2  1 50  3 4 ( ) ( )       0 1   1 0 1 2 24. (a) ( )  1 3 2 3    = ( 2) 1 2     3   1 1  3 4  1 = 2 4   2 6 3 = 8   26 6 14   0 1 9  1 1   =  180 18. C(BC) =  0 1 1 3  0 1         1 2  1 0   12 5 4 2  1 5 0  3 4  =  0 1 3 1 =  2 1 (b) A BC =        0 1    2   3  ( )  1        1 0  1  1  3  2 3 3     20. B(C + O) =  1 3 0 1 + 0 0   4 2  3 1  1     1 1  1 2 1 0 0 0  3  3 2 =  1 3  0 1 =  3 1 =  18 0          1 2  1 0  2 1  12 5 =  1 3 1 2 3 = 1 1 1 1  3 1  22. B(cA)  12( 2)01 1 26. AB 4 2 2 21 = 8 4 3     1 1 1  1    2 2  2 4 2 8  1 3  2 4 6  2 10 0 1 1 1 1   3 1  BA       AB  1 2  0 2 2  2 0 10 1 1  1 1 1 3 2 4  2 2 4 8  1 2 30 0 0 12 6 9 AC =      28. 0 5 40 0 0 = 16 8 12 3 214 23 4 2 3  4 =  5 6 30 0 0    4 4 0 0 0 = BC  1 014 But A  B 2 3 =

38 Chapter 2 Matrices 2 4 1 2 0 0 30. AB =   1  =   = O 2 4  2 1 0 0 But A  O and B  O

Section 2.2 Properties of Matrix Operations 39           ( )  =   2  32. AT =  1 2   1 0  =  1 2  36. A2 =  1 2   1 2  =  1 0  = I 2 0 10 1 0 1 0 10 1 0 1 1 2 1 0 1 2 So, A4 = (A2)2 = I 2 = I 1 0 34. A + IA = + 2 2 =  .        0 1 0 1 01 0 1 =  1 2  +  1 2  =  2 4 38. In general, AB  BA for matrices. 0 1 0 1 0 2  6 7 19T  6 7 19 40. DT = 7 0 = 7 0 23    23 1 2  3 1T  1 1T 1 4 42 (AB)T =     =   =   0 2 2 1  4 2 1 2   19 23 32    19 23 32  =  3 1T 1 2T =  3 2 1 0 = 1 4 BT AT            2 1 0 2  1 12 2 1 2 2 1 1 1 0 1T 4 0 7T  4 2 4 AB T = 44. 0 1 3 2 1 2 = 2 4 7 =  0 4 2  4 0 2   0 1 3  4 2 2  77 2 1 0 1T 2 1 1  1 2 0 2 0 4  4 2 4 BT AT = 1   1 2 2  0 3 =  0 1 1  1 1 0 =  0 4 2  0 1 3  4 0 2  1 2 3  132  77 2  1 1      46. (a) AT A =  1 3 0  10 11   1 4  3 20 4 = 2   11 21  1 1   1 3 0   2 1 2 (b) T AA = 3  4  1 4 = 1 25   8 0 2 2 2 8 4  4 2 14 6  4 3 2 0  252 8 168 104 48. (a) ATA =  3 0 2  2 11 12      82 0 11 1 8 77 70 50 5 14 2 12 9  168 70 294 139        0 1 9 4 6 8 5 4  104 50 139 98  4 3 2 0 4 2 14 6  29 30 86 10 0 11      (b) AAT =   1 3 0 2 8 =  30 126 169 47 14 2 12 9  2 11 12 5  86 169 425 28        6 8 5 4 0 1 9 4  10 47 28 141 50. (1)17 0 0 0 0   0 ( 1)17 0 0 0  1 0 0 0 0   T

40 Chapter 2 Matrices 17 0 1 0 0 A17 =    0 0 (1) 0 0   = 0 0 0 1 0 0  0 0 0 ( 1)17 0     0 0 0 0 (1)17 0 0 0 1 0 0 0 0 0 1  

(a) False. In general, for n

n matrices A and B it is not true that AB = BA For example, let

(c)

58. aX + A(bB) = b( AB + IB)

aX + ( Ab)B = b( AB + B)

aX + bAB = bAB + bB

Original equation

Associative property; property of the identity matrix

Propertyof scalar multiplication; distributive property

aX + bAB + ( bAB) = bAB + bB + ( bAB) Add bAB to both sides.

aX = bAB + bB + ( bAB) aX = bAB + ( bAB) + bB aX =

X = b B a

Additive inverse

Commutative property

Additive inverse

Divide by a

Section 2.2 Properties of Matrix Operations 41      52. (1)20 0 0 0 0  1 0 0 0 0  0 ( 1)20 0 0 0   1 0 0 0     0   A20 =  0 0 (1)20 0 0  = 0 0 1 0 0   0 0 0 ( 1)20    0 0 0 1 0 0     0 0 0 0 1  0 0 0 0 (1)20   8 0 0 23 0 0  2 0 0 A3 =    3    54. Because 0 1 0 =  0 ( 1) 0 , you have A = 0 1 0 0 0 27 0 0 (3)3  0 0 3 1 1 1 0



A =   , B =   0 0 1 0 2 0 Then AB =   11  = BA 0 0 1 1 (b) False. Let A = 11  , B = 1 0 , C = 2 0 Then AB = 20 = AC, but B  C      0 0 1 0 0 0

56.

