线性代数课后习题答案全解

第一章 行列式

1. 利用对角线法则计算下列三阶行列式: 201

(1)1 4 ;

183201

解 1 4

183

=2×( 4)×3+0×( 1)×( 1)+1×1×8 0×1×3 2×( 1)×8 1×( 4)×( 1) = 24+8+16 4= 4. abc

(2)bca;

cababc

解 bca

cab

=acb+bac+cba bbb aaa ccc =3abc a3 b3 c3.

111

(3)abc;

a2b2c2111

解 abc

a2b2c2

=bc2+ca2+ab2 ac2 ba2 cb2 =(a b)(b c)(c a).

xyx+y

(4)yx+yx.

x+yxyxyx+y

解 yx+yx

x+yxy

=x(x+y)y+yx(x+y)+(x+y)yx y3 (x+y)3 x3 =3xy(x+y) y3 3x2 y x3 y3 x3 = 2(x3+y3).

2. 按自然数从小到大为标准次序, 求下列各排列的逆序数:

(1)1 2 3 4; 解 逆序数为0 (2)4 1 3 2;

解 逆序数为4: 41, 43, 42, 32. (3)3 4 2 1;

解 逆序数为5: 3 2, 3 1, 4 2, 4 1, 2 1. (4)2 4 1 3;

解 逆序数为3: 2 1, 4 1, 4 3. (5)1 3 (2n 1) 2 4 (2n);

n(n 1)

: 解 逆序数为2 3 2 (1个) 5 2, 5 4(2个) 7 2, 7 4, 7 6(3个)

(2n 1)2, (2n 1)4, (2n 1)6, , (2n 1)(2n 2) (n 1个)

(6)1 3 (2n 1) (2n) (2n 2) 2. 解 逆序数为n(n 1) : 3 2(1个) 5 2, 5 4 (2个)

(2n 1)2, (2n 1)4, (2n 1)6, , (2n 1)(2n 2) (n 1个) 4 2(1个) 6 2, 6 4(2个)

(2n)2, (2n)4, (2n)6, , (2n)(2n 2) (n 1个) 3. 写出四阶行列式中含有因子a11a23的项. 解 含因子a11a23的项的一般形式为

( 1)ta11a23a3ra4s,

其中rs是2和4构成的排列, 这种排列共有两个, 即24和42. 所以含因子a11a23的项分别是

( 1)ta11a23a32a44=( 1)1a11a23a32a44= a11a23a32a44, ( 1)ta11a23a34a42=( 1)2a11a23a34a42=a11a23a34a42. 4. 计算下列各行列式:

41 (1)0125120214

2; 07

41 解 1252024c2 c342======10 1

23202 104 1 10

2=122×( 1)4+3 140117c4 7c30010

3 14 =4 110c2+c39910

123 142======c00 2=0.

1+2c31714

2 (2)31 114152203162; 22

42114 解 31 121=c=4 3=c22 1

40 1506

2==32r4 r22215

212036

=====3

2

12212021340 0

r4 r4 ====12

3 1=120

0210320=0. 00

(3) bdabacbf cfcdae

deef

;

解 bdabbf accfcdae deef=adf bbb ccce

ee

=adfbce 1111

1 11 1=4abcdef.

a1 (4) 001b 1001c 100. 1d

a1 解 001b 1001c 10r1+ar201+ab0===== 1b10 1d00a

1c 100 1d

+aba0c3+dc2+abaad

=( 1)( 1)2+1 1c1===== 1c1+cd

0 1d0 10

5. 证明:

abad=abcd+ab+cd+ad+1. =( 1)( 1)3+2+ 11+cd

a2abb2

(1)2aa+b2b=(a b)3;

