<s>[INST] Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$ 
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$

Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt:
$$A^3 -I_2 = \begin{bmatrix} 3 & 3\\ -3 & -3\end{bmatrix} = 9 \begin{bmatrix} 1 & 1\\ -1 & -1\end{bmatrix}$$
and 
$$B=\frac {A^3-I_2}{A-I_2}-I_2$$
$$(A-I_2)(A^2+A+I_2)=9\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$
But I get stuck here.
 [/INST] Although I prefer an algebraic approach, here is a "non-algebraic" way by directly determining $A$. Let 


*

*$C =A^3 \Rightarrow C^2=\begin{bmatrix}7& 6\\-6&-5\end{bmatrix}$


Looking at the pattern of the entries comparing $A^6$ with $A^3$ we find: 


*

*the entries in the first row go $3$ up

*the entries in the second row go $3$ down


So, applying this pattern "backwards" to $A^3$ we get a possible candidate for $A$:
$$A = \begin{bmatrix}2& 1\\-1&0\end{bmatrix}$$ 
Indeed, we get
$$A^3 = \begin{bmatrix}4&3\\-3&-2\end{bmatrix} \Rightarrow A^2+A = \begin{bmatrix}5&3\\-3&-1\end{bmatrix}$$
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