# 线性代数ch2习题解答1(肖蓬)

2 1/ 2 1/ 2 0 0 E1 (1/ 2) (1/ 2) E12 (1/ 2) E13 1/ 2 2 0 1/ 2 3 ( A, b) = → 1/ 2 0 2 1/ 2 3 0 1/ 2 1/ 2 2 0 0 0 1 1/ 4 1/ 4 0 0 E2 (4/7) 1 1/ 4 1/ 4 E23 (1/8) 1/14 2/ 7 12/ 7 0 7/ 4 1/8 1/ 2 3 E24 (1/ 2) 0 1 → 0 1/8 7/ 4 1/ 2 3 0 0 0 1/ 2 1/ 2 2 0 0 0

2,解: ,

1 -2 0 1 ( A, b) = 1 3 0 7 1 -2 3 4 0 1 1 1 0 5 3 1 0 7 3 1
E3 ( 1/ 2) E34 (4)

3 4 1 1

4 -3 E13 ( 1) → 0 3 1 3 1 3 4 4 1 - 2 3 4 E23 ( 5) -3 E24 (7) 0 1 1 1 -3 → 0 0 2 4 12 3 0 0 4 8 24 3 8 1 0 1 3 0 1 2 6 0 0 0 0 0 0 0

1 -2 3 4 4 E21 (2) 1 E31 ( 1) 0 1 1 1 -3 E32 (1) 0 → → 0 0 1 2 6 0 0 0 0 0 0 0

T

3,解: ,

2 3 ( A, b) = 1 7 1 2 0 2 0 1 0 5 1 E34 E3 (1/ 9) 0 → 0 0

3 E13 ( 3) 12 E13 (( 2) -2 E14 7) → 2 8 1 8 9 1 8 0 8 1 8 E2 ( 1/ 2) 1 E23 (1) 22 6 26 E24 (5) 0 → 0 11 3 13 0 55 15 56 2 8 1 8 E21 ( 2) 1 E31 (18) 1 11 3 13 E32 ( 13) 0 → 0 0 0 0 1 0 0 0 0 0 3 5 4 2 1 3

E

2 1 0 0 0 1 0 0

8 11 3 13 0 0 0 0 0 9 14 5 0 11 3 0 0 0 1 0 0 0 8

1

5,解: ,

p 1 1

1 p 1

1 1 p

E 1 13 → 1 p2 p 1 p p2

1 p 1

p 1 1

p 2 E12 ( 1) E13 ( p ) p → 1

1 1 p 0 p 1 1 p 0 1 p 1 p2 1 0 0 1 p

E23 (1) 2 p p → 1 p3 p2

2 p 1 1 p p p 0 2 p 2 p 1 p3 p 2 + p

1 1 = 0 p 1 0 0

p 1 p

( 2 + p )(1 p )
1 1 1 1 0 0 0 0 0 0 0 0,

2 (1 + p ) (1 p ) p2 p p2

x = (1 k1 k 2, k1, k 2 )

T

=-2 线性方程组的增广矩阵为: 当 P=- 时,线性方程组的增广矩阵为: =-

1 1 p p2 0 3 3 6 0 0 0 3 ,

p ≠ 1 且 p ≠ 2 时,线性方程组解唯一.解为: 线性方程组解唯一.解为:
2

x3

(1+ p) =
2+ p

p (1 p)( 2 + p) (1 p)(1+ p) (1+ p) p ( 2 + p) x2 = = 2+ p ( p 1)( 2 + p)
2 2

1 = 2+ p p (1+ p) 1 2 p2 + p3 p 2 p2 p3 1 2 x1 = p = 2+ p 2+ p 2+ p p +1 = 2+ p
2

p + 1 1 ( p + 1) x = , , p+2 p+2 p+2

2

T

.

