<p>Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(a, b, c, d \neq 0\). If \(A^2 = A\), then the value of \(|A|\) (determinant of \(A\)) is</p>
Step-by-Step Solution
Key Concept: If A² = A, then A(A - I) = 0, which means A is idempotent. For a non-zero matrix satisfying this, the determinant must equal 0 or 1. Since all entries are non-zero, A cannot be the identity matrix, forcing |A| = 0.
<p><strong>Step 1:</strong> Given A² = A with all entries non-zero.</p><p><strong>Step 2:</strong> Rewrite as A² - A = 0, or A(A - I) = 0.</p><p><strong>Step 3:</strong> Taking determinant: |A(A - I)| = |0| = 0, so |A| · |A - I| = 0.</p><p><strong>Step 4:</strong> This means either |A| = 0 or |A - I| = 0.</p><p><strong>Step 5:</strong> If |A - I| = 0, then A = I (the identity matrix), but this contradicts 'all entries non-zero and distinct' in typical JEE context, or we verify directly: if A = I, then a = d = 1, b = c = 0, violating b, c ≠ 0.</p><p><strong>Step 6:</strong> Therefore |A| = 0.</p><p>∴ Answer: A (|A| = 0)</p>
Correct Answer: A