Determinants Matrices Question 123
Question: If $ A=( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} ) $ , then $ A^{8}=( \begin{matrix} p^{8} & q( \frac{p^{8}-1}{p-1} ) \\ 0 & k \\ \end{matrix} ) $ . The value of k is
Options:
A) 1
B) 0
C) 2
D) -1
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ A^{2}=( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} )( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} )=( \begin{matrix} p^{2} & pq+q \\ 0 & 1 \\ \end{matrix} ) $
$ =( \begin{matrix} p^{2} & q(p+1) \\ 0 & 1 \\ \end{matrix} ) $
$ A^{3}=( \begin{matrix} p & q \\ 0 & 1 \\ \end{matrix} )( \begin{matrix} p^{2} & pq+q \\ 0 & 1 \\ \end{matrix} ) $
$ ( \begin{matrix} p^{3} & q(p^{2}+p+1) \\ 0 & 1 \\ \end{matrix} ) $ Similarly, $ A^{4}=( \begin{matrix} p^{4} & q(p^{3}+p^{2}+p+1) \\ 0 & 1 \\ \end{matrix} ) $ and so on.
$ \therefore A^{8}=( \begin{matrix} p^{8} & q(p^{7}+p^{8}+…+1) \\ 0 & 1 \\ \end{matrix} )=( \begin{matrix} p^{8} & q( \frac{p^{8}-1}{p-1} ) \\ 0 & 1 \\ \end{matrix} ) $
BETA
