Identify a formula for each entry of the matrix $\small \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$ Identify a formula for each entry of the matrix $\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$.
It's easy to find a solution by just looking at the first few results:
\begin{pmatrix} 7^{n-1} & 3\cdot 7^{n-1} \\ 2\cdot 7^{n-1} & 6\cdot 7^{n-1} \end{pmatrix}
But how would I do it if it wasn't so obvious?

Fleshing out a little what the comments hint you:
$$\det(tI-A)=\begin{vmatrix}t-1&-3\\-2&t-6\end{vmatrix}=t(t-7)\Longrightarrow$$
the matrix's eigenvalues are $\,0\,,\,7\,$ , and eigenvectors for these values can be found as follows:
$$(1)\,\,\lambda=0:\;\;\;\;\begin{cases}\;\,-x-3y=0\\{}\\-2x-6y=0\end{cases}\;\;\Longrightarrow x=-3y\Longrightarrow\,\,\text{for example}\,\,\binom{3}{\!\!-1}$$
$$(1)\,\,\lambda=t:\;\;\;\;\begin{cases}\;6x-3y=0\\{}\\-2x+y=0\end{cases}\;\;\Longrightarrow y=2y\Longrightarrow\,\,\text{for example}\,\,\binom{1}{2}$$
Form now the matrix 
$$P:=\begin{pmatrix}3&1\\\!\!\!-1&2\end{pmatrix}\Longrightarrow P^{-1}=\begin{pmatrix}\frac{2}{7}&\!\!\!-\frac{1}{7}\\{}\\\frac{1}{7}&\;\frac{3}{7}\end{pmatrix}$$
so that
$$P^{-1}AP=\begin{pmatrix}0&0\\0&7\end{pmatrix}=:D$$
And now it is a piece of cake:
$$A^n=\left(PDP^{-1}\right)^n=PD^nP^{-1}\;\;,\;\;\text{and}\;\;\;D^n=\begin{pmatrix}0&0\\0&7^n\end{pmatrix}\ldots$$
