Question

利用初等变换求逆矩阵A=3 -2 0 -1;0 2 -3 3;1 1 2 7;0 1 0 3 求大神帮忙

很着急，在线的大神们江湖救急

Answer 1

[A|E] =
3 -2 0 -1 1 0 0 0
0 2 -3 3 0 1 0 0
1 1 2 7 0 0 1 0
0 1 0 3 0 0 0 1

[A|E] =
1 1 2 7 0 0 1 0
0 2 -3 3 0 1 0 0
3 -2 0 -1 1 0 0 0
0 1 0 3 0 0 0 1

[A|E] =
1 1 2 7 0 0 1 0
0 2 -3 3 0 1 0 0
0 -5 -6 -22 1 0 -3 0
0 1 0 3 0 0 0 1

[A|E] =
1 1 2 7 0 0 1 0
0 1 0 3 0 0 0 1
0 -5 -6 -22 1 0 -3 0
0 2 -3 3 0 1 0 0

[A|E] =
1 1 2 7 0 0 1 0
0 1 0 3 0 0 0 1
0 0 -6 -7 1 0 -3 5
0 0 -3 -3 0 1 0 -2

[A|E] =
1 1 2 7 0 0 1 0
0 1 0 3 0 0 0 1
0 0 -6 -7 1 0 -3 5
0 0 18 18 0 -6 0 12

[A|E] =
1 1 2 7 0 0 1 0
0 1 0 3 0 0 0 1
0 0 -6 -7 1 0 -3 5
0 0 0 -3 3 -6 -9 27

[A|E] =
1 1 2 7 0 0 1 0
0 1 0 3 0 0 0 1
0 0 -6 -7 1 0 -3 5
0 0 0 1 -1 2 3 -9

[A|E] =
1 1 2 0 7 -14 -20 63
0 1 0 0 3 -6 -9 28
0 0 -6 0 -6 14 18 -58
0 0 0 1 -1 2 3 -9

[A|E] =
1 1 2 0 7 -14 -20 63
0 1 0 0 3 -6 -9 28
0 0 1 0 1 -7/3 -3 29/3
0 0 0 1 -1 2 3 -9

[A|E] =
1 1 0 0 5 -28/3 -14 131/3
0 1 0 0 3 -6 -9 28
0 0 1 0 1 -7/3 -3 29/3
0 0 0 1 -1 2 3 -9

[A|E] =
1 0 0 0 2 -10/3 -5 47/3
0 1 0 0 3 -6 -9 28
0 0 1 0 1 -7/3 -3 29/3
0 0 0 1 -1 2 3 -9

A^(-1) =
2 -10/3 -5 47/3
3 -6 -9 28
1 -7/3 -3 29/3
-1 2 3 -9