First slide

Operations of matrix

Question

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 $Q = [q_{i i}]$ ] is a matrix such that $P^{50} - Q = I$ then,
$\frac{9_{31} + q_{32}}{q_{21}}$ equal

Moderate

A

52
B

103
C

201
D

205

Solution

We have $P = [\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}]$
$\begin{array}{l} ∴ P^{2} = P P = [\begin{array}{lll} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}] [\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 16 + 32 & 8 & 1 \end{array}] \\ P^{3} = P^{2} P = [\begin{array}{ccc} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 16 + 32 & 8 & 1 \end{array}] [\begin{array}{ccc} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 12 & 1 & 0 \\ 16 + 32 + 48 & 12 & 1 \end{array}] \end{array}$
By observing the symmetry, we obtain
$\begin{array}{l} P^{50} = [\begin{array}{cccc} 4 \times 50 & 1 & 0 \\ 16 + 32 + 48 + \dots 50 & terms & 4 \times 50 & 1 \end{array}] \\ \Rightarrow P^{50} = [\begin{array}{ccc} 1 & 0 & 0 \\ 200 & 1 & 0 \\ \frac{16 \times 50 \times 51}{2} & 200 & 1 \end{array}] \end{array}$
$\begin{array}{l} ∴ P^{50} - Q = I \\ \Rightarrow Q = P^{50} - I = [\begin{array}{ccc} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 20400 & 200 & 0 \end{array}] \\ \Rightarrow q_{21} = 200, q_{31} = 20400 and q_{32} = 200 \\ ∴ \frac{q_{31} + q_{32}}{q_{21}} = \frac{20400 + 200}{200} = \frac{20600}{200} = 103 \end{array}$

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