
I tried equating -1*{1,-5} = {1,-5}A = {-1,5} and -4{-1,6} = {-1,6}*A = {4,-24} and think I'm on the right track but don't know what to do next...

I tried equating -1*{1,-5} = {1,-5}A = {-1,5} and -4{-1,6} = {-1,6}*A = {4,-24} and think I'm on the right track but don't know what to do next...
If all else fails, and you can't think of anything cleverer, you can fall back on brute force. Let $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\;.$$ You know that $$\begin{bmatrix}-1\\5\end{bmatrix}=-\begin{bmatrix}1\\-5\end{bmatrix}=A\begin{bmatrix}1\\-5\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1\\-5\end{bmatrix}=\begin{bmatrix}a-5b\\c-5d\end{bmatrix}$$ and $$\begin{bmatrix}4\\-24\end{bmatrix}=-4\begin{bmatrix}-1\\6\end{bmatrix}=A\begin{bmatrix}-1\\6\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}-1\\6\end{bmatrix}=\begin{bmatrix}-a+6b\\-c+6d\end{bmatrix}\;.$$
This gives you two systems of linear equations in two unknowns each:
$$\left\{\begin{align*}a-5b&=-1\\-a+6b&=4\end{align*}\right.$$ and
$$\left\{\begin{align*}c-5d&=5\\-c+6d&=-24\;.\end{align*}\right.$$
These are readily solved for $a,b,c$, and $d$.
Added: For example, if you add the two equations of the first system to each other, you get $(a-5b)+(-a+6b)=-1+4$, or $b=3$; substituting that into the first equation gives you $a-15=-1$, so $a=14$. For safety's sake you can check that the second equation is satisfied: $-14+6\cdot3=-14+18=4$.