Question

【线性代数】已知矩阵X与A满足关系式，AX=A+X，求X

请问答案是多少？过程如何？

Answer 1

Answer 2

AX=A+X
(A-E)X=A
| 0 2 1 |
| 3 3 2 |=0+4+6-3-0-6=1≠0
| 1 2 1 |
∴ X=(A-E)^(-1)A
[ 0 2 1 1 2 1 ]
[ 3 3 2 3 4 2 ] →
[ 1 2 1 1 2 2 ]
-------------------------------------------
[ 1 2 1 1 2 2 ]
[ 3 3 2 3 4 2 ] →
[ 0 2 1 1 2 1 ]
------------------------------------
[ 1 0 0 0 0 1 ]
[ 3 3 2 3 4 2 ] →
[ 0 2 1 1 2 1 ]
------------------------------------------
[ 1 0 0 0 0 1 ]
[ 0 3 2 3 4 -1 ] →
[ 0 2 1 1 2 1 ]
-----------------------------------------
[ 1 0 0 0 0 1 ]
[ 0 1 1 2 2 -2 ] →
[ 0 2 1 1 2 1 ]
------------------------------------------
[ 1 0 0 0 0 1 ]
[ 0 1 1 2 2 -2 ] →
[ 0 0 1 3 2 -5 ]
-----------------------------------
[ 1 0 0 0 0 1 ]
[ 0 1 0 -1 0 3 ] →
[ 0 0 1 3 2 -5 ]
--------------------------------------

[ 0 0 1 ]
X= [ -1 0 3 ]
[3 2 -5 ]