D2
2
即
D2
∏aidibiciD2
i2


而
D2
a1c1

b1d1
a1d1b1c1
得5aijij
D2
∏aidibici
i1
01
1012L
2101L
3210L
D
detaij
23L
L
1L
2L
3L
4LL
0
11

1
2
3
4L
f11r1r21r2r3L1L1111L
1111L
1111L
1111L
LLL
111c2c1c3c1c4c1L10
LLL

1
2
3
4L000L000L0222L22L02LLLL00L
1
1
12
2

12
32
41a1111a26D
LL11a100L0La2a2
00L00a30L00a3La4LLL00
2
5L
1L1L1c1c2c2c3LLc3c4LL1a
001001
00L00L11L
按最后一列展开由下往上展开由下往上
La
1a
11L0a
1a
a100L0La2a20a30L00a3La4LLL00
0000L
0000La
20
0000L0a

1a
a1a2La
10L00
La
2L0
12
fa1a20L00
0a2a3L00a2
00
LL
000L
000LLa
1a
000L
a3LLL000a3
La
1L0
a20
0L
LL
000L
a30L
a4LLL
000La
1a
1000L0a
1a
a1a2La
1a1a2La
3a
2a
La2a3La
1a1a2La
1∑i1ai
8用克莱姆法则解下列方程组用克莱姆法则解下列方程组
x1x2x3x45x12x2x34x4212x13x2x35x423x1x22x311x4015x16x20x15x26x32x25x36x40x35x46x50x45x51
1
解1D
1
1
1
1
1
1
1
1214012315053121102
233718
13
f11
1
1
11
1
1
0123012314200154001380051400014251115111
D1
05092214231523150121101211
15195190509012110133230509012110133231151915190121101211142001046001380023120000142
151115111214072322150123730211015185111511
D2
1
01320132284002311001190039310002841125124426232531011
D3
1
14
f1D4
1
1
5
121214223123120
∴
x1
D11D
x2
D22D
x3
D33D
x400
D41D
561560015按最后一行5D′2D0156001展开001560000015
55D′′6D′′′6D′′19D′′30D′′′65D′′′114D′′′′65×19114×5665
56000
605D′6D′′5016
′D′为行列式D中a11的余子式D′′为D′中a11的余子式D′′′D′′′′类推
600560056按第一列D101560D′15展开001560110015D′6419D′′′30′′′′6415071610600按第二列05D20056001展开0015600010155100r