<p>Let \(D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}\) and \(P = \begin{bmatrix} 7 & 0 & 2 \\ 0 & 1 & 0 \\ 2 & 0 & 5 \end{bmatrix}\). Consider \(A = P^{-1}DP\). Find \(\det.(A^2 + A)\).</p>
Step-by-Step Solution
Key Concept: Since A = P⁻¹DP, the determinant of A equals the determinant of D (similarity preserves determinant). Then use det(A² + A) = det(A(A + I)) = det(A)·det(A + I) by factoring out A.
<p><strong>Step 1:</strong> Since A = P⁻¹DP (similar matrices), det(A) = det(D) = 1·2·3 = 6</p><p><strong>Step 2:</strong> Find eigenvalues of A. Since A is similar to D, the eigenvalues of A are {1, 2, 3}. Therefore eigenvalues of (A + I) are {2, 3, 4}.</p><p><strong>Step 3:</strong> For A² + A = A(A + I), use det(A(A + I)) = det(A)·det(A + I) since A and (A + I) have no common eigenvalues (they commute: both come from similarity to D).</p><p><strong>Step 4:</strong> det(A + I) is the product of eigenvalues of (A + I) = 2·3·4 = 24</p><p><strong>Step 5:</strong> det(A² + A) = det(A)·det(A + I) = 6·24 = 144</p><p>∴ <strong>Answer: 144</strong></p>
Correct Answer: 144