Question

If $A=\left(\begin{array}{cc}p& q\\ 0& 1\end{array}\right)$, then show that
$A^8=\left(\begin{array}{cc}{p}^8& q\left(\frac{p^8-1}{p-1}\right)\\ 0& 1\end{array}\right)$

Answer 1

$A^2=\left[\begin{array}{cc}p& q\\ 0& 1\end{array}\right]\left[\begin{array}{cc}p& q\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{p}^2+0& pq+q\\ 0& 1\end{array}\right]$

$A^3=\left[\begin{array}{cc}{p}^2& q(p+1)\\ 0& 1\end{array}\right]\left[\begin{array}{cc}p& q\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{p}^3& p^{2}q+q\left(pq\right)\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{p}^3& q(p^2+p+1)\\ 0& 1\end{array}\right]$

-------- and so on

$A^8=\left[\begin{array}{cc}{p}^8& q\frac{p^8-1}{p-1}\\ 0& 1\end{array}\right]$

Hence proved