  0 0

True. See Theorem 2.6, part 2 on page 57.

60. f A = 1 0 0  2 1 1   2 1 12  21 13  ( ) 100 1 0 + 5 1 0 2 2 1 0 2 +  1 0 2 0 0 1  1 1 3  1 1 3  1 1 3 10 0 0  10 5 5  2 1 1 2 1 1  21 1 2 1 12 = 0 10  +     1 0 + 1 0  1 0   0  5 0 10 2 1 0 2 2  2 2  0 0 10   5 5 15    1 1 3   1 1 3    1 1 3   1 1 3   0 5 5  6 1 3  2 1 1 6 1 3 = 5    +  1 0  0 3   10 10 2 0 3 5  2 5   5 5 5   4 2 12   1 1 3   4 2 12   0 5 5  12 2 6  16 3 13 = 5    +  2 5   10 10  0 6 10  21   5 5 5   4 6 12 =  11 2    8 4 24   18 8 44

42 Chapter 2 Matrices  3 1  15 9 25

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

p, B has size

r, and C has size r

m. Then the entry in the kth row and the jth column of BC is r l =1 p

bklclj Therefore, the entry in ith row and jth column of A(BC) is r

aik  bklclj =  aik bklclj . k 1 l 1 k l l=1 r il lj il

The entry in the ith row and jth column of (AB)C is column. d c , where d is the entry of AB in the ith row and the lth

So, dil = r p

p k =1 aikbkl for each l = 1, , r So, the ijth entry of ( AB)C is  aikbkl clj i 1 k 1

=aikbklcij 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

in the ith row and jth column of AC + BC is

real numbers.

ai1b1 j + ai2b2 j + + ainbnj   The corresponding entry for (cA)B is (cai1)b1 j + (cai2 )b2 j + + (cain )bnj and the corresponding entry for A(cB) is ai1(cb1 j ) + ai2(cb2 j ) + + ain(cbnj ). Because these three expressions are equal, you have shown that c( AB) = (cA)B = A(cB)

(b) Let A and B besymmetric.

If AB = BA, then (AB)T = BT AT = BA = AB and AB is symmetric.

If (AB)T = AB, then AB = ( AB)

72. Because A =

AT , the matrix is symmetric.

74. Because A

= AT , the matrix is skew-symmetric.

is skew-symmetric.

Section 2.2 Properties of Matrix Operations 43 62. (cd )A = (cd ) aij   =  (cd )aij   =  c(daij )  = c daij   = c(dA) 64. (c + d)A = (c + d) aij  =  (c + d)aij   =  caij + daij  =  caij  +  daij   = c aij  + d  aij  = cA + dA

n 





(ail + bil )c1 j + (ai2 + bi2 )c2 j + + (ain + bin )cnj , whereas the

(ai1c1 j + + aincnj ) + (bi1c1 j + + bincnj ), which are equal

entry

by the distributive law for

c 

68. (2) (A + B)T = ( a   +  b  )T =  a + b  T = a + b  = a + b= AT + BT ij ij ij





ji (3) (cA)T = (c a  )T =  ca  T =  ca   = c a   = c(AT ) ij ij ji ji

ith row and jth

T is a b + a

ith row j1 1i j 2 2i jn ni and jth column of BT AT is b1iaj1 + b2ia j2 + + bnia jn , which is the same. 0 1 1 1  1 0 70.

:     =   1 0  1 0  1 1

ji ji

(4) The entry in the

column of (AB)

b + a b . On the other hand, the entry in the

(a) Answers will vary. Sampleanswer

2 1

= B

= BA and AB = BA

  = 1 3 0 2 1

2 0 3 

76.

AT  1 = A and BT  3 0 = B,

(A + B)T = AT + BT = A B = (A + B)

 

then

, which implies that

Section 2.3 The Inverse of a Matrix 41     a  2 3 2 1  01  7 2 4 3  A = = . 78. Let A = a11 a12 a13 a1n    a a a a21 22 23  ⁝ ⁝ ⁝ 2n     an1 an2 an3 A AT = a11 a12 a13 ann  a1n  a11 a21 a31 an1  a a a   a a a a a   21 22 23  2n   12   22 32 ⁝ n2     an1 an2 an3 ann   a1n a2n a3n ann   0 a12 a21 a13 a31 a1n an1  =a21 a12 0 a23 a32 a2n an2    ⁝ ⁝ ⁝ ⁝  an1 a1n an2 a2n an3 a3n 0   0 a12 a21 a13 a31 a1n an1 n2 =  (a12 a21) 0 a23 a32 a2n   ⁝ ⁝   (a1n an1) (a2n an2) So, A AT is skew-symmetric. Section 2.3 The Inverse of a M  1 1 2 1  2 1 1 1 (a3n an3) trix 1 0   0  8. Use the formula d b 2. AB =  1    =  2 + 2 1 +  =   1  2 1 1  2 0 1 A 1 =  , BA = 2 1  1 1 = 2 1 2 + 2 = 1 0 ad bc  c a     1    0 1 where 1 1 2 1 1 1 + 2   a b  2 2 1 1  3 1 1 0 4. AB =    5 5 =   5 5   3 1  1 1 1 0 A =   =  . c d 2 2 So, the inverse is  1 1  BA =  5 5   =   1 1  22  4 4  2 515  2 3 0 1 A =   =   2(2) ( 2)(2)  2 2  1 1   2 17 111 1 2 1 0 6. AB =  1 11    =  0 1  0 3 2   3 6 5   0 0 0   1 10. Use the formula  d   4 4  A 1 = 1 b ,  1 1 2  2 17 11 1 0 0 ad bc c a  BA =  2 4 3  1 11 7 = 01         0 where 3 6 5 3 6 2 00 1 a b  1 2 c d      2 3 0

42 Chapter 2 Matrices A 1 = 1  32  3 2 = So, the inverse is ( 1)( 3) ( 2)(2)     2 1  2 1

12. Usingthe formula

14. Adjoin the identity matrix

Using elementary row operations, reduce the matrix as follows.