AB BA = 3 1 1 1 1 2 1 2 2 1 1 2 3 1 0 6 2 4 6 = 6 1 4 4 8 1 4 4 1 1 1 0 2 1 1 1 0 4 6 0 1 0 = 2 3 4 4 1 3 1 1 0 2 1 2 1 1 2 3 2 2 0 4 4 0

16,解: ,

a11 ( x1 , x 2 ) a 21

a12 x1 a 22 x 2

x1 = ( a11 x1 + a 21 x 2, a12 x1 + a 22 x 2 ) x2 = ( a11 x1 + a 21 x 2 ) x1 + ( a12 x1 + a 22 x 2 ) x 2
2 = a11 x12 + ( a12 + a 21 ) x1 x 2 + a 22 x 2

21,解: ,

A = P ∧ Q, A 8 = ( P ∧ Q ) = P ∧ 8 Q
8

2 = 1 2 = 1 7 = 4

31 20
9

0 2 1 1 0 2 1 1

3 = I2 2 3 2 = 2 1 3 2 2 1 3 2

A9 = ( P ∧ Q ) = P ∧ 9Q 31 20 12 7

A2n = ( P∧Q ) A
2 n +1

2n

= P ∧ 2nQ = I 2 = P∧
2 n +1

= ( P∧Q )

2 n +1

7 Q= 4

12 7

22,解: ,

a

b

c

n

an =

( 1)

n

bn

cn

23,解: ,

cos sin

sin cos 2 = cos sin 2
2 k

sin 2 cos 2 sin k cos k sin cos

cos 设n=k时, sin 当n=k+1时,
k +1

sin cos k = cos sin k

sin cos cos k = sin cos sin k cos k cos sin k sin = sin k cos cos k sin sin ( k + 1) cos ( k + 1) = sin ( k + 1) cos ( k + 1)

sin k cos cos k sin

cos k sin + sin k cos cos k cos sin k sin

cos sin

sin cos n = cos sin n
n

sin n cos n

n 为任意自然数. 为任意自然数. 注意: 注意:本题计算过程用到三角函数恒等式. 本题计算过程用到三角函数恒等式.可以应用数学归 纳法进行矩阵的计算. 纳法进行矩阵的计算.

24.解: 解 (1)当 )
3

( A + B) = ( A + B) ( A + B) = ( A2 + B2 + 2AB) ( A + B)
2

AB = BA 时,等式成立.事实上, 等式成立.事实上,

= A3 + 3A2 B + 3AB2 + B3 等式不成立. 当 AB ≠ BA 时,等式不成立.事实上取
1 1 1 1 A= ,B = 1 1 1 1 ,则 AB ≠ BA ,

2 0 4 0 2 A+ B = ( ,A + B ) = , 2 0 4 0 8 0 3 ( A + B) = 8 0

0 0 A = =0, A3 = 3 A 2 B = 0, 0 0 2 2 2 B = = 2 B, B 3 = B 2 B = 2 B 2 = 4 B, 2 2 0 0 2 AB = = 0, 0 0
2

A3 + 3 A 2 B + 3 AB 2 + B 3 = 4 B ≠ ( A + B )

3

(2)当 )

AB = BA 时,等式成立.事实上, 等式成立.事实上,

( A + B)( A B) = A2 AB + BA B2 = A2 B2 .

AB ≠ BA 时,等式不成立.事实上取 等式不成立.

1 1 1 1 A= ,B = 1 1 1 1 ,则 AB ≠ BA ,

0 0 2 , A B = , 0 0 2 0 4 ( A + B )( A B ) = , 0 4 A 2 B 2 = 0 2 B = 2 B ≠ ( A + B )( A B )

2 A+ B = 2

25.证明:设 .证明:

A ∈ F m× n, B ∈ F u ×v,C ∈ F s×t , ∵ AB, BA, AC, CA有意义,

∴ n = u, v = m, n = s, t = m, ∴ n = u = s, v = m = t, ∵ AB ∈ F m×v, BA ∈ F u × n, AC ∈ F m×t, CA ∈ F s× n, AB = BA, AC = CA, ∴ m = u, v = n, m = s, t = n, ∴m = n = u = v = s = t

