帮忙做俩矩阵的初等变换,最好有过程
3个回答

3 -1 4 | 1 0 0

A|I = 1 0 0 | 0 1 0

2 1 -5 | 0 0 1

--> 交换行1,2

1 0 0 | 0 1 0

3 -1 4 | 1 0 0

2 1 -5 | 0 0 1

--> 行2 = 行2 - 行1 * 3,

行3 = 行3 - 行1 * 2

1 0 0 | 0 1 0

0 -1 4 | 1 -3 0

0 1 -5 | 0 -2 1

--> 行3 = 行3 + 行2

1 0 0 | 0 1 0

0 -1 4 | 1 -3 0

0 0 -1 | 1 -5 1

--> 行3 = 行3 * (-1)

1 0 0 | 0 1 0

0 -1 4 | 1 -3 0

0 0 1 | -1 5 -1

--> 行2 = 行2 * (-1)

1 0 0 | 0 1 0

0 1 -4 | -1 3 0

0 0 1 | -1 5 -1

--> 行2 = 行2 + 4 * 行3

1 0 0 | 0 1 0

0 1 0 | -5 23 -4

0 0 1 | -1 5 -1

故逆矩阵为:

0 1 0

-5 23 -4

-1 5 -1

3 6 1 | 1 0 0

B|I = 3 -3 2 | 0 1 0

6 9 2 | 0 0 1

-->行2 = 行2 - 行1,行3 = 行3 - 2 * 行1

3 6 1 | 1 0 0

0 -9 1 | -1 1 0

0 -3 0 | -2 0 1

-->行3 = 行3 * (-1)

3 6 1 | 1 0 0

0 -9 1 | -1 1 0

0 3 0 | 2 0 -1

-->交换 行3,行2

3 6 1 | 1 0 0

0 3 0 | 2 0 -1

0 -9 1 | -1 1 0

-->行1 = 行1 - 2 * 行2,行3 = 行3 + 3 * 行2

3 0 1 | -3 0 2

0 3 0 | 2 0 -1

0 0 1 | 5 1 -3

-->行1 = 行1 - 行3

3 0 0 | -8 -1 5

0 3 0 | 2 0 -1

0 0 1 | 5 1 -3

-->行1 = 行1 * 1/3,行2 = 行2 * 1/3

1 0 0 | -8/3 -1/3 5/3

0 1 0 | 2/3 0 -1/3

0 0 1 | 5 1 -3

逆矩阵为

-8/3 -1/3 5/3

2/3 0 -1/3

5 1 -3