3. 2 0 2 0
3 = 3 −3 6 0 3 = 12 15
( AB ) C = [1 −1 2] 0 3 8 10
2 1 4 2 −2 4 1 4
2 0
0 1 −1 1 0 −1 2
16. A ( BC ) = 3
3 −2 3 2 0 1
1
4
2 0 −4 −4 −2 2
= 0 −2 −2 −1 1 = −9 −12 −9 12
3
−3 −4 −3 4
4
1 −14 −18 −13 17
2 0
1 −1 1 0 −1 2
( AB ) C = 0 3
3 −2 3 2 0 1
1
4
2 −2 −4 −4 −2 2
1 0 −1 2
= 9 −6 = −9 −12 −9 12
3 2 0 1
−9
13 −14 −18 −13 17
Each of the homogeneous linear systems in Problems 17-22 is already in echelon form, so it
remains only to write (by back substitution) the solution, first in parametric form and then in
vector form.
17. x3 = s, x4 = t , x1 = 5s − 4t , x2 = − 2 s + 7t
x = s (5, −2,1, 0 ) + t ( −4, 7, 0,1)
18. x2 = s, x4 = t , x1 = 3s − 6t , x3 = − 9t
x = s (3,1, 0, 0 ) + t ( −6, 0, −9,1)
19. x4 = s, x5 = t , x1 = − 3s + t , x2 = 2 s − 6t , x3 = − s + 8t
x = s ( −3, 2, −1,1, 0 ) + t (1, −6,8, 0,1)
20. x2 = s, x5 = t , x1 = 3s − 7t , x3 = 2t , x4 = 10t
x = s (3,1, 0, 0, 0 ) + t ( −7, 0, 2,10, 0 )
4. 21. x3 = r , x4 = s, x5 = t , x1 = r − 2 s − 7t , x2 = − 2r + 3s − 4t
x = r (1, −2,1, 0, 0 ) + s ( −2,3, 0,1, 0 ) + t ( −7, −4, 0, 0,1)
22. x2 = r , x4 = s, x5 = t , x1 = r − 7 s − 3t , x3 = s + 2t
x = r (1,1, 0, 0,0 ) + s ( −7, 0,1,1, 0 ) + t ( −3, 0, 2, 0,1)
2 1 a b 1 0
23. The matrix equation = entails the four scalar equations
3 2 c d 0 1
2a + c = 1 2b + d = 0
3a + 2c = 0 3b + 2d = 1
that we readily solve for a = 2, b = −1, c = −3, d = 2. Hence the apparent inverse
2 −1
matrix of A, such that AB = I, is B = . Indeed, we find that BA = I
−3 2
as well.
3 4 a b 1 0
24. The matrix equation = entails the four scalar equations
5 7 c d 0 1
3a + 4c = 1 3b + 4d = 0
5a + 7c = 0 5b + 7 d = 1
that we readily solve for a = 7, b = −4, c = −5, d = 3. Hence the apparent inverse
7 −4
matrix of A, such that AB = I, is B = . Indeed, we find that BA = I
−5 3
as well.
5 7 a b 1 0
25. The matrix equation = entails the four scalar equations
2 3 c d 0 1
5a + 7c = 1 5b + 7 d = 0
2a + 3c = 0 2b + 3d = 1
that we readily solve for a = 3, b = −7, c = −2, d = 5. Hence the apparent inverse
3 −7
matrix of A, such that AB = I, is B = . Indeed, we find that BA = I
−2 5
as well.
5. 1 −2 a b 1 0
26. The matrix equation = entails the four scalar equations
−2 4 c d 0 1
a − 2c = 1 b − 2d = 0
−2a + 4c = 0 − 2b + 4d = 1.
But the two equations in a and c obviously are inconsistent, because ( −1)(1) ≠ 0, and
the two equations in b and d are similarly inconsistent. Therefore the given matrix A
has no inverse matrix.
a1 0 0 0 b1 0 0 0 a1b1 0 0 0
0 a2 0 0 0 b 0 0 0 ab 0 0
2 2 2
27. 0 0 a3 0 0 0 b3 0 = 0 0 a3b3 0
0
0 0 an
0 0
0
bn 0
0 0 anbn
Thus the product of two diagonal matrices of the same size is obtained simply by
multiplying corresponding diagonal elements. Then the commutativity of scalar
multiplication immediately implies that AB = BA for diagonal matrices.
28. The matrix power A n is simply the product AAA A of n copies of A. It follows
(by associativity) that parentheses don't matter:
A r A s = ( AAA A) ( AAA A) = ( AAA A) = A r + s ,
r copies s copies r + s copies
the product of r + s copies of A in either case.
a b 1 0
29. (a + d ) A − (ad − bc )I = (a + d ) − (ad − bc ) 0 1
c d
(a + ad ) − (ad − bc)
2
ab + bd a 2 + bc ab + bd
= =
ac + cd (ad + d 2 ) − (ad − bc) ac + cd bc + d 2
a b a b
= c d = A
2
c d
2 1
30. If A = then a + d = 4 and ad − bc = 3. Hence
1 2
2 1 1 0 5 4
A 2 = 4 A − 3I = 4 −3 = ;
1 2 0 1 4 5
6. 5 4 2 1 14 13
A 3 = 4 A 2 − 3A = 4 − 3 1 2 = 13 14 ;
4 5
14 13 5 4 41 40
A 4 = 4A3 − 3A 2 = 4 − 3 4 5 = 40 41 ;
13 14
41 40 14 13 122 121
A5 = 4A5 − 3A3 = 4 − 3 13 14 = 121 122 .
