哈工大版线性代数与空间解析几何课后题答案(3)

0102

( 1)t(p1p2 pn)a1p1a2p2 anpn (2) 0

n 1

n0

( 1)

t(23 n1)

a12a23 a(n 1)nan1

( 1)n 1 1 2 3 n ( 1)n 1n! 5．用行列式的定义证明：

a11a21

(1) 0

a12a22000a12a22a32a42

a13a2300000a33a44

a14a24a34a44a5400a34a45

a15a25

a35 0；

a45a55a11a21

a12a33

a22a43

a34a44

00a11

(2)

a21a31a41

.

a11a21

证：(1) D 0

a12a22000

a13a23000

a14a24a34a44a54

a15a25

a35 ( 1)t(p1p2p3p4p5)a1p1a2p2a3p3a4p4a5p5 a45a55

00

假设有a1P1a2P2a3P3a4P4a5P5 0，由已知p3,p4,p5必等于4或5，从而p3,p4,p5中至少有两个相等，这与p1,p2,p3,p4,p5是1,2,3,4,5的一个全排列矛盾，故所有项

a1P1a2P2a3P3a4P4a5P5 0，因此D 0.

a11

(2)

a12a22a32a42

00a33a43

00a34a44

( 1)t(p1p2p3p4)a1p1a2p2a3p3a4p4，由已知，只有当p1,p2

a21a31a41

取1或2时，a1p1a2p2a3p3a4p4 0，而p1,p2,p3,p4是1,2,3,4的一个全排列，故p3,p4取

3或4，于是

·3·