求解下列矩阵等式：
AX=A+2X，其中A=
(423
110
-123)

求解下列矩阵等式：
AX=A+2X，其中A=
(423
110
-123)

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)