40 41
2 −1 1 5
31. (a) If A = and B = 3 7 then
−4 3
3 4 1 −6 −25 −34
( A + B )( A − B ) = −7 −4 = −71 −34
−1 10
but
8 −5 16 40 −8 −45
A 2 − B2 = − 24 64 = −44 −51 .
−20 13
(b) If AB = BA then
( A + B )( A − B ) = A ( A − B) + B( A − B)
= A 2 − AB + BA − B 2 = A 2 − B 2 .
2 −1 1 5
32. (a) If A = and B = 3 7 then
−4 3
3 4 3 4 5 52
( A + B)2 = −1 10 = −13 96
−1 10
but
8 −5 2 −1 1 5 16 40
A 2 + 2 AB + B 2 = + 2 −4 3 3 7 + 24 64
−20 13
8 −5 −1 3 16 40 22 41
= + 2 5 1 + 24 64 = 14 79 .
−20 13
(b) If AB = BA then
( A + B )( A − B ) = A ( A − B) + B( A − B)
= A 2 − AB + BA − B 2 = A 2 − B 2 .
33. Four different 2 × 2 matrices A with A2 = I are
1 0 −1 0 1 0 −1 0
0 1 , 0 1 , 0 −1 , and 0 −1 .
7. 1 −1
If A = ≠ 0 then A 2 = (0) A − (0)I = 0.
1 −1
34.
2 −1
35. If A = ≠ 0 then A = (1) A − (0)I = A.
2
2 −1
0 1
If A = ≠ 0 then A 2 = (0) A − ( −1)I = I.
1 0
36.
0 1
37. If A = ≠ 0 then A = (0) A − (1)I = − I.
2
−1 0
0 1 0 1
38. If A = ≠ 0 is the matrix of Problem 36 and B = −1 0 ≠ 0 is the matrix
1 0
of Problem 37, then A + B = (I ) + (−I ) = 0.
2 2
39. If Ax1 = Ax2 = 0, then
A ( c1x1 + c2 x 2 ) = c1 ( Ax1 ) + c2 ( Ax 2 ) = c1 (0 ) + c2 ( 0 ) = 0.
40. (a) If Ax0 = 0 and Ax1 = b, then A ( x 0 + x1 ) = Ax0 + Ax1 = 0 + b = b.
(b) If Ax1 = b and Ax2 = b, then A ( x1 − x 2 ) = Ax1 − Ax 2 = b − b = 0.
41. If AB = BA then
(A + B)
3
(
= ( A + B )( A + B ) = ( A + B ) A 2 + 2AB + B 2
2
)
( ) (
= A A 2 + 2AB + B 2 + B A 2 + 2AB + B 2 )
= ( A + 2A B + AB ) + ( A B + 2AB
3 2 2 2 2
+B 3
)
= A + 3A B + 3AB + B .
3 2 2 3
(A + B) ( A + B ) = ( A + B )( A + B )
4 4 3
To compute , write and proceed similarly,
( A + B ) just obtained.
3
substituting the expansion of
![ 3 3 3 6 9
7. [1 2 3] 4 = [26];
4 1 2 3 = 4 8 12
[ ]
6
6
5 10 15
3 0 3 0 3 0 9
1 0 3 21 15 −1 4 1 0 3 = 7 −20 13
8. 2 −5 4 −1 4 =
; 2 −5 4
6 5 35 0 6 5 16 −25 38
0 −2 4 0 −2
3 1 3 = 7 but the product 3 3 1 is not defined.
9. −2 −2
−4 5 −22 −4 5
2 1 −1 0 4 1 −2 13
10. AB = 3 −2 5 = 5 −6 31 but the product BA is not defined.
4 3
2 7 5 6
11. AB = [3 −5] = [11 1 5 3] but the product BA is not defined.
−1 4 2 3
12. Neither product matrix AB or BA is defined.
3 1 2 5 0 1 3 1 10 17 32 51
13. A(BC) = −3 1 2 3 = −1 4 2 0 = −2 −17
−1 4
3 1 2 5 0 1 3 16 0 1 32 51
( AB)C = −3 1 2 3 = −14 −1 2 3 = −2 −17
−1 4
2 5 6 −13
14. A ( BC ) = [2 −1] −5 = [2 −1] −23 = [−3]
−3 1
2 5 6 6
( AB ) C = [2 −1] −5 = [7 9] −5 = [−3]
−3 1
2 0
3 3 12 15
15. A ( BC ) = [1 −1 2] 0 3 = [4 5] =
2 1 4 2 8 10
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