线性代数第五版答案(全)(2)

00

0a1 1

1

0a2

1

0a3

1

1an 1

1

11+an

100

=a1a2 an

0010 0001 0 000 0000 1

a1 1 1a2 1a3 1an 1

ni=1

000 001+∑ai 1

=(a1a2 an)(1+∑1.

i=1ai

8.用克莱姆法则解下列方程组: x1+x2+x3+x4=5 x+2x2 x3+4x4= 2(1) 1;

2x1 3x2 x3 5x4= 2 3x+x+2x+11x=0 1234解因为

1D=12

3

12 31

1 1 12

1

4= 142, 511

14= 142,D=2 52

113

5

2 20

1 1 12

1

4= 284, 511

n

52D1= 20D3=2

3

所以

12 311 1 12

12 315 2 20114= 426,D=1

42 5

1131

2 311 1 125

2=142, 20

x1=

DDDD

=1,x2==2,x3==3,x4== 1.DDDD

=1 5x1+6x2

=0 x1+5x2+6x3

(2) x2+5x3+6x4=0.

x3+5x4+6x5=0

x4+5x5=1 解因为

5

1D=0

0010D1=0

01D3=5D5=0

00

所以

65100651006510065100065100651010001065100065100651006510065100

0=665,65

05010=1507,D2=0605005010=703,D4=0605010

0=212,01

100016510006510065100065110001

00

0= 1145,6500

0= 395,65

x1=1507,x2= 1145,x3=703,x4= 395,x4=212.

665665665665665

λx1+x2+x3=0

9.问λ,µ取何值时,齐次线性方程组 x1+µx2+x3=0有非

x1+2µx2+x3=0零解？

解系数行列式为

λ1D=1µ=µ µλ.

12µ令D=0,得

µ=0或λ=1.

于是,当µ=0或λ=1时该齐次线性方程组有非零解.

(1 λ)x1 2x2+4x3=0

10.问λ取何值时,齐次线性方程组 2x1+(3 λ)x2+x3=0有

x1+x2+(1 λ)x3=0非零解？

解系数行列式为

λ 24 λ 3+λ4

D=23 λ1=21 λ1

111 λ101 λ=(1 λ)3+(λ 3) 4(1 λ) 2(1 λ)( 3 λ)=(1 λ)3+2(1 λ)2+λ 3.令D=0,得

λ=0,λ=2或λ=3.

于是,当λ=0,λ=2或λ=3时,该齐次线性方程组有非零解.

第二章

1.已知线性变换:

矩阵及其运算

x1=2y1+2y2+y3 x2=3y1+y2+5y3, x3=3y1+2y2+3y3

求从变量x1,x2,x3到变量y1,y2,y3的线性变换.

解由已知:

x1 221 y1

x = 315 y , x2 323 y2

2 3

故

1

y 1 221 x1 7 49 y1 y = 315 x = 63 7 y , y2 323 x2 32 4 y2

3 3 2

y1= 7x1 4x2+9x3

y2=6x1+3x2 7x3. y3=3x1+2x2 4x3

2.已知两个线性变换

x1=2y1+y3

x2= 2y1+3y2+2y3, x3=4y1+y2+5y3

解由已知

x1 201 y1 201 31 x = 232 y = 232 20 x2 415 y2 415 0 1

2 3

0 z1

1 z2 3 z3

y1= 3z1+z2

y2=2z1+z3, y3= z2+3z3

求从z1,z2,z3到x1,x2,x3的线性变换.

613 z1 = 12 49 z2 , 10 116 z 3

x1= 6z1+z2+3z3

所以有 x2=12z1 4z2+9z3.

x3= 10z1 z2+16z3

111 123

3.设A= 11 1 ,B= 1 24 ,求3AB 2A及ATB.

1 11 051 解

111 123 111 3AB 2A=3 11 1 1 24 2 11 1

1 11 051 1 11 058 111 21322

=3 0 56 2 11 1 = 2 1720 , 290 1 11 429 2 111 123 058 AB= 11 1 1 24 = 0 56 .

1 11 051 290

T

4.计算下列乘积: 431 7

(1) 1 23 2 ; 570 1 解

431 7 4×7+3×2+1×1 35 1 23 2 = 1×7+( 2)×2+3×1 = 6 . 570 1 5×7+7×2+0×1 49

3 (2)(123) 2 ;

1

解

(123) 3 2

=(1×3+2×2+3×1)=(10) 1 .

(3) 2 1

( 12); 3 解

2 4 1

2×( 1)2×23 ( 12)= 2 13××(( 11))13××2 = 12

2

36 .

(4) 2 1

3

11 1 114304 0 1 2 40

31 ; 2

31 2 1

解

1 114304 02

= 6 78 .

1 10

3 4

1 2

20 5 6 (5)(x a11a12a13 x1

1x2x3) a12a22a23 a x2 ;

13a23a33 x3 解

x a11a12a13 x(1

1x2x3) a a12a13a22a23a23 x2

33 x3 =(a11x1+a12x2+a13x3

a12x1+a22x2 x+a1

13x123x2+a33x3) x2

x 3

=a11x12+a22x22+a33x32

+2a12x1x2+2a13x1x3+2a23x2x3.

+a23x3

a