If \(A=\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix}\) then value of \(\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix}^{2007}\)
\(\begin{bmatrix}1 & 0\\4014 & 1\end{bmatrix}\)
Write \(A=I+N\), where \(I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(N=\begin{bmatrix}0&0\\2&0\end{bmatrix}\).
Compute \(N^2=\begin{bmatrix}0&0\\2&0\end{bmatrix}\begin{bmatrix}0&0\\2&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\), so N is nilpotent with \(N^2=0\).
Since I and N commute, by the binomial theorem \(A^n=(I+N)^n=I+nN+\binom{n}{2}N^2+\cdots=I+nN\) (all higher terms vanish because \(N^2=0\)).
So \(A^n=I+nN=\begin{bmatrix}1&0\\2n&1\end{bmatrix}\).
For \(n=2007\): \(A^{2007}=\begin{bmatrix}1&0\\2\times2007&1\end{bmatrix}=\begin{bmatrix}1&0\\4014&1\end{bmatrix}\).
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