设矩阵A=(2,3,-1)(0,-1,1)(0,1,0)求A的-1次方

2 3 -1
0 -1 1
01 0
取左列,只有一个2其他两个是0
detA=2der(-1 1 ;1 0)=2*(0-1)=-2
找adjA
-1 1 0 10 -1
1 0 0 0 01
---------------------------
3 -12 -1 2 3
1 0 00 01
------------------------------
3 -12 -123
-1 1010-1
求9个det + - +
- + -
+ - +
-1 00
-1 0-2
2-2-2
然後转置得adjA
-1 -12
00 -2
0 -2 -2
A^-1=adjA/detA

=0.50.5-1;
00 1;
01 1;
请采纳答案,支持我一下.