20. Adjoin the identity matrix toform

Using elementary row operations, you cannot form the

identity matrix on the left side. Therefore, the matrix has

no inverse.

22.

Using elementary row operations, you cannot form the identity matrix on the left side. Therefore, the matrix has no inverse.

26. Adjoin the identity matrix to form

Using elementary row operations, reduce the matrix as follows.

Therefore, the inverseis

Section 2.3 The Inverse of a Matrix 43 12 5      2     0 5

1 1  d b  5 1 11 1 0 0 A =  ,  6 3 6  ad bc c a A I  =  0 2 2 0 1 0. where  1 3 1 5 0 0 1  a b  1 1 A =   =   c d 3 2 2 

elementary row operations,

cannot

identity matrix on the left

Therefore, the matrix has    3 you see that ad bc = ( 1)( 3) (1)(3) = 0. So, the matrix has no inverse.

Using

you

form the

side.

no inverse.

to form

A I =  0.1 0.2 0.3 1 0 0    0.3 0.2 0.2 0 1 0 A I =  1 2 2 10 0  0.5 0.5 0.5 0 0 1    3 7 9 0 1 0  Using elementary row operations, reduce the matrix as  1 4 7 0 0 1 follows. Using elementary row operations, reduce the matrix as follows. I A 1 = 1 0 0 0 2 0.8   0 1 0 10 4 4.4 I A 1 =  1 0 0 13 6  4 0 1 0 3 0 0 1 5 2 1 0 0 1 10 2 3.2 Therefore, the inverse is  0 2 0.8

Adjoin the identity matrix to form

A 1 =  10 4 4.4 10 5 7 1 0 0  10 2 3.2 A I  =  5 1 4 0 1 0 24.

matrix to form 32 2001

16. Adjoin the identity matrix to form

Adjoin the identity

1 0 0 10 4 27 0 0 1 0 0 1  I A 1  =  1 0 2 1  001 13 535

Therefore, the inverse is

 10 427  1 0 0 0 1 0 0 0 A 1 = 2 1 5     0 2 0 0 0 1 0 0   13 5 35 A I  = 0 0 2 0 0 0 1 0 

matrix to form 0 0 0 3 0 0 0 1 A I =  3    2 2 5 1 0 2 4 0 1 0 0

18. Adjoin the identity

 4 4 0 0 0 1 1 0 0 0 1 0 0 0  

I A 1 = 0 1 0 0 0 1  0 0 1 0

0 0 1 0 0 0 2 

0 0 0 1 0 0 0 1

1 0 0 0 3 A I = 1 0 0 1   3 0 0 0 2 5 5 0

44 Chapter 2 Matrices 1 0 3  0 1  A 1 =  2 0 0 0 0 2  0 0 0 1

Section 2.3 The Inverse of a Matrix 45     4 9 4 3 36 = 143 1    ( ) A 1 =    0 28. Adjoin the identity matrix to form  1 2 4 8 7 14 1 0 0 0 32. A =  3  A I = 25 0 2   2 4 6 0 1 0 0 . ad bc = (1)(2) ( 2)( 3) = 4  1 7 0 0 1 0  2 2  1 1   3 6 5 10 0 0 0 1 A 1 = 1  4 3 Using elementary row operations, reduce the matrix as 1=  2 3 2 1  4 4  follows.  12 3 1 0 0 0 27 10 4 29 34. A =    5 2 A I  = 0 1 0 0 0 0 1 0 16 5 17 4 2 18 2 20 ad bc = ( 12)( 2) 3(5) = 24 15 = 9   0 0 0 1 7 2 1 8 A 1 = 1  2 3  2 1  5 12 =  9 3 Therefore the inverse is 9    5 4  9 3  27 10 4 29 16 5 2  1 9  A 1 =   2 18 36. A =  4 4  17 4 20  5 8   3 9  7 2 1 8 ad bc = ( 1)(8) (9)(5) = 143 30. Adjoin the identity matrix to form  8 9  32 81  1 3 2 0 1 0 0 0   A 1 36  9 4 =  143 143  5 60 9   3 4  143 143 A I  = 0 2 4 6 0 1 0 0 0 0 2 1 0 0 1 0 38. A 2 = A 1 =  1 6 7 2   = 1  1 56     ( )  47  2209 0 0 0 5 00 0 1  5 2  40 31 Using elementary row operations, reduce the matrix as 1  31 56 1  1 56 A 2 = A2 =   = 1   follows. 1 0 0 0 1 1.5 4 2.6  40 1 The results are equal. 2209 40 31    I A 1  = 0 1 0 0 0 0.5 1 0 1 0 0 0 0.5 0.80.1   0 0 0 1 0 0 0 0.2 Therefore, the inverse is 1 1.5 4 2.6   0 0.5 1 0.8 0 0 0.5 0.1   0 0 0 0.2   15 4 28 2  873 228 1604 40. A 2 = (A 1)2 =  1  1 0   = 1  61 16   2  23 6 422 4  1317 344 112    2420    48 4 32 1  873 228 1604 A 2 = (A2) 1 =  29 48 17 = 1  61 16    4 112  22 9 15  1317 344 2420 The results are equal.  2 

