A) 10
B) 135
C) 9
D) 15
Correct Answer: A
Solution :
Here, \[P=\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{matrix} \right]\]and\[Q={{P}^{5}}+{{I}_{3}}\] \[\therefore \]\[{{P}^{2}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \\ \end{matrix} \right]\] \[{{P}^{3}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 9 & 1 & 0 \\ 54 & 9 & 1 \\ \end{matrix} \right]\] Similarly, \[{{P}^{5}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \\ \end{matrix} \right]\] \[\therefore \]\[Q=\left[ \begin{matrix} 2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\frac{{{q}_{21}}+{{q}_{31}}}{{{q}_{32}}}=\frac{15+135}{15}=10\]
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