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?
$A^{-1} = \begin{pmatrix} 5 &-3 &-1 \\ 4 & -2&-1 \\ -12 & 7& 3 \end{pmatrix}$
Now $$AC= \begin{pmatrix} 1 &2\\ 0 & 1\\ 4 & 1 \end{pmatrix}\implies C=A^{-1} \begin{pmatrix} 1 &2\\ 0 & 1\\ 4 & 1 \end{pmatrix} =\begin{pmatrix} 5 &5\\ 4 & 4\\ 12 & 11 \end{pmatrix}$$