Using the algorithm to invert a matrix, you find that the

Using a graphing utilityor software program, you have Ax = b

46 Chapter 2 Matrices   3 3  3 3  0 1 3 2 2    2 2  0 = 2 42. (a) (AB) 1 = B 1A 1 48. The coefficient matrix for each system is  5 2  2 1  1 1 2 = 11 11  7  7 A =   3 1 3 2 1 2 1  11 11     4 9 = 177  7 7 1 1 1

 9 1 2 1 T inverse is  2 3  1 1 1 (b) (AT ) 1 = ( A 1)T =  7 7 =  7 7  1 2 1  3 2   1 2 A = 3 3 1  7 7  7 7  1 2 1 1  2 1  1 1   3 3  (c) (2A) = 1 A = 1  7 7 =  7 14    1 1 1 0 1  2 2  3 2  3 1 1 2 1    7 7  14 7 (a) x = A b = 3 3 1  0 = 1 44. (a) (AB) 1 = B 1A 1  1 2 1 1 1  6 5 3 1 4 2 The solution is: x1 = 1, x2 = 1, and x3 = 1. = 2 4     1 1 1 1 1  10 1 3        1 3 44 2 1 (b) x = A 1b = 2 1  1  2 = 0  6 25 24  1 3 3 2 1   0   1  =  6 10 7 The solution is: x1 = 1, x2 = 0, and x3 = 1.  17 1 T  7 15 1 4 2T  1 0 4

(b) (AT ) = (A 1) =   =    1 0 1 3  4 1 2 4 2 1  2 3 1    2 1 4 2 1 2 2 1 x = A 1b =  1 (c) (2A) 1 = 1 A 1 = 1    1 3  2 2 0   4 2 1 2 1 1 where  1

The coefficient matrix for each system is 1 1 1 3 1  x1  3 2 1    x    A =   2 1 1 1 1  2   4 2 1 and the formula for the inverse of a 2  2 matrix A = 1 1 1 2 1 , x = x3 , and b =  3  2 1 4 1   x     produces 1  4       1   1  1 1 =  1 1  3 1 1 2 1 x5  5 A 1 =   4 4 . The solution is: x = 1, x = 2, x = 1, x = 0, and 2 + 2  2 2  1 1  x5 = 1. 1 2 3 4 1  1 1  3 = 1 (a) x = A b =  4 4     1

50.

46.

Section 2.3 The Inverse of a Matrix 47 2 2 2 2  1 1  7 5 The solution is: x = 1 and y = 5. 1  1 1  1 =  1 (b) x = A b =  4 4      1 1  3  1 The solution is: x = 1 and y = 1.

52. Using a graphing utilityor software program, you have

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

Using elementary row operations, reduce the matrix as follows.

48 Chapter 2 Matrices   So,  3 1 4  3 0

Ax = b

 1 A 1 = 1  d b    2 x A 1b =  1 ad bc c a  = 1 2  2  sec  tan   =   sec  3  tan  tan  sec       sec  tan   where  0 =    1  tan  sec   62. Adjoin the identity matrix to form  4 2 4 2 5 1  x1 0.017 0 3 6 5 6 3   2  0.010 0.008 1 0   3 x  F I  = 0.010  2 3 1 3 1 2 x3 0.008 A = , x =  , and   1 4 4 6 2 4 x4  1 5 2 3    5 x5 

 2 3 4 6 1 2 x  1 0 0 115.56 100 4.44  111  6  I F 1  =  1 0 100 250 10  0   0 0 1 4.44 100 115.56  0 b =  . 115.56 100 4.44  9 1     So, F =  100 250 100 and  1  4.44 100 115.56  12   The solution is: x1 = 1, x2 = 2, x3 = 1, x4 = 3, x = 0, and x = 1. 115.56 100 4.44 0  15 100 250   =   5 6  100 0.15 37.5  4.44 100 115.56  0  15 54. The inverse of A is given by A 1 1  2  x. 64. 1 T = (A 1A)T = I Tn = I a nnd = x 4  1 2 A (A ) 1 T T = (AA 1) = I T = I T Letting A 1 = A, you find that 1 = 1. (A ) A 1 T n n T 1 So, x = 3. x 4 (A ) = (A ) 66. (I 2 A)(I 2 A) = I 2 2IA 2 AI + 4 A2 56. The matrix  x 2  will be singular if = I 4 A + 4 A2  3 4 ad bc = (x)(4) ( 3)(2) = 0, which implies that 4x = 6 or x = 3 2 = I 4 A + 4 A = I So, (I 2A) 1 = I 2A (because A = A2) 58. First, find 4 A 68. Because ABC = I , A is invertible and A 1 = BC  1 1 1 2 4  1 1  4 A = (4A) =   =  8 4 So, ABC A = A and BC A = I   4 + 12 3 2 Then, multiply by 1 to obtain T w = 1  F d = 0.012 0.010 0 1 01 0.010 0.017 0 0 

Section 2.3 The Inverse of a Matrix 49 4   3 1  3 1  1( 2) = 1 16 8 So, B 1 = CA 70. Let A2 = A and suppose A is nonsingular. Then, A 1 A = 1 (4 A) = 4  1 1  1 1  exists, and you have the following. 1  8 4  = 32 16 A A A A 16 8 64 32 (A 1A)A = I A = I

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

(b) False. For example, consider the matrix

but

78. Let C be the inverse of (I AB), that is C = (I AB)

1 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, 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 –

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

Section 2.4 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,

50 Chapter 2 Matrices    a   4        nn

74. A

an inverse if aii  0 for all i = 1 n and

1 1  1 0 0  0 0, a11 0     

1 0 0 = 1  0. A 1 =  0 1 a22  0 

72.

