Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$

I tried plugging in: $A= \begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}$ but that became messy very quickly. I got the equations:

$7a-b = 6$

$7c-d = 2$

$a^2+bc = a$

$ab+bd = b$

$ac + cd = c$

$bc + d^2 = d$ from trying that method. What should I do?
As $A^2=A$, you also know that $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$. Since $\begin{pmatrix}7\\-1\end{pmatrix},\begin{pmatrix}6\\2\end{pmatrix}$ are linearly independant, this determines $A$ completely. We conclude that both columns of $A$ are multiples of $\begin{pmatrix}6\\2\end{pmatrix}$, so
$$ A=\begin{pmatrix}6x&amp;6y\\2x&amp;2y\end{pmatrix}$$
for certain $x,y$. Now $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ just means that $6x+2y=1$, and $A\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ means that $7x-y=1$. Can you find $x,y$ now?
Write out the equations and use a simple linear solver to find:

${1 \over 10}\left(
\begin{array}{cc}
 9 &amp; 3 \\
 3 &amp; 1 \\
\end{array}
\right)
$