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?
 A: We have $AB=0$ for all $B$ of the form
$$
B=\begin{pmatrix} 3\alpha & 3\beta \cr \alpha & \beta \end{pmatrix}
$$
On the other hand, $AC=0$ for $C=0$. Now choose $\alpha,\beta$ non-zero. Then we have $AB=AC$ but $B\neq C$.
A: There's no unique answer.

But the task of finding some valid answer is easy.

Since the determinant of $A$ is zero, there exists a nonzero vector $v$ such that $Av = 0$.

Find such a vector $v$, and let $B$ be the $2\,\times\,2$ matrix with both columns equal to $v$. It follows that $AB=0$.

Then just let $C$ be the zero matrix.
A: $B=\left(
\begin{array}{ll}
 3 & 0 \\
 2 & 1 \\
\end{array}
\right);\;C=\left(
\begin{array}{ll}
 -6 & 9 \\
 -1 & 4 \\
\end{array}
\right)$
$AB=AC=\left(
\begin{array}{ll}
 3 & 3 \\
 6 & 6 \\
\end{array}
\right)$
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
