TITLE: For some given $3\times3$ matrix $A^{-1}$ and $3\times2$ matrix $B$, solve $AC=B$ for $C$
QUESTION [0 upvotes]: Let $A^{-1}$ = $\begin{pmatrix}
1 &2  &1 \\ 
0 & 3 & 1\\ 
4 & 1 & 2
\end{pmatrix}$. 
  Find matrix $C$ such that $AC = \begin{pmatrix}
1 &2\\ 
0 & 1\\ 
4 & 1
\end{pmatrix}$.

How do I do this problem?
The only thing I could think of doing finding $A$, which is $(A^{-1})^{-1}$
$A = \begin{pmatrix}
5 &-3  &-1 \\ 
4 &  -2&-1 \\ 
-12 &  7& 3
\end{pmatrix}$
I know $C = 3 \times 2$, since $AC = 3 \times 2$.
So, 
\begin{pmatrix}
5 &-3  &-1 \\ 
4 &  -2&-1 \\ 
-12 &  7& 3
\end{pmatrix}
times
\begin{pmatrix}
 a& b\\ 
c & d\\ 
e & f
\end{pmatrix}
is $AC$.
How do I proceed from here? This seems very long. Is there an easier way?

REPLY [0 votes]: $A^{-1} = \begin{pmatrix}
5 &-3  &-1 \\ 
4 &  -2&-1 \\ 
-12 &  7& 3
\end{pmatrix}$
Now
$$AC= \begin{pmatrix}
1 &2\\ 
0 & 1\\ 
4 & 1
\end{pmatrix}\implies C=A^{-1} \begin{pmatrix}
1 &2\\ 
0 & 1\\ 
4 & 1
\end{pmatrix} =\begin{pmatrix}
5 &5\\ 
4 & 4\\ 
12 & 11
\end{pmatrix}$$
{"set_name": "stack_exchange", "score": 0, "question_id": 2130706}