Solve $AX=B+AXC$ for the missing matrix $X$

Question

imgur link in case I mess up the formatting

Solve the following matrix equation for $X$

$$AX=B+AXC$$

where non-symmetric matrix $A$ is $2 \times 2$, matrix $B$ is $2 \times 3$, and non-invertible matrix $C$ is $3 \times 3$.

$$\begin{bmatrix} 2&3\\1&2 \end{bmatrix}X=\begin{bmatrix}1&1&0\\0&1&1\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}X\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}$$

This one has me stumped. The only way I managed to find an answer for $X$ was by doing this:

$$\begin{align}X&=A^{-1}(B+AXC)\\\\ X&=A^{-1}B+XC\\\\ X-XC&=A^{-1}B\\\\ X(I-C)&=A^{-1}B\\\\ X&=A^{-1}B(I-C)^{-1}\end{align}$$

obviously, by plugging in this answer for $X$ into the original question, the answer is false. its probably because by doing $X(I-C)$, you create an identity matrix that doesn't have a clear dimension. Your help would be much appreciated.

EDIT: turns out my solution was correct. I must have made a miscalculation somewhere along the way. Thanks everyone!

Answer 1

Your answer is correct, if you substitute $X$ back into the equation you get a true result: $$AX = B + AXC$$ $$A(A^{-1}B(I-C)^{-1}) = B + A(A^{-1}B(I-C)^{-1})C$$ $$B(I-C)^{-1} = B + B(I-C)^{-1}C$$ $$0 = B + B(I-C)^{-1}C - B(I-C)^{-1}$$ $$0 = B + B(I-C)^{-1}(C - I)$$ $$0 = B - B(I-C)^{-1}(I - C)$$ $$0 = B - B$$ $$0 = 0$$

Answer 2

You are solving for $X$ in $AX = B + AXC.$ Now, let $Y = AX,$ so you are trying to solve for $Y$ in $Y = B + Y C.$ Therefore $Y(I-C) = B,$ so $Y = B(I-C)^{-1},$ assuming the inverse exists. Otherwise, treat the equation as a linear equation in the coefficients of $Y$ and see what you can find.