6. −6 10 2
= 15 −21 −6
−1 1
38. det A = 6, A
6
12 −18 −6
−21 −1 −13
1
39. det A = 37, A −1
= 4 9 6
37
−6 5 −9
9 12 −13
−1 1
40. det A = 107, A = 11 −21 −4
107
−15 −20 −14
a1
41. If A = and B = [b1 b 2 ] in terms of the two row vectors of A and the two column
a 2
a1b1 a1b 2
vectors of B, then AB = , so
a 2b1 a 2b 2
a b a 2b1 b1 T
T
( AB ) = 1 1 = T a1 a1 = BT AT ,
T T
a1b 2 a 2b 2
b 2
because the rows of A are the columns of AT and the columns of B are the rows of BT.
a b x ax + by ax by
42. det AB = det = = +
c d y cx + dy cx + dy cx + dy
x x y y x y
= ac + ad + bc + bd = ad + bc
x y x y y x
x x x
= ad − bc = (ad − bc) = (det A)(det B)
y y y
8. 46. We expand the left-hand determinant along its first column:
a1 + kb1 b1 c1
a2 + kb2 b2 c2
a3 + kb3 b3 c3
= ( a1 + kb1 )(b2c3 − b3c2 ) − ( a2 + kb2 )(b1c3 − c3b3 ) + ( a3 + kb3 )(b1c2 − b2 c1 )
= a1 (b2 c3 − b3c2 ) − a2 (b1c3 − c3b3 ) + a3 (b1c2 − b2c1 )
+ k b1 (b2 c3 − b3c2 ) − b2 (b1c3 − c3b3 ) + b3 (b1c2 − b2 c1 )
a1 b1 c1 b1 b1 d1 a1 b1 c1
= a2 b2 c2 + k b2 b2 d 2 = a2 b2 c2
a3 b3 c3 b3 b3 d3 a3 b3 c3
47. We illustrate these properties with 2 × 2 matrices A = aij and B = bij .
T
a a21 a a12
(A )
T
T
= 11 = 11 = A
a22 a22
(a)
a12 a22
T
ca ca12 ca ca21 a11 a21
(cA ) = 11 = 11 = c a = cA
T T
(b)
ca21 ca22 ca12 ca22 12 a22
a + b a12 + b12 a11 + b11 a21 + b21
T
(A + B) = 11 11 = a + b
T
(c)
a21 + b21 a22 + b22 12 12 a22 + b22
a a21 b11 b21
= 11 + b b = A + B
T T
a12 a22 12 22
48. The ijth element of ( AB)T is the jith element of AB, and hence is the product of the jth
row of A and the ith column of B. The ijth element of BT AT is the product of the ith row
of BT and the jth column of AT . Because transposition of a matrix changes the ith row to
the ith column and vice versa, it follows that the ijth element of BT AT is the product of the
jth row of A and the ith column of B. Thus the matrices ( AB)T and BT AT have the
same ijth elements, and hence are equal matrices.
a1 b1 c1 a1 a2 a3
49. If we write A = a2 b2 c2 and A = b1
T
b2 b3 , then expansion of A along its
a3 b3 c3 c1 c2 c3
first row and of AT along its first column both give the result
a1 (b2 c3 − b3c2 ) + b1 ( a2 c3 − a3c2 ) + c1 ( a2b3 − a3b2 ).
9. If A 2 = A then A = A , so A − A = A ( A − 1) = 0, and hence it follows
2 2
50.
immediately that either A = 0 or A = 1.
If A n = 0 then A = 0, so it follows immediately that A = 0.
n
51.
−1
If AT = A −1 then A = AT = A −1 = A . Hence A
2
52. = 1, so it follows that A = ±1.
−1
53. If A = P −1BP then A = P −1BP = P −1 B P = P B P = B.
54. If A and B are invertible, then A ≠ 0 and B ≠ 0. Hence AB = A B ≠ 0, so it
follows that AB is invertible. Conversely, AB invertible implies that AB = A B ≠ 0,
so it follows that both A ≠ 0 and B ≠ 0, and therefore that both A and B are
invertible.
55. If either AB = I or BA = I is given, then it follows from Problem 54 that A and B are
both invertible because their product (one way or the other) is invertible. Hence A–1 exists.
So if (for instance) it is AB = I that is given, then multiplication by A–1 on the right
yields B = A −1.
