<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] Presumably $A$ is real. Observe that $A^3-I$ is nonzero but nilpotent (because $(A^3-I)^2=0$). Therefore $A^3$ and in turn $A$ are not diagonalisable. Hence $A$ has repeated eigenvalues. So, if $A$ is real, its eigenvalues must be real (or else the trace of $A$ would become non-real) and equal to $1$ (because $A^3-I$ is nilpotent). Hence the Jordan form of $A$ is $\pmatrix{1&1\\0&1}$. Since
$$
\pmatrix{1&1\\0&1}^2+\pmatrix{1&1\\0&1}=\pmatrix{2&3\\0&2}
=\pmatrix{1&1\\0&1}^3+I,
$$
we conclude that $A^2+A=A^3+I$.
Remark. Note that the above conclusion does not hold when $A$ can be non-real. (Thus the other answers here are either wrong or incomplete.) E.g. suppose $w=\exp(2\pi i/3)$ and
$$
A=\pmatrix{2w&w\\ -w&0}.
$$
Then $A^3$ is indeed equal to $\pmatrix{4&3\\ -3&-2}$ but $(A^2+A)-(A^3+I)$ is non-real.
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