<s>[INST] How can I prove the following matrix equality? Let $n\in\mathbb{N}$ be arbitrary. Why is the following equality true?
$$
\left(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
+ 3 \begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}\right)^n
= \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
+ (4^n-1)\begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}
.
$$
Also, it might help that:
$$\begin{pmatrix}4 & 6\\\ -2 & -3\end{pmatrix}^n=\begin{pmatrix}4 & 6\\\ -2 & -3\end{pmatrix}$$
Thanks in advance!
 [/INST] Just use binomial formula and the fact that $A^2 = A \Rightarrow A^k = A$
$$(I + 3A)^n = \sum_{k=0}^n \binom{n}{k}3^kA^k =I + A\sum_{k=1}^n \binom{n}{k}3^k = I + A ((1+3)^n-1) = I +(4^n-1)A$$
Note that you can apply the binomial formula since $IA = AI$.
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