A) 1
B) 0
C) 2
D) -1
Correct Answer: A
Solution :
[a] \[{{A}^{2}}=\left( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} \right)\left( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} \right)=\left( \begin{matrix} {{p}^{2}} & pq+q \\ 0 & 1 \\ \end{matrix} \right)\] \[=\left( \begin{matrix} {{p}^{2}} & q(p+1) \\ 0 & 1 \\ \end{matrix} \right)\] \[{{A}^{3}}=\left( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} \right)\left( \begin{matrix} {{p}^{2}} & pq+q \\ 0 & 1 \\ \end{matrix} \right)\] \[\left( \begin{matrix} {{p}^{3}} & q({{p}^{2}}+p+1) \\ 0 & 1 \\ \end{matrix} \right)\] Similarly, \[{{A}^{4}}=\left( \begin{matrix} {{p}^{4}} & q({{p}^{3}}+{{p}^{2}}+p+1) \\ 0 & 1 \\ \end{matrix} \right)\] and so on. \[\therefore {{A}^{8}}=\left( \begin{matrix} {{p}^{8}} & q({{p}^{7}}+{{p}^{8}}+...+1) \\ 0 & 1 \\ \end{matrix} \right)=\left( \begin{matrix} {{p}^{8}} & q\left( \frac{{{p}^{8}}-1}{p-1} \right) \\ 0 & 1 \\ \end{matrix} \right)\]
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