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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...

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If $M = \pmatrix{1 & -1\cr -5 & 6\cr}$, then what do you suppose $AM$ is? –  Robert Israel May 2 '12 at 1:28
    
@RobertIsrael I'm interested in where you're going with this. Is there some sort of theorem that relates $A$ and $M$? –  StickFigs May 2 '12 at 3:59
    
Once you know $AM = B$, you can recover $A$ as $B M^{-1}$. –  Robert Israel May 2 '12 at 4:02
    
@RobertIsrael This worked for this question but when I tried this method on a similar question it failed to produce the correct answer, why? –  StickFigs May 2 '12 at 5:07
    
How should I know what went wrong without seeing the question and your answer? –  Robert Israel May 2 '12 at 8:00

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up vote 2 down vote accepted

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: $-14+6\cdot3=-14+18=4$.

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How exactly do I solve those two systems of equations? Do I solve them together simultaneously? –  StickFigs May 2 '12 at 4:06
    
No, you solve the first to find $a$ and $b$, and you solve the second to find $c$ and $d$. –  Brian M. Scott May 2 '12 at 4:12
    
Can you give me an example. like maybe solve the first system? –  StickFigs May 2 '12 at 4:31
    
@StickFigs: I will if you absolutely insist, but it's basic high school algebra. And if you're now studying eigenvalues in linear algebra, you must already have spent a fair bit of time on solving systems of linear equations. Are you sure that you aren't imagining some complexity that isn't really there? –  Brian M. Scott May 2 '12 at 4:36
    
Nevermind then. –  StickFigs May 2 '12 at 4:52

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