Let P = [ 1 0 0 4 1 0 16 4 1 ] and I be the identity matrix of order 3. If Q = [ q i j ] is a matrix such that P 50 - Q = I , then q 31 + q 32 q 21 equals [2016]

Question

Let $P = [\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{matrix}]$ and $I$ be the identity matrix of order 3. If $Q = [q_{i j}]$ is a matrix such that $P^{50} - Q = I,$ then $\frac{q_{31} + q_{32}}{q_{21}}$ equals [2016]

Smart Achievers · Accepted Answer

(2)

$P = [\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{matrix}] = I + [\begin{matrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{matrix}] = I + A,$

where $A = [\begin{matrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{matrix}]$

and $A^{3} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}],$ $∴$ $A^{n} = 0, \forall n \geq 3$

Now, $P^{50} = {(I + A)}^{50} = {}^{50}C_{0} I^{50} + {}^{50}C_{1} I^{49} A + {}^{50}C_{2} I^{48} A^{2} + 0$

$= I + 50 A + 25 \times 49 A^{2}$

$∴$ $Q = P^{50} - I = 50 A + 25 \times 49 A^{2}$

$\Rightarrow q_{21} = 50 \times 4 = 200,$ $q_{31} = 50 \times 16 + 25 \times 49 \times 16 = 20400,$

and $q_{32} = 50 \times 4 = 200$

$∴$ $\frac{q_{31} + q_{32}}{q_{21}} = \frac{20600}{200} = 103$

Solution

Q.
Let $P = [\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{matrix}]$ and $I$ be the identity matrix of order 3. If $Q = [q_{i j}]$ is a matrix such that $P^{50} - Q = I,$ then $\frac{q_{31} + q_{32}}{q_{21}}$ equals [2016]

1 52

2 103

3 201

4 205

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