Question

利用矩阵的初等变换，求方阵的逆阵 3 -2 0 -1 0 2 2 1 1 -2 -3 -2 0 1 2 1

要有详细过程

Answer 1

3 -2 0 -1 1 0 0 0
0 2 2 1 0 1 0 0
1 -2 -3 -2 0 0 1 0
0 1 2 1 0 0 0 1

r1-3r3
0 4 9 5 1 0 -3 0
0 2 2 1 0 1 0 0
1 -2 -3 -2 0 0 1 0
0 1 2 1 0 0 0 1

r1-2r2, r3+r2,r2-2r4
0 0 5 3 1 -2 -3 0
0 0 -2 -1 0 1 0 -2
1 0 -1 -1 0 1 1 0
0 1 2 1 0 0 0 1

r1+2r2, r4+r2
0 0 1 1 1 0 -3 -4
0 0 -2 -1 0 1 0 -2
1 0 -1 -1 0 1 1 0
0 1 0 0 0 1 0 -1

r2+2r1, r3+r1
0 0 1 1 1 0 -3 -4
0 0 0 1 2 1 -6 -10
1 0 0 0 1 1 -2 -4
0 1 0 0 0 1 0 -1

r1-r2
0 0 1 0 -1 -1 3 6
0 0 0 1 2 1 -6 -10
1 0 0 0 1 1 -2 -4
0 1 0 0 0 1 0 -1

交换行得
1 0 0 0 1 1 -2 -4
0 1 0 0 0 1 0 -1
0 0 1 0 -1 -1 3 6
0 0 0 1 2 1 -6 -10

满意请采纳 ^_^

Answer 2

直接用matlab 编程 a=[3 -2 0 -1;0 2 2 1; 1 -2 -3 -2;0 1 2 1];
a*b=ones(1,4);
b