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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?

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    Multiply by $A^{-1}$ on both sides of the desired equality. What do you get on the LHS? – 2017-02-05
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    ohh! I get that $C = \text {matrix AC times } A^{-1}$ – 2017-02-05
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    or I get that $C = A^{-1} \text { times matrix AC} $ – 2017-02-05
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    Let $B$ equal $\begin{pmatrix} 1 &2\\ 0 & 1\\ 4 & 1 \end{pmatrix}$. You wish to find $C$ such that $AC=B$. I asked you multiply on the left by $A^{-1}$, so you get $A^{-1}(AC)=A^{-1}B$. Do you understand this? Can you proceed? – 2017-02-05
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    yes thanks it worked out! – 2017-02-05

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$A^{-1} = \begin{pmatrix} 5 &-3 &-1 \\ 4 & -2&-1 \\ -12 & 7& 3 \end{pmatrix}$

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