Question

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

• A. $\left(\begin{array}{cc}{p}^8& q^8\\ 0& 1\end{array}\right)$
• B. $\left(\begin{array}{cc}{p}^8& \left(\frac{p^8-1}{p-1}\right)\\ 0& 1\end{array}\right)$
• C. $\left(\begin{array}{cc}{p}^8& q\left(\frac{p^8-1}{p-1}\right)\\ 0& 1\end{array}\right)$
• D. $\left(\begin{array}{cc}\left(\frac{p^8-1}{p-1}\right)& p\left(\frac{q^8-1}{q-1}\right)\\ 0& 1\end{array}\right)$

Answer 1

The correct option is C $\left(\begin{array}{cc}{p}^8& q\left(\frac{p^8-1}{p-1}\right)\\ 0& 1\end{array}\right)$

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

$=\left(\begin{array}{cc}{p}^3& q(p^2+p+1)\\ 0& 1\end{array}\right)$
Similarly,
$A^4=\left(\begin{array}{cc}{p}^4& q(p^3+p^2+p+1)\\ 0& 1\end{array}\right)$ and so on.
$\therefore A^8=\left(\begin{array}{cc}{p}^8& q(p^7+p^6+....+1)\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}{p}^8& q\left(\frac{p^8-1}{p-1}\right)\\ 0& 1\end{array}\right)$