Let P = [ \beginmatrix 1 & 0 & 0 4 & 1 & 0 16 & 4 & 1 \endmatrix ] and I be the identity matrix of order 3. If Q = [q_ij] is a matrix such that P^50 - Q = I and q_31 + q_32 \over q_21 equals

Matrices and Determinants PYQ — JEE Advanced 2016 · Shift 2 · Q32 with Solution

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Correct answer: B

We have $P=\begin{bmatrix}1&0&0\\4&1&0\\16&4&1\end{bmatrix},\qquad P^{50}-Q=I.$ So, $Q=P^{50}-I.$ Hence we need entries of $P^{50}$ .

Let $N=P-I=\begin{bmatrix}0&0&0\\4&0&0\\16&4&0\end{bmatrix}.$ Then $P=I+N.$

Since $N$ is strictly lower triangular of order $3$ , we have $N^3=0.$ So by binomial theorem, $P^{50}=(I+N)^{50}=I+\binom{50}{1}N+\binom{50}{2}N^2.$

$N=\begin{bmatrix}0&0&0\\4&0&0\\16&4&0\end{bmatrix}.$ Now, $N^2=N\cdot N=\begin{bmatrix}0&0&0\\0&0&0\\16&0&0\end{bmatrix}.$

(Only the $(3,1)$ entry survives: $4\cdot 4=16$ .)

Using $P^{50}=I+50N+\binom{50}{2}N^2,$ and $\binom{50}{2}=\frac{50\cdot49}{2}=1225,$ we get

$50N=\begin{bmatrix}0&0&0\\200&0&0\\800&200&0\end{bmatrix},$

$1225N^2=\begin{bmatrix}0&0&0\\0&0&0\\19600&0&0\end{bmatrix}.$

Therefore,

=\begin{bmatrix}1&0&0\\200&1&0\\20400&200&1\end{bmatrix}.$$ --- 5. **Find $Q$** Since $$Q=P^{50}-I,$$ we have $$Q=\begin{bmatrix}0&0&0\\200&0&0\\20400&200&0\end{bmatrix}.$$ Thus, $$q_{21}=200,\qquad q_{31}=20400,\qquad q_{32}=200.$$ --- 6. **Compute the required value** $$\frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=\frac{20600}{200}=103.$$ --- 7. **Check options** The value is $$103,$$ which matches **Option B**. --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.