A power of a matrix I am trying to find 
$$\begin{bmatrix}t&1-t\\1-t&t\\\end{bmatrix}^n$$ in $\mathbb{Z}_n[t^{\pm 1}]/(t-1)^2$, where $n$ is a positive integer. My guess is that the result is the identity matrix but am not sure about a proof. Thanks for any help or hints!
 A: A variant of avs's strategy in comments: Write
$$ \begin{bmatrix} t & 1-t \\ 1-t & t \end{bmatrix}  =
I_{2\times 2} + (t-1)\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$
and use the binomial theorem (which applies because the $I$ commutes with everything) to raise this to the $n$th power.
All but two of the resulting terms contain two or more factors of $t-1$ and therefore vanish. One the the remaining terms has a factor of $\binom n1$ which equals $0$ modulo $n$, so it vanishes too. The last one is $\binom 11 I^n = I$.
A: If $$M = \pmatrix{t & 1-t\cr 1-t & t}$$
show by induction that 
$$ M^j = \pmatrix{j t - (j-1) & j - j t\cr
j - jt & jt - (j-1)\cr} $$
and then take $j=n$.
A: $\newcommand{\bbx}[1]{\,\bbox[10px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mbox{Note that}\quad
\pars{\begin{array}{cc}
\ds{t} & \ds{1 - t}
\\
\ds{1 - t} & \ds{t}
\end{array}}^{n} =
t^{n}\pars{\begin{array}{cc}
\ds{1} & \ds{\Lambda}
\\
\ds{\Lambda} & \ds{1}
\end{array}}^{n} =
t^{n}\pars{\sigma_{0} + \Lambda\sigma_{x}}^{n}
\label{1}\tag{1}
\\[2mm]
&\mbox{where}\quad\Lambda \equiv {1 - t \over t}\label{2}\tag{2}
\end{align}

$\ds{\sigma_{0}}$ is the $\ds{2 \times 2\ identity\ matrix}$ and $\ds{\sigma_{x}}$ is a Pauli Matrix.

\begin{equation}
\bbx{\ds{\mbox{Note that}\quad \sigma_{x}^{2} = \sigma_{0}}}
\label{3}\tag{3}
\end{equation}

\begin{align}
\pars{\begin{array}{cc}
\ds{t} & \ds{1 - t}
\\
\ds{1 - t} & \ds{t}
\end{array}}^{n} & =
t^{n}\,n!\bracks{z^{n}}\exp\pars{\bracks{\sigma_{0} + \Lambda\sigma_{x}}z} =
t^{n}\,n!\bracks{z^{n}}\bracks{\exp\pars{z\sigma_{0}}\exp\pars{\Lambda z\sigma_{x}}}
\\[5mm] & =
t^{n}\,n!\bracks{z^{n}}\braces{\exp\pars{z\sigma_{0}}
\bracks{\cosh\pars{\Lambda z}\sigma_{0} + \sinh\pars{\Lambda z}\sigma_{x}}}
\qquad\pars{~\mbox{see}\ \eqref{3}~}
\\[5mm] & =
t^{n}\,n!\bracks{z^{n}}\bracks{%
{\exp\pars{\bracks{1 + \Lambda}z} \over 2}\,\pars{\sigma_{0} + \sigma_{x}} +
{\exp\pars{\bracks{1 - \Lambda}z} \over 2}\,\pars{\sigma_{0} - \sigma_{x}}}
\\[5mm] & =
{1 \over 2}\,t^{n}\bracks{%
\pars{1 + \Lambda}^{n}
\pars{\begin{array}{cc}\ds{1} & \ds{1} \\ \ds{1} & \ds{1}\end{array}} +
\pars{1 - \Lambda}^{n}\pars{\begin{array}{rr}\ds{1} & \ds{-1} \\ \ds{-1} & \ds{1}\end{array}}}
\\[5mm] & =
{1 \over 2}\,\bracks{%
\pars{\begin{array}{cc}\ds{1} & \ds{1} \\ \ds{1} & \ds{1}\end{array}} +
\pars{2t - 1}^{n}\pars{\begin{array}{rr}\ds{1} & \ds{-1} \\ \ds{-1} & \ds{1}\end{array}}}\qquad\pars{~\mbox{see}\ \eqref{2}~}
\\[5mm] & =
\bbx{\ds{{1 \over 2}
\pars{\begin{array}{cc}
\ds{1 + \bracks{2t - 1}^{n}} & \ds{1 - \bracks{2t - 1}^{n}}
\\[2mm]
\ds{1 - \bracks{2t - 1}^{n}} & \ds{1 + \bracks{2t - 1}^{n}}
\end{array}}}}
\end{align}