111 证明

a2abb2c2 c1a2ab a2b2 a2

2aa+b2b=====2ab a2b 2a

00111c3 c11

222

ab ab aab+a=(a b)3 . =(b a)(b a)1 =( 1)2b a2b 2a

3+1

ax+byay+bzaz+bxxyz

(2)ay+bzaz+bxax+by=(a3+b3)yzx;

az+bxax+byay+bzzxy 证明

ax+byay+bzaz+bx

ay+bzaz+bxax+by

az+bxax+byay+bz

xay+bzaz+bxyay+bzaz+bx

=ayaz+bxax+by+bzaz+bxax+by

zax+byay+bzxax+byay+bzxay+bzzyzaz+bx

=a2yaz+bxx+b2zxax+by

zax+byyxyay+bzxyzyzx

=a3yzx+b3zxy

zxyxyzxyzxyz

=a3yzx+b3yzx

zxyzxyxyz

=(a3+b3)yzx.

zxy

a2b2

(3)2

cd2 证明 a2b2

2

cd2

(a+1)2(b+1)2(c+1)2(d+1)2

(a+2)2(b+2)2(c+2)2(d+2)2

(a+3)2(b+3)2

(c c, c c, c c得) (c+3)2433221(d+3)2

(a+1)2(b+1)2(c+1)2(d+1)2

(a+2)2(b+2)2(c+2)2(d+2)2

(a+3)2(b+3)2

=0; (c+3)2(d+3)2

a22b =c2d22a+12b+12c+12d+12a+32b+32c+32d+32a+5

2b+5(c c, c c得) 2c+543322d+5

a22b =c2d2 1a (4)a2a4

1bb2b4

2a+12b+12c+12d+11cc2c4

1d d2d4

22222

2=0. 22

=(a b)(a c)(a d)(b c)(b d)(c d)(a+b+c+d); 证明 1a a2a4

1bb2b4

1cc2c4

1d d2d4

1110b ac ad a

=0b(b a)c(c a)d(d a)

0b2(b2 a2)c2(c2 a2)d2(d2 a2)

111

=(b a)(c a)(d a)b cd

222

(b+a)c(c+a)d(d+a)

11

=(b a)(c a)(d a) c bd b

c(c b)(c+b+a)d(d b)(d+b+a)

1 =(b a)(c a)(d a)(c b)(d b)c(c+1b+a)d(d+b+a) =(a b)(a c)(a d)(b c)(b d)(c d)(a+b+c+d). x

(5)

0an

1x 0an 1

0 1 0an 2

0000

=xn+a1xn 1+ +an 1x+an . x 1a2x+a1

证明 用数学归纳法证明.

x 1=x2+ax+a, 命题成立. 当n=2时, D2=a122x+a1

假设对于(n 1)阶行列式命题成立, 即 Dn 1=xn 1+a1 xn 2+ +an 2x+an 1, 则Dn按第一列展开, 有 1

Dn=xDn 1+an( 1)n+1 x

1

0 1 1 00 x00 1

=xD n 1+an=xn+a1xn 1+ +an 1x+an . 因此, 对于n阶行列式命题成立.

6. 设n阶行列式D=det(aij), 把D上下翻转、或逆时针旋转90°、或依副对角线翻转, 依次得

an1 annann a1na1n ann

D1= , D2= , D3= ,

a11 a1nan1 a11a11 an1

证明D1=D2=( 1)

n(n 1)

2

D, D3=D .

a1nann a2n

证明 因为D=det(aij), 所以 a11

an1 ann

D1= =( 1)n 1an1

a11 a1n

a21

a11a21

=( 1)n 1( 1)n 2an1

a31

a1na2n

ann= a3n

n(n 1)2

=( 1)1+2+ +(n 2)+(n 1)D=( 1) 同理可证 D2=( 1) D3=( 1)

n(n 1)2

D.

a11 an1n(n 1)n(n 1)

=( 1)2DT=( 1)2D. a1n ann

n(n 1)2

D2=( 1)

n(n 1)2

( 1)

n(n 1)2

D=( 1)n(n 1)D=D.

7. 计算下列各行列式(Dk为k阶行列式): (1)Dn=都是0; 解 a0

Dn=0

01

0a0 00

00a 00

000 a0

10

0(按第n行展开) 0a

a1 1a

, 其中对角线上元素都是a, 未写出的元 …… 此处隐藏：4398字，全部文档内容请下载后查看。喜欢就下载吧 ……