67.

has

which is not invertible,

a    ⁝ ⁝ ⁝ ⁝  nonsingular matrix.  0 0 0 1    76.  1 2 A =  2 1 (a) A2 2 A + 5I =  3 4  2 4 + 50 = 0 0  3  4 2 0 5 0 0 (b) A(1(2I A)) = 1(2 A A2) = 1(5I ) = I 5 5 5 Similarly, (1(2I A)) A = I Or, 1(2I A) = 1 2 = A 1 directly. 5 5 5   2 1 

––I

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

1 0 1   E = 0 1 0. 0 0 1 0 1

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

14. Answers will vary. Sample answer:

Section 2.4 Elementary Matrices 47 2   2   7   7  2 7

1 0 1   E = 0 1 0 0 0 1

Matrix Elementary Row Operation Elementary Matrix 1 1 2 2 R1  R2 0 1 0     0 3 3 6 1 0 0 0 0 2 2 001 1 1 2 2 1 0 0  1 1  (1)R2 → R2 0 1 0 0 2 3  3  0 0 2 2 001 1 1 2 2 1 0 0     0 1 0 (1)R3 → R3 0 0 1  So,  1 0 01 0 0 0 1 00 3 3 6 1 1 2 2        0 1 0 0 3 0 1 0 0 1 1 2 2 = 0 1 1 2 0 0 1 0 0 10 0 10 0 2 2 0 0 1 1 16. Answers will vary. Sample answer: Matrix Elementary Row Operation Elementary Matrix  1 3 0    1 0 0  0 1 1 ( 2)R1 + R2 → R2  2 1 0 3 2 4 001 1 3 0  1 0 0     0 1 1  0 1 0 0 11 4 ( 3)R1 + R3 → R3   3 0 1   1 3 0   1 0 0  0 1 1 ( 1)R2 → R2 0 1 0 0 11 4 0 01 1 3 0 1 0 0  0 1 1 (11)R2 + R3 → R3   0 1 0 0 0 7 0 11 1 1 3 0 1 0 0   1 R → R   0 1 1 ( ) 3 3 0 1 0 0 0 1 0 0 1  1 0 01 0 01 0 0  1 0 0  1 0 0 1 3 0 1 3 0 So,               0 1 00 1 0 0 1 0  0 1 0  2 1 02 5 1 = 0 1 1. 0 0 1  0 11 1   0 0 1    3 0 1    0 0 1   3 2 4   0 0 1  1 0 1 1 2 0 0 1 1

48 Chapter 2 Matrices   3 2 2  3      0 0 16 48 0 0 1 0      18. Matrix Elementary Row Operations Elementary Matrix 1 6 0 2   1 0 0 0 0 3 3 9   0 1 0 0  0 17 1 3  R3 + ( 2)R1 → R3  2 0 1 0  4 8 5 1 0 0 0 1     1 6 0 2   1 0 0 0 0 3 3 9 0 1 0 0     0 17 1 3 R + 4 R  0 0 1 0 0 32 5 7 4 ( ) 2 → R4 4 0 0 1     1 6 0 2  1 0 0 0 0 1 1 3 ( 1 )R → R 0 1 0 0 0 17 1 3 0 0 1 0 0 32 5 7     0 0 0 1 1 6 0 2  1 0 0 0 0 1 1 3 0 1 0 0 0 0 16 48 R3 + ( 17)R2 → R3 0 17 1 0  0 32 5 7 0 0 0 1     1 6 0 2  1 0 0 0 0 1 1 3 0 1 0 0     0 0 27 89 R4 + ( 32)R2 → R2 0 32 0 1     1 6 0 2  1 0 0 0 0 1 1 3 0 1 0 0 0 0 1 3  ( 1 )R3 → R3 0 0 1 0 0 0 27 89 16  16    000 1 1 6 0 2  1 0 0 0 0 1 1 3 0 1 0 0     0 0 1 3 0 0 0 8 R4 + ( 27)R3 → R4 0 0 1 0 0 0 27 1     1 6 0 2  1 0 0 0 0 1 1 3  0 1 0 0 0 0 1 3   0  0 0 0 1  1  0 1 0 R → R 0 1   (8) 4 4  0 0 8 So, 1 0 01 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0         0 0 0 0 1 0 0 0 1 0 0 0 1 0  1    16    0 8 0 0 27 1 0 0 0 1 0 32 0 1 1 0 0 0 1 0 0 0  1 0 0 0  1 0 0 0 0 1 0 0 0 1 0 0  0 1 0 0  0 1 0 0 0 17 1 0  3   0 0 1 0 2 0 1 0   0 0 1 0     0 0 0 1 0 0 0 1  4 0 0 1  0 0 0 1 1 6 0 2 1 6 0 2 0 3 3 9 0 1 1 3   =     0 0 1 0 0 1 0 0

