用初等变换求A=(1 -1 2,-1 1 0 ,2 1 2 )的逆矩阵

解：
1 -1 2 1 0 0
（A，E）= -1 1 0 0 1 0
2 1 2 0 0 1

1 -1 2 1 0 0
r2+r1得 0 0 2 1 1 0
2 1 2 0 0 1

1 -1 2 1 0 0
r3-2r1得 0 0 2 1 1 0
0 3 -2 -2 0 1

1 -1 2 1 0 0
r2<->r3得 0 3 -2 -2 0 1
0 0 2 1 1 0

1 -1 2 1 0 0
1/3·r2 0 1 -2/3 -2/3 0 1/3
1/2·r3得 0 0 1 1/2 1/2 0

1 0 4/3 1/3 0 1/3
r1+r2得 0 1 -2/3 -2/3 0 1/3
0 0 1 1/2 1/2 0

1 0 0 -1/3 -2/3 1/3
r1-4/3·r3得 0 1 -2/3 -2/3 0 1/3
0 0 1 1/2 1/2 0

1 0 0 -1/3 -2/3 1/3
r2+2/3·r3得 0 1 0 -1/3 1/3 1/3
0 0 1 1/2 1/2 0

-1/3 -2/3 1/3
所以逆矩阵为：-1/3 1/3 1/3
1/2 1/2 0