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