Product of Matrices I Given the matrix
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\end{align}
consider the first few powers of $A^{n}$ for which
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\hspace{15mm} A^{2} = \left( \begin{matrix} 7 & 6 \\ 9 & 10 \end{matrix} \right) \hspace{15mm}
A^{3} = \left( \begin{matrix} 25 & 26 \\ 39 & 38 \end{matrix} \right).
\end{align}
Notice that the first rows have the values $(1,2)$, $(7,6)$, $(25,26)$ of which the first and second elements are rise and fall in a cyclical pattern.
The same applies to the bottom rows. 


*

*Is there an explanation as to why the numbers rise and fall in order of compared columns?

*What is the general form of the $A^{n}$ ? 

 A: It is often possible to diagonalize a square matrix, that is find a diagonal matrix $D$ and another matrix $P$ such that $$A=PDP^{-1}.$$ This is a nice property because $$A^n=PDP^{-1}PDP^{-1}...PDP^{-1}=PD^nP^{-1}.$$
The $P$ and $P^{-1}$ matrices cancel except on the ends, so we only have to take a power of the diagonal matrix, which is easy. If 
$$D=\left(\begin{array}{cc}
a & 0\\
0 & b\\ \end{array}\right),$$
then
$$D^n=\left(\begin{array}{cc}
a^n & 0\\
0 & b^n\\ \end{array}\right).$$
In fact, it is possible to diagonalize your matrix as
$$A=\left(\begin{array}{cc}
2 & -1\\
3 & 1\\ \end{array}\right)
\left(\begin{array}{cc}
4 & 0\\
0 & -1\\ \end{array}\right)
\left(\begin{array}{cc}
1/5 & 1/5\\
-3/5 & 2/5\\ \end{array}\right).$$
Thus,
$$A^n=\left(\begin{array}{cc}
2 & -1\\
3 & 1\\ \end{array}\right)
\left(\begin{array}{cc}
4^n & 0\\
0 & (-1)^n\\ \end{array}\right)
\left(\begin{array}{cc}
1/5 & 1/5\\
-3/5 & 2/5\\ \end{array}\right)=\left(
\begin{array}{cc}
 \frac{3 (-1)^n}{5}+\frac{1}{5} 2^{2 n+1} & \frac{1}{5} (-2) (-1)^n+\frac{1}{5} 2^{2 n+1} \\
 \frac{1}{5} (-3) (-1)^n+\frac{3\ 4^n}{5} & \frac{2 (-1)^n}{5}+\frac{3\ 4^n}{5} \\
\end{array}
\right).$$
As for the top row business, you can see that each term in the top row has two terms. The second of each is the same. In $n$ is odd, the first terms are $-3/5$ and $2/5$ respectively. If $n$ is even, the two terms are $3/5 $ and $-2/5$. Can you see why this relationship causes the behavior you noticed?
A: In answer to part 2, the general form is $$M^n=\frac{1}{5}\begin{pmatrix} 2\times 4^n+3 \times(-1)^n & 2\times 4^n-2(-1)^n\\3\times 4^n+3(-1)^{n+1} & 3\times 4^n+2(-1)^{n}\\\end{pmatrix}$$
But you know this already because it has already been posted!
(it just took me longer to write out in MathJax)
