Question

行列式 4 1 2 4 1 2 0 2 10 5 2 0 0 1 1 7 怎么做好难

四阶行列式

Answer 1

r1-2r4,r3-2r4
4 -1 0 -10
1 2 0 2
10 3 0 -14
0 1 1 7
按第3列展开 = (-1)*
4 -1 -10
1 2 2
10 3 -14
c3-c2,c2-2c1
4 -9 -9
1 0 0
10 -17 -17
2,3列相同,行列式等于0

Answer 2

你好很高兴回答你的问题：
先初等行变换，再降成3阶，用对角线法则算
解:
r3-2r1-2r2, r1-4r2
0 -7 2 -4
1 2 0 2
0 -1 -2 -12
0 1 1 7

r1<->r2
1 2 0 2
0 -7 2 -4
0 -1 -2 -12
0 1 1 7

r2-7r3,r4+r3
1 2 0 2
0 0 16 80
0 -1 -2 -12
0 0 -1 -5

r2+16r4
1 2 0 2
0 0 0 0
0 -1 -2 -12
0 0 -1 -5

行列式 = 0

Answer 3

r1-2r4,r3-2r4
4 -1 0 -10
1 2 0 2
10 3 0 -14
0 1 1 7
按第3列展开 = (-1)*
4 -1 -10
1 2 2
10 3 -14
c3-c2,c2-2c1
4 -9 -9
1 0 0
10 -17 -17
2,3列相同,行列式等于0望采纳~~~^_^