<p><strong>760.</strong> Let \(A = P^{-1}DP\) where \(D = \text{diag}(1,2,3)\). Find \(\det(A^2 + A)\).</p>
Step-by-Step Solution
Key Concept: Since A = P⁻¹DP, the matrix A is similar to D, so they share the same eigenvalues (1,2,3). The determinant of any polynomial in A equals the product of that polynomial evaluated at each eigenvalue.
<p><strong>Step 1:</strong> Since A = P⁻¹DP, matrices A and D are similar and share the same eigenvalues.</p><p>Eigenvalues of D are λ₁ = 1, λ₂ = 2, λ₃ = 3.</p><p><strong>Step 2:</strong> For a polynomial f(x) in matrix A, if A has eigenvalues λ₁, λ₂, λ₃, then det(f(A)) = f(λ₁)·f(λ₂)·f(λ₃).</p><p><strong>Step 3:</strong> Here f(x) = x² + x. Compute f at each eigenvalue:</p><p>f(1) = 1² + 1 = 2</p><p>f(2) = 2² + 2 = 6</p><p>f(3) = 3² + 3 = 12</p><p><strong>Step 4:</strong> Therefore, det(A² + A) = 2 × 6 × 12 = 144</p><p>∴ Answer: <strong>144</strong></p>
Correct Answer: 144