Find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$ 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!

In problems such as these one of the most efficient ways to calculate powers of a matrix is through diference equations. For the matrix of this problem the following is obtained. 
Given
\begin{align}
A = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)
\end{align}
it is found that
\begin{align}
A^{2} = \left( \begin{array}{cc} 9 & 16 \\ 8 & 17 \end{array} \right)
\hspace{10mm} 
A^{3} = \left( \begin{array}{cc} 41 & 84 \\ 42 & 83 \end{array} \right). 
\end{align}
The problem asks for the values of $A^{n}$. This is done by letting 
\begin{align}
A^{n} = \left( \begin{array}{cc} a_{n} & b_{n} \\ c_{n} & d_{n} \end{array} 
\right).
\end{align}
With this and $A^{n} = A \cdot A^{n-1}$ then
\begin{align}
\left( \begin{array}{cc} a_{n} & b_{n} \\ c_{n} & d_{n} \end{array} \right)
= \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)
\left( \begin{array}{cc} a_{n-1} & b_{n-1} \\ c_{n-1} & d_{n-1} \end{array} \right).
\end{align}
From this equation it is quickly determined that
\begin{align}
a_{n} &= a_{n-1} + 4 c_{n-1} \hspace{12mm} b_{n} = b_{n-1} + 4 d_{n-1} \\
c_{n} &= 2 a_{n-1} + 3 c_{n-1} \hspace{10mm} d_{n} = 2 b_{n-1} + 3 d_{n-1}.
\end{align}
These equations are reducible to the second order difference equation, $\phi_{n} \in \{ a_{n} , b_{n}, c_{n}, d_{n} \}$,
\begin{align}
\phi_{n+2} = 4 \phi_{n-1} + 5 \phi_{n}
\end{align}
which has the solution 
\begin{align}
\phi_{n} = A \, 5^{n} + B \, (-1)^{n}.
\end{align}
Now returnig to the first few powers of $A$ the conditions 
\begin{align}
a_{1} &= 1, a_{2} = 9, a_{3} = 41 \\
b_{1} &= 4, b_{2} = 16, b_{3} = 84 \\
c_{1} &= 2, c_{2} = 8, c_{3} = 42 \\
d_{1} &= 3, d_{2} = 17, d_{3} = 83
\end{align}
are obtained. The second order difference equation equation can be solved for each of the condition sets and is determined that
\begin{align}
a_{n} &= \frac{1}{3} \left( 5^{n} + 2 (-1)^{n} \right) \hspace{10mm} 
b_{n} = \frac{2}{3} \left( 5^{n} - (-1)^{n} \right) \\
c_{n} &= \frac{1}{3} \left( 5^{n} - (-1)^{n} \right) \hspace{12mm}
d_{n} = \frac{1}{3} \left( 2 \cdot 5^{n} + (-1)^{n} \right).
\end{align}
Putting this together the $n^{th}$-power of $A$ is given by
\begin{align}
A^{n} = \frac{1}{3} \left( \begin{array}{cc} 5^{n} + 2 (-1)^{n} & 2(5^{n} - (-1)^{n}) \\ 5^{n} - (-1)^{n} & 2 \cdot 5^{n} + (-1)^{n} \end{array} 
\right).
\end{align}
