I'm trying to calculate the powers of a 2 by 2 matrix with entries in $\mathbb{Z} \left[ t,t^{-1} \right]$.


The matrix is \begin{bmatrix}
0 & 1 \\
1 & t
\end{bmatrix}


I tought of writing my matrix in the following form 
$$ \begin{bmatrix}
0 & 1 \\
1 & t
\end{bmatrix} = t \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix} + \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix} $$ 
Now we get 
$$ \begin{bmatrix}
0 & 1 \\
1 & t
\end{bmatrix} ^n = \sum\_{k=0}^n \binom{n}{k} t^{n-k} \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}^{n-k} \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}^{k} $$


The only issue is that if $\mathcal{M}\_n(\mathbb{Z} \left[ t,t^{-1} \right])$ is not isomorphic to $\mathcal{M}\_n(\mathbb{Z}) \left[ t,t^{-1} \right] $ then my logic will be wrong.


I also thought of diagonalizing the matrix, but I found no algorithme to get it done.


Thanks in advance for your help.



Powers of small square matrices over the Laurent polynomial ring with integer coefficients
This is not much different from Robert Israel's answer, but here goes:


Let $A(t)=\begin{pmatrix}0&1\\1&t\end{pmatrix}$ and write $A^n(t)=\begin{pmatrix}c\_n&b\_n\\b\_n&a\_n\end{pmatrix}$. Then, 
$$\begin{pmatrix}c\_{n+1}&b\_{n+1}\\ b\_{n+1}&a\_{n+1}\end{pmatrix}=A^{n+1}=\begin{pmatrix}c\_n&b\_n\\b\_n&a\_n\end{pmatrix}\begin{pmatrix}0&1\\1&t\end{pmatrix}
=\begin{pmatrix}b\_n&c\_n+tb\_n\\a\_n&b\_n+ta\_n\end{pmatrix}$$
implies $c\_{n+1}=b\_n, b\_{n+1}=a\_n, a\_{n+1}=b\_n+ta\_n$. That means, $A^n=\begin{pmatrix}a\_{n-2}&a\_{n-1}\\a\_{n-1}&a\_n\end{pmatrix}$ and
$a\_n=ta\_{n-1}+a\_{n-2}$ with initial conditions $a\_{-1}:=0$ and $a\_0=1$. This recurrence can be utilized to prove the following explicit formula
$$a\_n(t)=\sum\_{k=0}^{\lfloor\frac{n}2\rfloor}\binom{n-k}kt^{n-2k}.$$