56. The matrix A–1 in part (a) and the solution vector x in part (b) have only integer entries
because the only division involved in their calculation — using the adjoint formula for the
inverse matrix or Cramer's rule for the solution vector — is by the determinant A = 1.
a d f bc −cd de − bf
If A = 0 b e then A −1 =
1
57. 0 ac −ae .
abc
0 0
b 0
0 ab
58. The coefficient determinant of the linear system
c cos B + b cos C = a
c cos A + a cos C = b
b cos A + a cos B = c
in the unknowns {cos A, cos B, cos C} is
10. 0 c b
c 0 a = 2abc.
b a 0
Hence Cramer's rule gives
a c b
1 ab 2 − a 3 + ac 2 b2 − a 2 + c2
cos A = b 0 a = = ,
2abc 2abc 2bc
c a 0
whence a 2 = b 2 + c 2 − 2bc cos A.
59. These are almost immediate computations.
60. (a) In the 4 × 4 case, expansion along the first row gives
2 1 0 0
2 1 0 1 1 0 2 1 0
1 2 1 0 2 1
= 21 2 1 −0 2 1 = 21 2 1− ,
0 1 2 1 1 2
0 1 2 0 1 2 0 1 2
0 0 1 2
so B4 = 2 B3 − B2 = 2(4) − (3) = 5. The general recursion formula Bn = 2 Bn −1 − Bn − 2
results in the same way upon expansion along the first row.
(b) If we assume inductively that
Bn −1 = (n − 1) + 1 = n and Bn − 2 = (n − 2) + 1 = n − 1,
then the recursion formula of part (a) yields
Bn = 2 Bn −1 − Bn − 2 = 2(n ) − (n − 1) = n + 1.
61. Subtraction of the first row from both the second and the third row gives
1 a a2 1 a a2
1 b b 2 = 0 b − a b 2 − a 2 = (b − a)(c 2 − a 2 ) − (c − a )(b 2 − a 2 )
1 c c2 0 c − a c2 − a2
= (b − a)(c − a)(c + a) − (c − a)(b − a)(b + a)
= (b − a)(c − a) [(c + a) − (b + a)] = (b − a)(c − a)(c − b).
11. 62. Expansion of the 4 × 4 determinant defining P ( y ) along its 4th row yields
1 x1 x12
P( y ) = y 1 x2
3
x2 + = y 3V ( x1 , x2 , x3 ) + lower-degree terms in y.
2
2
1 x3 x3
Because it is clear from the determinant definition of P ( y ) that
P( x1 ) = P( x2 ) = P( x3 ) = 0, the three roots of the cubic polynomial P( y ) are x1 , x2 , x3 .
The factor theorem therefore says that P( y ) = k ( y − x1 )( y − x2 )( y − x3 ) for some constant
k, and the calculation above implies that
k = V ( x1 , x2 , x3 ) = ( x3 − x1 )( x3 − x2 )( x2 − x1 ).
Finally we see that
V ( x1 , x2 , x3 , x4 ) = P( x4 ) = V ( x1 , x2 , x3 ) ⋅ ( x4 − x1 )( x4 − x2 )( x4 − x1 )
= ( x4 − x1 )( x4 − x2 )( x4 − x1 )( x3 − x1 )( x3 − x2 )( x2 − x1 ),
which is the desired formula for V ( x1 , x2 , x3 , x4 ) .
63. The same argument as in Problem 62 yields
P( y ) = V ( x1 , x2 , , xn −1 ) ⋅ ( y − x1 )( y − x2 ) ⋅ ⋅ ( y − xn −1 ).
Therefore
V ( x1 , x2 , , xn ) = ( xn − x1 )( xn − x2 ) ⋅ ⋅ ( xn − xn −1 )V ( x1 , x2 , , xn−1 )
n −1
= ( xn − x1 )( xn − x2 ) ⋅ ⋅ ( xn − xn −1 ) ∏ ( xi − x j )
i j
n
= ∏ ( x − x ).
i j
i j
64. (a) V(1, 2, 3, 4) = (4 – 1)(4 – 2)(4 – 3)(3 – 1)(3 – 2)(2 – 1) = 12
(b) V(–1, 2,–2, 3) =
3 − ( −1) [3 − 2] 3 − ( −2 ) ( −2 ) − ( −1) ( −2 ) − 2 2 − ( −1) = 240
