<s>[INST] Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$
I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything. 
thanks. 
 [/INST] $$
A=\pmatrix{{2}&{3}\\{-1}&{-1}}.$$
Note that $a^3+b^3=(a+b)^3-3ab(a+b)$. 
Let $a,b$ be eigenvalues of $A$. Then $trA=a+b$, $detA=ab$, and they are integers.
Let $X=trA$. 
We see from $A(A^2-3I)$ has determinant $13$, we have $detA | 13$. 
From the identity in the beginning, we have $X^3-3(ab+1)X+5=0$ , which we can obtain from taking traces of original equation.
Just plug in the four possible values for $ab$, namely $\pm 1, \pm 13$, we find that only possible value is $ab=1$, and consequently 
$X^3-6X+5=0$. 
Solving for integers, we obtain $X=1$. 
Thus, $A$ should satisfy $trA=1$, $detA=1$. 
By Cayley-Hamilton, we have $A^2-A+I=0$, this forces $A^3=-I$. 
Now, plug in to original equation to get 
$$
-3A=\pmatrix{{-6}&{-9}\\{3}&{3}}.$$
Then we arrive at the answer.
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