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