Section 2.4 Elementary Matrices 49 2 3    4 1  5 1 1 0 0  1 8 5 1 0 0 0

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

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

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

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

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

50 Chapter 2 Matrices 25   2   1 1 1 1

1 0 E 1 = 5  0 1

1 0 0   E 1 = 0 1 0 031

 1 0 0 0  E 1 = 0 1 k 0 . 0 0 1 0   0 0 0 1

1 0 (1)R1 → R1 1 0   2 E1 = 2  1 1 1 0   E2=  0 1   1 0   0 1 R2 R1 → R2  1 1 Use the elementary matrices to find the inverse.  1 01 0  1 20 A 1 = E2E1 =  2  =    1 1 0 1   1 1

 1 0 2 1 1 E =  1 0 10  0 1 2 (2)R2 → R2 1 0 2 0 0 0 1 1 0 0 R1 + 2R3 → R1 0 0 1 1 0 2 0 1 1  E    2  2 = 0 1 0 0 0 1 1 0 0 001 1 0 0 0 1  21  R2 (1)R3 → R2 E3 = 0 1  21 2 0 0 1 0 0 1 Use the elementary matrices to find the inverse. 1 0 0 1 0 2 1 0 0 A 1 = E E E =  0 1 1   0 1 0 0  1 0  3 2 1 2   2 0 0 1001001 1 0 2 1 0 0 1 0 2 =  0 1      2 0 2 0 = 0 2 2  0 0 10 0 1 0 0 1

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

The matrix

is itself an elementary matrix, so the factorization is

Section 2.4 Elementary Matrices 51 

0 1

1 0 

0 1 A =   1 0   1 1 32. Reduce the matrix A = 2 1  as

  Matrix Elementary Row Operation Elementary Matrix 1 1   Add 2 times row one to row two. E1  1 0   0 1  2 1  1 1  0 1 1 0 Multiplyrow two by 1. E2 = 1 0 0 1 1 1   0 1 Add 1 times row two to row one. E3 =   0 1 So, one way to factor A is A = E 1E 1E 1 =1 0 1 0 1 1 1 2 3      2 1 0 1 0 1 1 2 3   34. Reduce the matrix A = 2 5 6 as follows.  1 3 4  Matrix Elementary Row Operation Elementary Matrix 1 2 3  1 0 0 1  0 1 0 0 1 0 0 1 1  1 2 3  Add 1 times row one to row three. E2 =  0 1 0  1 0 1  1 0 0  0 1 0 Add 1 times row two to row three. E3 = 0 1 0 001 0 11 1 2 0 1 0 3   0 1 0 0 0 1 Add 3 times row three to row one.   E4 = 0 1 0 0 0 1 1 0 0 1 2 0 0 1 0  Add 2 times row two to row one. E5 = 0 1     0 0 0 1 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 = E 1E 1E 1E 1E 1 =           1 2 3 4 5 2 1 0 0 1 0 0 1 0 0 1 0 0 1 0.  0 0 1   1 0 1   0 1 1   0 0 1   0 0 1  =  0 1  0 Add 2 times row one to row two. E =  2 1  1 3 4  0 0  1 2 3  1 0

30.

A =

follows.

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

52 Chapter 2 Matrices 0   3   2 =    0 2 4 E =     1 2 3 4 5 6 7 2    2  2 0

1 0 0 1  (1)R1 → R1 1 0 0 0  2  4 4  0 1 0 1 E = 0 1 0 0 0 0 1 2 1 0 0 1 0  1 0 0 2    0 0 0 1  1 0 0 1  1 0 1  1 0 0 0      E = 0 1 0 0 00 0 1 0 0 2 5R4 R1 → R4 2  0 0 1 0 1 0 0 1 2    1 0 0 1   2 0 1 0 1 0 0 1 2 1 0 0 0   0 1 0 0 E = 0 0 1 0 0 0 0 1 ( 2)R4 → R4  0 0 0 2 5  5  1 0 0 1  2 1 0 0 0   0 1 0 1 0 0 1  R → R 0 1 0 0 E4 0 0 1 0   3 3 0 0 0 1   0 0 0 1 1 0 0 0 R1 (1)R4 → R1 1 0 0 1    2  2  0 1 0 1 E = 0 1 0 0 0 0 1 2 5  0 0 1 0 0 0 0 1 0 0 0 1 1 0 0  0 1 0 0 0 1 0  R R → R 2 1 0  0 1 6 0 0 0 0 0 1 1 0  0 0 0 1 1 0 0 0   0 0 0 1 0 1 0  E7 =  0 0 1 0 R 3 + 2R 4 → R3 0 0 0 1 So, one way to factor A is A = E 1E 1E 1E 1E 1E 1E 1 4 0 0 0 1 0 0 0 1 0 0 01 0 0 0 1 0 0       1  1 0 0 01 0 0 0 = 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0  0 1 0 1 0 1 0 0 0 0 1 0 00 0 0 0 1 0 1  0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 2            0 0 01  1 0 0 1 0 0 0 50 0 01000100 0 10 0 0 1    2 1 0 0 0 0 1 0 0 0 0 1 2 0 0 0 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

