Question

逆矩阵的问题

矩阵B为5 5 -3 -5 3
-6 -5 4 5 -3
4 3 -3 -3 2
2 -4 0 2 -1
-2 -6 2 5 -3
1.用初等变化将B化为最简矩阵，并求其逆矩阵
2.求 ∣B∣

谢谢！！！

Answer 1

解: (B,E)=
5 5 -3 -5 3 1 0 0 0 0
-6 -5 4 5 -3 0 1 0 0 0
4 3 -3 -3 2 0 0 1 0 0
2 -4 0 2 -1 0 0 0 1 0
-2 -6 2 5 -3 0 0 0 0 1

r1+r2
-1 0 1 0 0 1 1 0 0 0
-6 -5 4 5 -3 0 1 0 0 0
4 3 -3 -3 2 0 0 1 0 0
2 -4 0 2 -1 0 0 0 1 0
-2 -6 2 5 -3 0 0 0 0 1

r2-6r1,r3+4r1,r4+2r1,r5-2r1
-1 0 1 0 0 1 1 0 0 0
0 -5 -2 5 -3 -6 -5 0 0 0
0 3 1 -3 2 4 4 1 0 0
0 -4 2 2 -1 2 2 0 1 0
0 -6 0 5 -3 -2 -2 0 0 1

r2+2r3
-1 0 1 0 0 1 1 0 0 0
0 1 0 -1 1 2 3 2 0 0
0 3 1 -3 2 4 4 1 0 0
0 -4 2 2 -1 2 2 0 1 0
0 -6 0 5 -3 -2 -2 0 0 1

r3-3r2,r4+4r2,r5+6r2
-1 0 1 0 0 1 1 0 0 0
0 1 0 -1 1 2 3 2 0 0
0 0 1 0 -1 -2 -5 -5 0 0
0 0 2 -2 3 10 14 8 1 0
0 0 0 -1 3 10 16 12 0 1

r1-r3,r4-2r3
-1 0 0 0 1 3 6 5 0 0
0 1 0 -1 1 2 3 2 0 0
0 0 1 0 -1 -2 -5 -5 0 0
0 0 0 -2 5 14 24 18 1 0
0 0 0 -1 3 10 16 12 0 1

r1*(-1),r5*(-1),r2+r5,r4+2r5
1 0 0 0 -1 -3 -6 -5 0 0
0 1 0 0 -2 -8 -13 -10 0 -1
0 0 1 0 -1 -2 -5 -5 0 0
0 0 0 0 -1 -6 -8 -6 1 -2
0 0 0 1 -3 -10 -16 -12 0 -1

r4*(-1), r1+r4,r2+2r4,r3+r4,r5+3r4
1 0 0 0 0 3 2 1 -1 2
0 1 0 0 0 4 3 2 -2 3
0 0 1 0 0 4 3 1 -1 2
0 0 0 0 1 6 8 6 -1 2
0 0 0 1 0 8 8 6 -3 5

r4<->r5
1 0 0 0 0 3 2 1 -1 2
0 1 0 0 0 4 3 2 -2 3
0 0 1 0 0 4 3 1 -1 2
0 0 0 1 0 8 8 6 -3 5
0 0 0 0 1 6 8 6 -1 2

所以B的最简形即4阶单位矩阵, B^-1 =
3 2 1 -1 2
4 3 2 -2 3
4 3 1 -1 2
8 8 6 -3 5
6 8 6 -1 2

注意到对(B,E)实施初等行变换时, ri+krj 不改变B的行列式的值
而其中 r1*(-1),r5*(-1),r4*(-1), r4<->r5 使行列式变符号
所以 |B|=1.