Q.

Let $P = [\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}]$ and I be the identity matrix of order 3. If Q = [q_ij] is a matrix such that $P^{50} - Q = I, then \frac{q_{31} + q_{32}}{q_{21}} =$

Moderate

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By Expert Faculty of Sri Chaitanya

answer is 103.

(Detailed Solution Below)

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Detailed Solution

$\begin{array}{l} P^{2} = [\begin{array}{lll} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}] [\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 48 & 8 & 1 \end{array}] \\ P^{3} = [\begin{array}{lll} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 48 & 8 & 1 \end{array}] [\begin{array}{lll} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 12 & 1 & 0 \\ 96 & 12 & 1 \end{array}] \\ ∴ P^{n} = [\begin{array}{ccc} 1 & 0 & 0 \\ 4 n & 1 & 0 \\ 8 (n^{2} + n) & 4 n & 1 \end{array}] \\ ∴ P^{50} = [\begin{array}{ccc} 200 & 1 & 0 \\ 8 \times 50 (52) & 200 & 1 \end{array}] \Rightarrow p^{50} - Q = I \\ ∴ Equating we get q_{21} = 200 \\ \Rightarrow q_{31} = 400 \times 51 \Rightarrow q_{32} = 200 \\ \Rightarrow \frac{q_{31} + q_{32}}{q_{21}} = \frac{400 \times 51 + 200}{200} = 2 (51) + 1 = 103 \end{array}$

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