Question

线性代数求逆矩阵

Answer 1

解: (A,E) =

1 2 3 4 1 0 0 0

2 3 1 2 0 1 0 0

1 1 1 -1 0 0 1 0

1 0 -2 -6 0 0 0 1

r1-r3,r2-2r3,r4-r3

0 1 2 5 1 0 -1 0

0 1 -1 4 0 1 -2 0

1 1 1 -1 0 0 1 0

0 -1 -3 -5 0 0 -1 1

ri-r4,i=1,2,3

0 0 -1 0 1 0 -2 1

0 0 -4 -1 0 1 -3 1

1 0 -2 -6 0 0 0 1

0 -1 -3 -5 0 0 -1 1

r1*(-1),r2+4r1,r3+2r1,r4+3r1

0 0 1 0 -1 0 2 -1

0 0 0 -1 -4 1 5 -3

1 0 0 -6 -2 0 4 -1

0 -1 0 -5 -3 0 5 -2

r2*(-1),r3+6r2,r4+5r2

0 0 1 0 -1 0 2 -1

0 0 0 1 4 -1 -5 3

1 0 0 0 22 -6 -26 17

0 -1 0 0 17 -5 -20 13

r4*(-1), 交换行得

1 0 0 0 22 -6 -26 17

0 1 0 0 -17 5 20 -13

0 0 1 0 -1 0 2 -1

0 0 0 1 4 -1 -5 3

所以 A^-1 =

22 -6 -26 17

-17 5 20 -13

-1 0 2 -1

4 -1 -5 3