<p>Let <em>P</em> and <em>Q</em> be 3 × 3 matrices \(P \neq Q\). If \(P^3 = Q^3\) and \(P^2Q = Q^2P\), then determinant of \((P^2 + Q^2)\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the commutation relation P²Q = Q²P to establish that P and Q satisfy (P-Q)(P²+PQ+Q²) = 0, combined with P³ = Q³ to deduce P² + PQ + Q² is singular, making det(P² + Q²) = 0.
<p><strong>Step 1:</strong> From P³ = Q³, we have P³ - Q³ = 0, which factors as (P - Q)(P² + PQ + Q²) = 0.</p><p><strong>Step 2:</strong> Since P ≠ Q, we cannot have P - Q = 0. Therefore, P² + PQ + Q² = 0 (null matrix).</p><p><strong>Step 3:</strong> From the commutation relation P²Q = Q²P, multiply both sides: if P² + PQ + Q² = 0, then P² = -PQ - Q². This is consistent with our factorization.</p><p><strong>Step 4:</strong> From P² + PQ + Q² = O, we get P² + Q² = -PQ, so det(P² + Q²) = det(-PQ) = (-1)³·det(P)·det(Q) = -det(P)·det(Q).</p><p><strong>Step 5:</strong> However, more directly: since P² + PQ + Q² = O (null matrix), we have P² + Q² = -PQ. Taking determinants: det(P² + Q²) = det(-PQ). But examining the constraint structure: P²(P - Q) + Q²(P - Q) = 0 with P ≠ Q forces det(P² + Q²) = <strong>0</strong>.</p><p>∴ Answer: C (det(P² + Q²) = 0)</p>
Correct Answer: C