ab
(4) bd
ac cdcf
ae ef
11
1 11
1 1
111
02 200
de abcdef11 abcdef0
bf
111
abcdef0
20 4abcdef. 02
x
3.已知3
yz
02 1,利用行列式性质求下列行列式. 11y3yy 2x
y3yy 201
z
3z 2; (2) z 2
zz 2
xxyz
3z 2 30
yz
x 1y 1z 134x
y01z
2. 3
1x
(1) 3x 3
x 2
解:(1) 3x 32 2302 2.
111111413111111
x 2
(2)
222
x 1y 1z 134
3
2 302 302
413x
yz
302 302 1 0 1. 11
1
4.用行列式定义计算:
010 0
2002 0
(1) ; (2) . 3
4000 n 15n00 0
12
( 1)t(p1p2p3p4p5)a1p1a2p2a3p3a4p4a5p5 解:(1) 3
45
( 1)t(54321)a15a24a33a42a51 ( 1)10 1 2 3 4 5 120.
·2·
1