发布时间 : 星期五 文章工程数学线性代数同济大学第五版课后习题答案更新完毕开始阅读
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
解
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