Question

计算行列式1 1 2 2,3 -1 -1 1,2 2 1 -1,1 2 3 0

Answer 1

行列式D=
1 1 2 2
3 -1 -1 1
2 2 1 -1
1 2 3 0 r2-3r1,r3-2r1,r4-r1
=
1 1 2 2
0 -4 -7 -5
0 0 -3 -5
0 1 1 -2 r2+4r4
=
1 1 2 2
0 0 -3 -13
0 0 -3 -5
0 1 1 -2 r2-r3，交换r2和r4
=
1 1 2 2
0 1 1 -2
0 0 -3 -5
0 0 0 -8 *(-1)

得到三角形行列式，
所以得到D= 1*1*(-3) *(-8) *(-1) = -24

Answer 2

1.r2-r1,r3-3r1,r4-2r11 1 2 30 1 1 -40 -4 -7 -110 1 -5 -7r3+4r2,r4-r21 1 2 30 1 1 -40 0 -3 -270 0 -6 -3r4-2r31 1 2 30 1 1 -40 0 -3 -270 0 0 51= -3*51 = -153.