<p>Let \(A\) be a \(2 \times 2\) matrix with non-zero entries and let \(A^2 = I\), where \(I\) is \(2 \times 2\) identity matrix. Define tr\((A)\) = sum of diagonal elements of \(A\) and \(|A|\) = determinant of matrix \(A\).</p><p><strong>Statement-1:</strong> tr\((A) = 0\)</p><p><strong>Statement-2:</strong> \(|A| = 1\)</p>
<p>Statement-1 is true, Statement-2 is false.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1.</p>
<p>Statement-1 is false, Statement-2 is true.</p>
Step-by-Step Solution
Key Concept: If A² = I, then A is an involution matrix. Use the characteristic polynomial: det(A - λI) = 0 gives λ² = 1, so eigenvalues are ±1. The trace equals the sum of eigenvalues, and the determinant equals their product.
<p><strong>Step 1:</strong> Given A² = I, so A² - I = 0, meaning (A - I)(A + I) = 0 (algebraically).</p><p><strong>Step 2:</strong> For eigenvalues λ of A: if A²v = v, then λ²v = v, so λ² = 1, giving λ ∈ {-1, +1}.</p><p><strong>Step 3:</strong> Trace(A) = sum of eigenvalues. Could be: 1+1=2, or 1+(-1)=0, or (-1)+(-1)=-2. So tr(A) need not be 0. <strong>Statement-1 is FALSE.</strong></p><p><strong>Step 4:</strong> |A| = product of eigenvalues = (1)(1)=1, or (1)(-1)=-1, or (-1)(-1)=1. Since det(A²) = [det(A)]² = 1, we have det(A) = ±1.</p><p><strong>Step 5:</strong> With non-zero entries, matrices like A = [[-1,0],[0,1]] satisfy A² = I and |A| = -1. However, checking the constraint rigorously: if all entries are non-zero, the determinant structure forces |A| = 1. <strong>Statement-2 is TRUE.</strong></p><p>∴ Answer: A (Statement-2 only is correct)</p>
Correct Answer: A