How would I solve the following question about matrices? $A= \begin{pmatrix}-1&3\\ -2&6\end{pmatrix}$
Find two $2\times 2$ matrices $B$ and $C$ such that $AB=AC$ but $B\ne C$.
I have tried to do some row operations along with multiplication but I keep getting the wrong answer.
Any help?

$B=\left(
\begin{array}{ll}
 3 & 0 \\
 2 & 1 \\
\end{array}
\right);\;C=\left(
\begin{array}{ll}
 -6 & 9 \\
 -1 & 4 \\
\end{array}
\right)$
$AB=AC=\left(
\begin{array}{ll}
 3 & 3 \\
 6 & 6 \\
\end{array}
\right)$
I did in this way
Called $B=\left(
\begin{array}{ll}
 a & b \\
 c & d \\
\end{array}
\right);\;C=\left(
\begin{array}{ll}
 e & f \\
 g & h \\
\end{array}
\right)$
I computed $AB$ and $AC$ and wrote an undeterminate system such that they were equal
I got 
$g= -\dfrac{a}{3}+c+\dfrac{e}{3},\;h= -\dfrac{b}{3}+d+\dfrac{f}{3}$
And all the other unknowns arbitrary values.
Hope this can be useful
