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.