So, multiplying E by itself interchanges the rows twice and E 2 = In

Section 2.4 Elementary Matrices 53

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 = E E E , where each E is an

matrix, then A isinvertible (because

every elementary matrix is) and

44. Matrix Elementary Matrix

54 Chapter 2 Matrices 1  1 A = E E E   E =   = =  U E 1 2 3   =      1 0 0  1 a 0 1 0 0 11 0 0 11 a 0 1    1 0 0 1 a 0   40. A 1 =                0 1 0 b 1 0 0 1 0 = 0 1 0  b 1 0 0 1 0 =  b 1 + ab 0 0 0 c 0 0 1 0 0 1    0 0 10 0 1   0 0   0 0   c  c

48. Matrix Elementary Matrix  2 0 0 0  2 1 1 0 = A

1 2 k i  6 2 1 0 elementary

 

 0 0 0 1 1 1 1 1 2 0 0 0 1 0 0 0 k 2 1    

0 1 1 0 6 2 1 0   0 0 0 1 1 1 0 0 1 0 0 1 0   0 0 0 1  2 1 2 0 0 0  1 0 0 0   = A 0 1 1     64  0 E =  0 1 0 0  2 1 E =  1 0 0 2 1 0 2  3 0 1 0  = U 1     0 0 0  0 0 0 1   1    0 1  3 1 1 1 0 2 1 2 0 0 0  1 0 0 0  E1A = U  A = E1 U =    = LU 0 1 1 0 0 1 0 0 3 1 0 1 0 0 3 0 3 0 2 1 0     46. Matrix Elementary Matrix 0 0 0 1 0 00 1 2 00 E3E2E1A = U  A = E 1E 1E 1U 0 3 1 = A  1 0 0 0 2 0 0 0 10 12 3       1 1 0 0 0 1 1 0 2 0 0  1 0 0  3 2 1 0 0 0 3 0   E =       0 3 1 1  0 1 0  0 0 0 1 0 0 0 1 0 12 3  5 0 1 = LU 2 0 0 E = 1 0 0  1 0 0 0  y1    4 0 3 1 = U 2 0 1 0 Ly = b:  1 1 0 0y2 =  4 0 0 7 E2E1A = U  A = E 1E 1 1U 2 041  3 2 1 0 y3  15        0 0 0 1    y4    1  1 0 02 0 0 y1 = 4, y1 + y2 = 4  y2 = 0, =    3y1 + 2 y2 + y3 = 15  y3 = 3, and y4 = 1. 0 1 0 0 3 1 5 4 10 0 7  2 0 0 0  x1    4  = LU Ux = y : 0 1 1 0 x2 0 0 3 0x  =  0  3   3    0 0 0 1   x4   1 x4 = 1, x3 = 1, x2 x3 = 0  x2 = 1, and x1 = 2.

Section 2.4 Elementary Matrices 55 So, the solution to the system Ax = b is: x1 = 2, x2 = x3 = x4 = 1.

, A is not idempotent.

54. Assume A is idempotent. Then

58. If A is row-equivalent to B, then

A = Ek E2E1B, A

where E1, …, Ek are elementary matrices.

So, B = E 1E 1 E 1A, 1 2 k which shows that B is row equivalent to A

which means that AT is idempotent. Now assume AT is idempotent. Then

60. (a) When an elementary row operation is performed on a matrix A, perform the same operation on I toobtain

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 AA = A which means that A is idempotent.

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

A Then solve the system Ly 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.

= b for y and

Section 2.5 Markov Chains 53 0 1 0 1 1 0 =     =    A. 56 (AB)2 = ( AB)( AB) 1 01 0 0 1 = A(BA)B Because A2  A, A is not idempotent. = A( AB)B = ( AA)(BB) 0 1 0 0 1 0 1 0 0 = AB 52. A2 = 1 0 0 10  0 = 0 1 0         2 0 0 10 0 1 001 So, (AB) = AB, and AB is idempotent. Because A2 

A2)

AT AT )

2 = A (

T = A

(

= A

T AT (AT AT )T

T =

AT )T

= A

(

50. A2

The matrix of transition probabilities is shown.

The initial state matrix represents the amounts of the physical states is shown.

To represent the amount of each physical state after the catalyst is added, multiply P by X 0 to obtain

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.

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

(a) The state matrix for next week is

54 Chapter 2 Matrices   3  5      8. 60% 70%

––From ––G L S 0.60 0 0  G    P = 0.400.70 0.50 L To  0 0.30 0.50 S  

0.20(10,000) 2000 X0 = 0.60(10,000) =     6000 0.20(10,000) 2000

0.60 0 0  2000 1200 PX =     =   0 0.40 0.70 0.50 6000 6000  0 0.30 0.50 2000 2800

10. 0.6 0.2 0 1  4  0.26   3 15  

X1 = PX0 = 0.2 0.7 0.1 31  = 31  =  0.3 noninfected.  0.2 0.1 0.9      0.4    1  2   From A B 0.6 0.2 0  4    15  17   75  0.226 P = 0.2 0.1 A  To X 2 = PX1 = 0.20.70.1  1 =349 = 0.326 0.8 0.9 B  0.2 0.10.9   3   1 50   67   0 446  The state matrix representing the current population is 5  150 0.6 0.2 0  17  151 0.2013    75  750   X 3 = PX 2 = 0 2 0 7 01 15409 = 7 2350 9 = 0 3186 0.3 A 0 =   0.7 B 0.2 0.1 0.9 67 12 150  25  0.48  

0.2 0.1 0.3 0.13 X1 = PX 0 =     =   0.8 0.9 0.7 0.87

next week 0.13(1000) = 130 mice will be infected. 0.2 0.1 0.13 0.113 (b) X 2 = PX1 =     =   0.8 0.9 0.87 0.887 0.2 0.1 0.113 0.1113 X 3 = PX 2 =     =   0.8 0.9 0.887 0.8887 In 3 weeks, 0.1113(1000)  111 mice will be infected. Gas (G) 40% Liquid (L) 50% 30% Solid (S) 50% X

So,

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

The state matrix representing the students is

(a) The state matrix for tomorrow is

tomorrow

(

) = 90 students will swim and 0.64(250) = 160 students will playbasketball.

