4阶行列式的运算过程,1)4 1 2 41 2 0 210 5 2 00 1 1 72)a 1 0 0-1 b 1 00

1个回答

  • 1)

    D=|4 1 2 4|

    1 2 0 2

    2 3 -2 -8 r3-r1*2

    0 1 1 7

    = - |1 2 0 2| r1与r2交换

    4 1 2 4

    2 3 -2 -8

    0 1 1 7

    =-|1 2 0 2|

    0 -7 2 -4 r2-r1*4、r3-r1*2

    0 -1 -2 -12

    0 1 1 7

    =|1 2 0 2|

    0 1 1 7 r2与r4交换

    0 -1 -2 -12

    0 -7 2 -4

    =|1 2 0 2|

    0 1 1 7

    0 0 -1 -5 r3+r2、r4+r2*7

    0 0 9 45 【其实,这里就可以得出结果了:r4:r3=-9,D=0】

    =|1 2 0 2|

    0 1 1 7

    0 0 -1 -5

    0 0 0 0 r4+r3*9

    =1*1*(-1)*0=0

    2)按列展开降阶

    D4=a*|b 1 0| + |1 0 0|

    -1 c 1 -1 c 1

    0-1 d 0-1 d

    =ab|c 1|+ a|1 0| + |c 1| 【第三个二阶行列式是前面按第一行展开】

    -1 d -1 d -1 d

    =abcd+ab+ad+0+cd+1

    =abcd+ab+ad+cd+1