A( B + C) = AB + AC = BA + CA = ( B + C) A , A( BC) = ( AB) C = ( BA) C = B ( AC) = B( CA) = ( BC) A

26.解:设所求的矩阵是 解

x y A= s t ,

x2 + ys xy + yt A2 = =0 2 sx + ts sy + t x2 + ys = 0......(1) xy + yt = 0......(2) ∴ sx + ts = 0......(3) sy + t 2 = 0......(4)

x2 = t 2 ,

y( x + t) = s( x + t) = 0.

= 0 ,即 x = t 时, sy = t

2

,那么

A2 = 0 ,

x y A= , sy = x2 . s x 如果 x + t ≠ 0 ,则 s = y = 0 ,由(1)和(4)得 ) ) x2 = t 2 = 0 ,这与 x + t ≠ 0 矛盾,所以这种情况是不可能的. 矛盾,所以这种情况是不可能的.

27.解:设 解

x y 1 1 1 1 x y = s t 0 1 0 1 s t ,

x x + y x + s y +t = s s +t s t

s = 0,x = t,

1 1 x y 于是得与 0 1 可交换的矩阵形如 0 x ,x,y 为任意常

36.证明: 证明: 证明 (1) ) 由于 ( A + A )
T T T

= A +( A
T T T

T T

)

= AT + A = A + AT ,

T

T T T

= A ( A

)

= AT A = ( A AT ) ,所

1 1 A + AT ) + ( A AT ) ( 2 2 1 1 T ( A+ A ) 是对称矩阵, 2 ( A AT ) 是反对称 是对称矩阵, 由 (1)知, 2 ) A=

37.证明: 证明: 证明 必要性, 都是对称的, 必要性,已知矩阵 A,B,AB 都是对称的,则 , ,

( AB) = AB = AT BT = ( BA)
T

T

,

( AB)

T

= BT AT = BA = AB ,

38.证明:设矩阵 证明: 证明

A = ( aij ) ∈ Rn×n,aij = aji,i, j = 1,2,..., n .

2

= 0 ,所以 A2 的对角元全为零.即 的对角元全为零.

2 ai2 + ai22 + ... + ain = 0,i = 1,2,.., n 1

= 0,i, j = 1,2,..., n .即 A= 0 .

39. )解: . (1) ( 是对称的. 因为 ( A ) = ( A ) = A ,所以 A 是对称的. 因为 ( A ) = ( A ) = ( A) = ( 1) A , 所以当
k T T k k
k k T T k k k k

k 为奇数时, 为奇数时,

Ak 是反对称的.当 k 为偶数时, Ak 不是反对称 是反对称的. 为偶数时,

( AB + BA)

T

= ( AB ) + ( BA)
T

T

= BT AT + AT BT = BA AB = ( AB + BA)

40.(3)解: ( )
2) 1 2 2 1 E12 ((2) 1 2 2 1 E13 2 1 -2 1 → 0 3 -6 -2 1 2 -2 1 1 1 0 -6 -3 -2

1 E23 ( 2) → 0 0 E2 ( 1/3) 1 E21 ( 2) → 0 0

2 0

2 9

1 2

3 -6 -2 0 0 1/9 1 0 2/9 0 1 2/9
1

1 2 0 5/9 4/9 -2/9 1 → 0 3 0 -6/9 -3/9 6/9 0 0 1 2/9 -2/9 1/9 -2 1 2/9 2/9 1/9 -2/9 -2/9 1/9

E3 (1/9) E32 (6) E31 ( 2)

1 2 2 1 2 2 1 2 1 2 = 2 1 2 9 2 2 1 2 2 1

A11 A12 A* = A1, n 1 A 1,n

A21 A22 A2, n 1 A2 n

... ...

An 1,1 An 1,2 ... An 1, n 1 ... An 1,n

An1 An 2 An ,n 1 Ann