求解下列矩阵等式：
AX=A+2X，其中A=
(423
110
-123)
【正确答案】：AX=A+2X ⇒(A-2E)X=A ⇒X=(A-2E)-1A，
(A-2E，E)=
(2 2 3 1 0 0
1 -1 0 0 1 0
-1 2 1 0 0 1)
→
(1 -1 0 0 1 0
0 -1 -1 0 -1 -1
0 3 5/2 1/2 0 1)
→
(1 -1 0 0 1 0
0 1 1 0 1 1
0 1 5/6 1/6 0 1/3)
→
(1 -1 0 0 1 0
0 1 1 0 1 1
0 0 -1/6 1/6 -1 -2/3)
→
(1 -1 0 0 1 0
0 1 1 0 1 1
0 0 1 -1 6 4)
→
(1 0 0 1 -4 -3
0 1 0 1 -5 -3
0 0 1 -1 6 4)
(A-2E)-1=
(1 -4 -3
1 -5 -3
-1 6 4)
X=(A-2E)-1A=
(1 -4 -3
1 -5 -3
-1 6 4)
(4 2 3
1 1 0
-1 2 3)
=
(3 -8 -6
2 -9 -6
-2 12 9)
求解下列矩阵等式： AX=A+2X，其中A= (423 110 -123)
- 2024-11-07 03:12:17
- 线性代数（工）(13175)