Question

Let $A=\left[\begin{array}{cc}2& q\\ 0& 1\end{array}\right]$ and $A^8=\left[\begin{array}{cc}x& yq\\ 0& 1\end{array}\right]$, find $x-y$.

Answer 1

Given, $A=\left[\begin{array}{cc}2& q\\ 0& 1\end{array}\right]$
$A^2=\left[\begin{array}{cc}2& q\\ 0& 1\end{array}\right]\left[\begin{array}{cc}2& q\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{2}^2& 3q\\ 0& 1\end{array}\right]$
$A^3=\left[\begin{array}{cc}{2}^2& 3q\\ 0& 1\end{array}\right]\left[\begin{array}{cc}2& q\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{2}^3& 7q\\ 0& 1\end{array}\right]$
Hence, $A^n=\left[\begin{array}{cc}{2}^n& (2^n-1)q\\ 0& 1\end{array}\right]\forall n\in N$
$\Rightarrow A^8=\left[\begin{array}{cc}{2}^8& {(2}^8-1)q\\ 0& 1\end{array}\right]$
Given $A^8=\left[\begin{array}{cc}x& yq\\ 0& 1\end{array}\right]$
So, $x=2^8,y=2^8-1$
$\Rightarrow x-y=1$