Let \(P = \begin{bmatrix} 1 & 0 & 0\\ 3 & 1 & 0\\ 9 & 3 & 1\end{bmatrix}\) and \(Q = [q_{ij}]\) be two \(3 \times 3\) matrices such that \(Q - P^5 = I_3\). Then \(\frac{q_{21} + q_{31}}{q_{32}}\) is equal to :
Show Hint
- 15
- 9
- 135
- 10
The Correct Option is D
Solution and Explanation
\(P^5 = (I+N)^5 = I + 5N + 10N^2\). So \(Q = I + P^5 = 2I + 5N + 10N^2\).
\(q_{21} = 5\cdot 3 = 15\), \(q_{32} = 5\cdot 3 = 15\), \(q_{31} = 5\cdot 9 + 10\cdot 9 = 135\).
\(\frac{q_{21} + q_{31}}{q_{32}} = \frac{15 + 135}{15} = 10\).
Answer: (4) 10
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