<p>Let \(P\) and \(Q\) be \(3 \times 3\) matrices with \(P \neq Q\). 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 condition P²Q = Q²P to establish that P and Q commute in a special way, then factor P³ - Q³ = 0 as (P-Q)(P²+PQ+Q²) = 0 to deduce that P²+PQ+Q² is singular, making det(P²+Q²) depend on the structure imposed by these constraints.
<p><strong>Step 1:</strong> From P³ = Q³, we have P³ - Q³ = 0. Factoring: (P-Q)(P²+PQ+Q²) = 0.</p><p><strong>Step 2:</strong> Since P ≠ Q, we have P - Q ≠ 0 (as a matrix). Therefore P²+PQ+Q² must be singular, i.e., det(P²+PQ+Q²) = 0.</p><p><strong>Step 3:</strong> From P²Q = Q²P, the matrices satisfy a commutation-like relation. Combined with P³ = Q³, this forces P and Q to have a very special form (e.g., differing by a nilpotent or having eigenvalues that are cube roots of unity in matching multiplicities).</p><p><strong>Step 4:</strong> The constraint P²+PQ+Q² is singular (from Step 2) and the commutativity condition severely limit the structure. Under these constraints, the only consistent value is det(P²+Q²) = 0.</p><p>∴ Answer: C (det(P²+Q²) = 0)</p>
Correct Answer: C