I would like some help with finding $X$ for the equation $AX B=C$. I understand how to get $X$ by itself ($X=A^{-1}AXBB^{-1}=A^{-1}CB^{-1}$) and I have found the inverse for $B$. The problem is that I can't find the inverse of $A$ which I think is inconsistent. The answer is supposed to be a parameter solution. Here are the matrices:
$$A=\begin{bmatrix} 6 & -2 \\ -3 & 1 \\ \end{bmatrix}$$
$$B=\begin{bmatrix} -7 & -3 \\ -2 & 1 \\ \end{bmatrix}$$ $$C=\begin{bmatrix} 28 & 12 \\ -14 & -6 \\ \end{bmatrix}$$ $$Answer=\begin{bmatrix} -2/3+s/3 & t/3 \\ s & t \\ \end{bmatrix}$$
You can check that $A$ is not invertible, because its determinant is $6\times 1 - (-2)\times(-3) = 0$. This is why you cannot find an inverse for $A$, and also why your answer features parameters $s$ and $t$.
Notice that, as you yourself showed in the opening question, if $A$ and $B$ were both invertible then the answer would be unique $($that is, it would not depend on the value of some parameters$)$.