Question No. 37
Let $P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$ and I be the identity matrix of order 3. If $Q=[q_{ij}]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals
(A) 52
(B) 103
(C) 201
(D) 205

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Solution:

Step 1: Express $P$ as $I+A$

Let $A=P-I=\left[\begin{array}{ccc}0& 0& 0\\ 4& 0& 0\\ 16& 4& 0\end{array}\right]$.
Then $P=I+A$.

Step 2: Compute powers of $A$

• $A^2=\left[\begin{array}{ccc}0& 0& 0\\ 4& 0& 0\\ 16& 4& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 4& 0& 0\\ 16& 4& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 16& 0& 0\end{array}\right]$

• $A^3=A^2\cdot A=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 16& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 4& 0& 0\\ 16& 4& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$
  So, $A^n=0$ for $n\ge 3$.

Step 3: Compute $P^{50}$ using binomial expansion

Since $A$ is nilpotent with $A^3=0$,

$$P^{50}=(I+A{)}^{50}=I+\left(\genfrac{}{}{0px}{}{50}{1}\right)A+\left(\genfrac{}{}{0px}{}{50}{2}\right)A^2$$$$=I+50A+\frac{50\cdot 49}{2}{A}^2=I+50A+1225A^2$$

Step 4: Write $P^{50}-I=Q$

Given $P^{50}-Q=I⟹Q=P^{50}-I$.
So,

$$Q=50A+1225A^2$$

Step 5: Compute $50A$ and $1225A^2$

• $50A=50\left[\begin{array}{ccc}0& 0& 0\\ 4& 0& 0\\ 16& 4& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 200& 0& 0\\ 800& 200& 0\end{array}\right]$

• $1225A^2=1225\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 16& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 19600& 0& 0\end{array}\right]$

So,

$$Q=\left[\begin{array}{ccc}0& 0& 0\\ 200& 0& 0\\ 800+19600& 200& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 200& 0& 0\\ 20400& 200& 0\end{array}\right]$$

Step 6: Find the required ratio

We need:

$$\frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=\frac{20600}{200}=103$$

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Final Answer:

$$\boxed{{\displaystyle 103}}$$