<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 matrix A is similar to D, so det(A) = det(D). Use the property that det(A² + A) = det(A(A + I)) = det(A)·det(A + I), and compute det(A + I) using similarity: A + I is similar to D + I.
<p><strong>Step 1:</strong> Recognize that A = P⁻¹DP means A is similar to D. Similar matrices have the same determinant and the same characteristic polynomial.</p><p><strong>Step 2:</strong> Therefore det(A) = det(D) = 1·2·3 = 6</p><p><strong>Step 3:</strong> Factor: det(A² + A) = det(A(A + I)) = det(A)·det(A + I)</p><p><strong>Step 4:</strong> Since A is similar to D, the matrix (A + I) is similar to (D + I). Therefore:</p><p>det(A + I) = det(D + I) = det<span style='display:inline-block'><span style='border: 1px solid; padding: 2px 4px;'>2 & 0 & 0<br/>0 & 3 & 0<br/>0 & 0 & 4</span></span> = 2·3·4 = 24</p><p><strong>Step 5:</strong> Thus det(A² + A) = det(A)·det(A + I) = 6·24 = 144</p><p>∴ <strong>Answer: 144</strong></p>
Correct Answer: 144