工程数学线性代数同济大学第五版课后习题答案 下载本文

ax by ay bz az bx ay bz az bx ax by az bx ax by ay bz x ay bz az bx y ay bz az bx a y az bx ax by b z az bx ax by z ax by ay bz x ax by ay bz

x ay bz z y z az bx a2 y az bx x b2 z x ax by z ax by y x y ay bz x y z y z x a3 y z x b3 z x y z x y x y z x y z x y z a3 y z x b3 y z x z x y z x y (a3

x y z b3) y z x z x y 1)2 1)2 1)2 1)2

(a (b (c (d

2

a2 (a 2 b(3) 2(b c (c d 2 (d 证明

2)2 2)2 2)2 2)2 (a (b (c (d 3)2

3)2 0 ; 3)2 3)2 3)

3)

(c4 c3 c3 c2 c2 c1 得) 3)3)

2 2 2 2

ab2 c2 d 2

2 (a (b (c (d

2

1) 1)2 1)2 1)2

(a

(b (c (d

2) 22) 2 2) 22)

2

(a

(b (c (d

a2a b2 2b c2 2c d 2 2d a2 b2c2 d 2

2a 2b 2c 2d

1 2a 1 2b 1 2c 1 2d

3 2a 3 2b 3 2c 3 2d

5 5 (c c c c 得) 5 4 3 3 2 5 1 2 2 1 2 2 01 2 2 1 2 2

1 1 1 1 a b c d (4) a2222 b c d a4 b4 c4 d 4 (a b)(a c)(a d)(b c)(b d)(c d)(a b c d); 证明

1 1 a b a2 b2 a4 b4

1 1 1 1 0 b a c a d a 0 b(b a) c(c a) d (d a) 0 b2(b2 a2 ) c2(c2 a2) d 2(d 2 a2) 1 1 1

d c (b a)(c a)(d a) 2b 22

b(b a) c(c a) d (d a) 1 1 1

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

0 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) (5) x 1 0

0 x 1

1 1 c d c2 d 2 c4 d 4

0 0 0 0

0 0 0 an an 1 an 2 x 1 a2 x a1 xn a1xn 1

an 1x an

证明 用数学归纳法证明 当 n 2 时 D2

x 1 x 2

a1x a2 命题成立 a2 x a1

假设对于(n 1)阶行列式命题成立 即

Dn 1 xn 1 a1 xn 2

则 Dn 按第一列展开 有

an 2x an 1

Dn xDn 1 an ( 1)n 1

1 0 x 1 1

1

0 0 x

0 0 1

xD n 1 an x n 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

D1

an1 a11

ann

a1n

n(n 1) 2

D2

a1n a11

ann

an1 D3

ann an1

a1n

a11

证明 D1 D( 1) 2

D D3 D

证明 因为 D det(aij) 所以

D1

an1 a11

ann

a1n

a11

( 1)n 1 an1

a1n ann

a21 a2n a11 a21

n 1

( 1)n 2 an1 ( 1)

a1n

a2n ann

a31

a3n

( 1)同理可证

D2

1 2

(n 2) (n 1)

D ( 1)

n(n 1) 2

D

T

n(n 1) a11 ( 1) 2 a1n

n(n 1) 2

an1

ann

( 1)

n(n 1) 2

D ( 1)

n(n 1) 2

D

D3 ( 1)

D2 ( 1) n(n 1) 2

( 1) n(n 1)

2

n(n 1)

D ( 1) D D

7 计算下列各行列式(Dk 为 k 阶行列式)

(1) Dn

是 0

a 1

1 , 其中对角线上元素都是 a 未写出的元素都 a Dn

a 0 0

0 a 0 0 0 a 0 0 0 1 0 0

0 1 0 0

0 0 (按第 n 行展开)

a 0 0 a

0 0 0 a 0 0 ( 1)n 1 0 a 0

0 0 0

a

( 1)n 1 ( 1)n

0 1 0 0 0 0 0 (n 1) (n 1)

a

( 1)2n a

a

a (n 1) (n 1)

a (n 2)(n 2)

a a ; x

an an an 2 an 2(a2 1)

(2) Dn

x a a x a a

解 将第一行乘( 1)分别加到其余各行 得

Dn

x a a a x x a 0 a x 0 x a

a 0 0

a x 0

0 0 x a

再将各列都加到第一列上 得

x (n 1)a a a

0 x a 0 0 0 x a

Dn

a

0 0 0 x a

[x (n 1)a](x a)n 1

0 0 0