<p>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 \(Q = [q_{ij}]\) is a matrix such that \(P^{50} - Q = I\), then \(\dfrac{q_{31} + q_{32}}{q_{21}}\) equals</p>
Step-by-Step Solution
Key Concept: Recognize P as a lower triangular matrix with a pattern: P = I + N where N is nilpotent (N³ = 0). Use binomial expansion for P⁵⁰ = (I + N)⁵⁰ to find specific entries without computing the full matrix.
<p><strong>Step 1:</strong> Express P in the form P = I + N where N is strictly lower triangular.</p><p>$$N = \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix}$$</p><p>Verify: N² = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 64 & 16 & 0 \end{bmatrix} and N³ = 0 (nilpotent of index 3)</p><p><strong>Step 2:</strong> Apply binomial expansion for P⁵⁰ = (I + N)⁵⁰:</p><p>$$P^{50} = I + 50N + \binom{50}{2}N^2 + 0 + ...$$</p><p>$$P^{50} = I + 50N + 1225N^2$$</p><p><strong>Step 3:</strong> Calculate the entries of P⁵⁰:</p><p>$$P^{50} = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 800 + 78400 & 200 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 79200 & 200 & 1 \end{bmatrix}$$</p><p><strong>Step 4:</strong> Find Q from P⁵⁰ - Q = I, so Q = P⁵⁰ - I:</p><p>$$Q = \begin{bmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 79200 & 200 & 0 \end{bmatrix}$$</p><p>Thus: q₂₁ = 200, q₃₁ = 79200, q₃₂ = 200</p><p><strong>Step 5:</strong> Calculate the ratio:</p><p>$$\frac{q_{31} + q_{32}}{q_{21}} = \frac{79200 + 200}{200} = \frac{79400}{200} = 397$$</p><p>∴ Answer: B (397)</p>
Correct Answer: B