Solve the following equation for matrix $X$. $$ \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} X = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}+\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}X\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0& 0\end{bmatrix}\ $$
So this has the form of $AX=B+AXC$. I've tried moving the $A$ on the left side to the right side with the inverse. Then you can move the $XC$ to the left side. So you get something like $X−XC=A^{−1}B$. After this point I'm stuck and uncertain what to do or if I did the above correctly.
Edit: So the answer I got with the commented suggestion was [(2,1,0),(-1,0,1)]. Is that correct?
$AX=B+AXC$
$AX-AXC=B$
$A(X-XC)=B$
$(X-XC)=A^{-1}B$
$X(1-C)=A^{-1}B$
Note that $1$ represents the $3\times3$ identity matrix.
$X=A^{-1}B(1-C)^{-1}$
$$ X=\begin{pmatrix} 2 & 3 \\ 1&2 \end{pmatrix}^{-1} \begin{pmatrix} 1 & 1 &0 \\ 0&1 & 1 \end{pmatrix}\left[\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}-\begin{pmatrix} 0&1&1 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}\right]^{-1}$$
Now by simplifying this, you will get the value of X. the X will be a $2*3$ matrix