TITLE: Find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$
QUESTION [7 upvotes]: We want to find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$ for an arbitrary "n".
I have tried writing out a few elements of the sequence as $n \to \infty$:

$A^2 =  \left( \begin{array}{cc} 9 & 16 \\ 8 & 17 \end{array} \right)$
$A^3 = \left( \begin{array}{cc} 41 & 84 \\ 36 & 51 \end{array} \right)$

However, a pattern doesn't seem to appear.
This is where I want to ask my question: if we put this matrix into reduced-row echelon form, would an expression of the $(reduced matrix)^n$ work as an expression for the original matrix $A$? 
i.e. reduced-row matrix $ = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)^n$. Then, we know that any diagonal matrix to the $n^{th}$ is just the diagonal entries to the $n^{th}$ and this would make an expression easy to come up with. 
Thank you!

REPLY [5 votes]: you can also find $A^n$ using cayley-hamilton theorem which says that a matrix satisfies its own characteristic equation. the characteristic polynomial of $A$ is $$det (A-xI) = (x-5)(x+1).$$  
by polynomial division algorithm, we can find constants $a$ and $b$ so that $$x^n = q(x)(x-5)(x+1) + ax + b \to a = \frac16\left(5^n - (-1)^n\right), b = \frac16\left(5^n + 5(-1)^n \right) $$
by cayley-hamilton, $$A^n = aA + bI. $$