Solving AXB=C equation 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:

$$A=\begin{bmatrix}
    6 &amp; -2 \\
    -3 &amp; 1 \\
   \end{bmatrix}$$

$$B=\begin{bmatrix}
    -7 &amp; -3 \\
    -2 &amp; 1 \\
   \end{bmatrix}$$
$$C=\begin{bmatrix}
    28 &amp; 12 \\
    -14 &amp; -6 \\
   \end{bmatrix}$$
$$Answer=\begin{bmatrix}
    -2/3+s/3 &amp; t/3 \\
    s &amp; t \\
   \end{bmatrix}$$
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$)$.
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$)$.