Question: $A^{2}-2A=\begin{bmatrix}
5 & -6 \\ 
-4 & 2 
\end{bmatrix}$

Can someone help me solve this? I've been trying to solve it for a while, but no matter what I try, the only information that I manage to get about A is that if $A=\begin{bmatrix}
a & b \\ 
c & d 
\end{bmatrix}$ then $c=\frac{2b}{3}$. Any help would be appreciated, thanks!

Answer: 
Denote by $I$ the identity matrix. Then, completing squares you can write
$$A^2 - 2A = A^2 -2IA + I^2 -I^2 = (A-I)^2 -I^2.$$
Hence, your equation is equivalent to
$$(A-I)^2 = X + I$$
since $I^2 = I$. Denote by $Y=X+I$ the new matrix (which is known). You want to find $B$ such that
$B^2=Y.$
Here, I recommend to diagonalize $Y$, i.e. find $U$ and diagonal $D$ such that
$$Y=UDU^{-1}.$$
Thus,
$$B= Y^{1/2} = UD^{1/2}U^{-1}.$$
See [this link](https://en.wikipedia.org/wiki/Square_root_of_a_matrix#By_diagonalization) for more information.

Once you have found $B$, $A$ is given by
$$A=B+I.$$
Remember that you may have more than one square root of the matrix $Y$.
$\newcommand{\Tr}{\mathrm{Tr}\,}$
Let $Y=A-I$, $X=B+I$, $B$ for the rhs matrix. Then $\Tr X = 9$, $\det X=-6$, and we need to solve
$$Y^2=X$$
for $Y$. Write $\alpha=\Tr Y$, $\beta = \det Y$. Then
$$Y^2-\alpha Y + \beta I=0$$
or
$$\alpha Y = \beta I + X$$
so that finding allowed values of $\alpha$, $\beta$ solves the problem.

Take the trace and determinant of both sides of $Y^2=X$. Then
$$\begin{align}
\alpha^2 - 2\beta &= \Tr X = 9 \\
\beta^2 &= \det X = -6
\end{align}$$
which means that
$$\begin{align}
\alpha A &= (\alpha+\beta +1)I + B \\
\alpha^2 & = 9 +2\beta \\
\beta^2 &= -6\text{.}
\end{align}$$
Note that there are four solutions.