急问一道简单的线性代数题目设 A= [ 4 2 3 ],且矩阵X满足AX=A+2X,求X.[ 1 1 0 ][-1 2

搞错了 重来
因为 AX=A+2X
所以 (A-2E)X = A
(A-2E,A) =
2 2 3 4 2 3
1 -1 0 1 1 0
-1 2 1 -1 2 3
r1-2r2,r3+r2
0 4 3 2 0 3
1 -1 0 1 1 0
0 1 1 0 3 3
r1-3r3,r2+r3
0 0 -1 2 -12 -9
1 0 1 1 4 3
0 1 1 0 3 3
r2+r1,r3+r1,r1*(-1)
0 0 1 -2 12 9
1 0 0 3 -8 -6
0 1 0 2 -9 -6
交换行
1 0 0 3 -8 -6
0 1 0 2 -9 -6
0 0 1 -2 12 9
X =
3 -8 -6
2 -9 -6
-2 12 9

由AX=A+2X得(A-2E)X=A
A-2E=2 2 3
1 -1 0
-1 2 1
X=(A-2E)的逆*A
A-2E的逆=1 -4 -3
1 -5 -3
-1 6 4
X=(A-2E)的逆*A=3 -8 -6
2 -9 -6
-2 12 9