Question

设矩阵A=(2,3,-1)(0,-1,1)(0,1,0)求A的-1次方

Answer 1

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

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

-1 1 0 1 0 -1

求9个det + - +
- + -
+ - +
-1 0 0
-1 0 -2
2 -2 -2
然後转置得adjA
-1 -1 2
0 0 -2
0 -2 -2

A^-1=adjA/detA

=0.5 0.5 -1;
0 0 1;
0 1 1;
请采纳答案，支持我一下。