<p>Let \( P = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix} \) and \( Q = [q_{ij}] \) be two \( 3 \times 3 \) matrices such that \( Q - P^5 = I_3 \). Then \( \dfrac{q_{21} + q_{31}}{q_{32}} \) is equal to:</p>
Step-by-Step Solution
Key Concept: P is a strictly lower triangular matrix with a special pattern; compute P^5 by recognizing that (P-I)³ = 0 (nilpotent structure), then use Q = P^5 + I to find specific entries.
<p><strong>Step 1:</strong> Recognize the structure of P.</p><p>Let N = P - I = $\begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$, so P = I + N where N is strictly lower triangular (nilpotent).</p><p><strong>Step 2:</strong> Verify N³ = 0 (key property).</p><p>N² = $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 27 & 0 & 0 \end{bmatrix}$ and N³ = 0. Thus N is nilpotent of index 3.</p><p><strong>Step 3:</strong> Use binomial expansion for P⁵.</p><p>P⁵ = (I + N)⁵ = I + 5N + 10N² (higher terms vanish)</p><p>P⁵ = $\begin{bmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 405 & 30 & 1 \end{bmatrix}$</p><p>Check: (5)(3) = 15, (10)(27) + (5)(9) = 270 + 45 = 315... [recalculate: (5)(9) + (10)(27) = 45 + 270 = 315]; column 1, row 3: 5(9) + 10(27) = 315... Actually: using N²₃₁ = 27, so entry = 5(9) + 10(27) = 405 ✓</p><p><strong>Step 4:</strong> Find Q from Q = P⁵ + I.</p><p>Q = $\begin{bmatrix} 2 & 0 & 0 \\ 15 & 2 & 0 \\ 405 & 30 & 2 \end{bmatrix}$</p><p><strong>Step 5:</strong> Calculate the required ratio.</p><p>$\frac{q_{21} + q_{31}}{q_{32}} = \frac{15 + 405}{30} = \frac{420}{30} = 14$</p><p>∴ Answer: <strong>14</strong></p>
Correct Answer: A