<p>Let \(A = \begin{vmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{vmatrix}\). If \(|A^2| = 25\) then \(|\alpha|\) equals</p>
Step-by-Step Solution
Key Concept: For an upper triangular matrix, the determinant is the product of diagonal elements. Use this to find |A|, then apply the property |A²| = |A|² to solve for |α|.
<p><strong>Step 1:</strong> Find determinant of upper triangular matrix A.</p><p>For upper triangular matrix, det(A) = product of diagonal elements</p><p>|A| = 5 × α × 5 = 25α</p><p><strong>Step 2:</strong> Apply the property |A²| = |A|².</p><p>|A²| = (25α)² = 625α²</p><p><strong>Step 3:</strong> Use given condition |A²| = 25.</p><p>625α² = 25</p><p>α² = 25/625 = 1/25</p><p><strong>Step 4:</strong> Solve for |α|.</p><p>|α| = √(1/25) = 1/5</p><p>∴ Answer: C</p>
Correct Answer: C