Question

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

• A. $52$
• B. $103$
• C. $201$
• D. $205$

Answer 1

Answer: (B)

$P=\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} \, \, \Rightarrow P^{2}=\begin{bmatrix} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 16+32 & 8 & 1 \end{bmatrix}$
So, $P^{3}=\begin{bmatrix} 1 & 0 & 0 \\ 12 & 1 & 0 \\ 16+32+48 & 12 & 1 \end{bmatrix}$
Using symmetry, we can write
$P^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ \frac{16.50.51}{2} & 200 & 1 \end{bmatrix}$
As, $P^{50}-Q=I \Rightarrow Q=P^{50}-I=\left[\begin{array}{ccc}0 & 0 & 0 \\ 200 & 0 & 0 \\ \frac{16.50 .51}{2} & 200 & 0\end{array}\right]$
$q_{31}=\frac{16.50.51}{2},$ $q_{32}=200$ and $q_{21}=200$
$\therefore $ $\frac{q_{31} + q_{32}}{q_{21}}=\frac{16.50.51}{2.200}+1=102+1=103$