(b) The state matrix for two days from now is

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.

In 2 months, the distribution of users will be

The state matrix representing the current product usage is

Section 2.5 Markov Chains 55   11 0.75 0.15 0.10  11

–From –S B 0.30 0.40 P =   S   To 0.70 0.60 B

0.4 S X 0 =   . 0.6 B

0.30 0.400.4 0.36 X1 = PX 0 =    =  . 0.70 0.600.6 0.64 So,

0.36

250

0.37 0.360.4 0.364 X 2 = P2 X 0 =    =   0.63 0.640.6 0.636 So, two days from now 0.364

250

= 91

basketball.

(

)

students will swim and 0.636(250) = 159 students will play

0.363637 0.3636370.4 0.36364 X 4 = P4 X 0 =    =   0.636363 0.6363630.6 0.63636 So, four days from now, 0.36364(250)  91 students will swim and 0.63636(250)  159 students will play basketball.

––From ––(b) X 2 = P2 X 0 0.2511   0.3330  0.325 A B N 0.75 0.15 0.10 A    P = 0.20 0.75 0.15 B To 0.05 0.10 0.75 N 

0.2511

0.3330 110,000 = 36,625

0.325  110,000 = 35,750 for

0.3139

110,000 = 27,625 for Brand A,

for Brand B, and

neither.

 2  A 11 (c) X18 = P18 X 0    0.3801  0.2151  X 0 =  3  B  5 11 (a) The state matrix for next month is  2  0.2227  11   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. X1 = P1X0 = 0.20 0.75 0.15131 =  0.309. 0.05 0.10 0.75  5   0.372 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.  N

18.

regular because every power of P has a zero in the second column.

56 Chapter 2 Matrices   = = 7    1    

Thestochastic matrix

The stochasticmatrix P = 00.3 2 7  P = 5 10  1 0.7 is regular because P2 has only positive entries. 3 3 5 10  is regular because P1 has onlypositive entries. PX = X 0 0.3 x1  1 0.7 x   x1  2 7  x   x  x PX = X  5 10 1 = 1   2  2  3 3  x  x  0.3x = x  5 10  2  2  2 1 2 7 x1 + 0.7x2 = x2  5 x1 + 10 x2 = x1 Because x + x = 1, the system of linear equations is 3 x1 + 3 x2 = x2 1 2 5 10 as follows. Because x1 + x2 = 1, the system of linear equations is x1 + 0.3x2 = 0 as follows. x1 0.3x2 = 0 3 x1 + 7 x2 = 0 5 10 x1 + x2 = 1 3 x1 7 x2 = 0 5 10 The solution to the system is x2 = 10 and x1 + x2 = 1 x = 1 10 13 = 3 The solution of the system is x2 = 6 and 1 13 13 3   x = 1 6 13 13 13 So, X 13. 10 13 So, X  7  = 13  6  20.

0.2 0 P =   0.8 1 is not

0.2 0 x1 x1 13 24. The stochastic matrix 2 1 1 9 4 3 P = 1 1 1 3 2 3 4 1 1 9 4 3 PX = X     =   0.8 1 x x is regular because P1 has

positive entries.  0.2x1  2 =  2  x1 2 1 1 x   x  9 4 3 1   1  0.8x1 + x2 = x2  1 1 1 x2  = x2 PX = X 3 2 3     Because x1 + x2 = 1,the system of linear equations is 4 1 1 9 4 3 x 3  x3  as follows. 2 x + 1 x + 1 x = x 0.8x = 0 9 1 4 2 3 3 1 1 1 x + 1 x + 1 x = x 0.8x1 = 0 3 1 4 2 2 3 3 2 x1 + 1 x2 + 1 x3 = x3 x1 + x2 = 1 9 4 3 The solution of the system is x1 = 0 and x2 = 1. Because x1 + x2 + x3 = 1, the system of linear equationsis as follows. = 0 7 x + 1 x + 1 x = 0 So, X 1 9 1 4 2 3 3   1 x1 1 x2 + 1 x3 = 0 3 2 3 4 x1 + 1 x2 + 2 x3 = 0 9 4 3  = 

22.

The stochastic matrix

only

Section 2.5 Markov Chains 57 x2 + x3 = 1 The solution of the system is x3 = 0.33, x2 x1 = 1 0.4 0.33 = 0.27. 0.27   = 0.4, and So, X =  0.4  0.33

Solution manual for elementary linear algebra 8th edition by larson isbn 1305658000 9781305658004

Solution Manual for Elementary Linear Algebra 8th Edition by Larson

C H A P T E R 2

Matrices Section 2.1 Operations with Matrices

Section 2.2 Properties of Matrix Operations

Section 2.4 Elementary Matrices

Section 2.5 Markov Chains