Matrices & Determinants
Types of Matrices
Grade 12
Question:
<p>Let \(A\) be a \(2 \times 2\) matrix with real entries. Let \(I\) be the \(2 \times 2\) identity matrix. Denote by tr\((A)\), the sum of diagonal entries of \(A\). Assume that \(A^2 = I\).</p><p><strong>Statement-1:</strong> If \(A \neq I\) and \(A \neq -I\), then \(\det A = -1\).</p><p><strong>Statement-2:</strong> If \(A \neq I\) and \(A \neq -I\), then \(\text{tr}(A) \neq 0\).</p>
<p>Statement-1 is true, Statement-2 is false.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.</p>
<p>Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: Since A² = I, we have det(A²) = det(I) = 1, which gives (det A)² = 1, so det A = ±1. The Cayley-Hamilton theorem combined with the constraint A ≠ ±I determines which value must hold.
<p><strong>Step 1: Analyze A² = I</strong></p><p>From A² = I, we get (det A)² = det(A²) = det(I) = 1, so det A = ±1.</p><p><strong>Step 2: Apply Cayley-Hamilton Theorem</strong></p><p>The characteristic polynomial of A is: χ(λ) = λ² - tr(A)·λ + det(A) = 0</p><p>Since A² = I, eigenvalues λ satisfy λ² = 1, so λ ∈ {-1, 1}.</p><p><strong>Step 3: Consider cases when A ≠ I and A ≠ -I</strong></p><p>• If both eigenvalues are 1: A = I (contradiction)</p><p>• If both eigenvalues are -1: A = -I (contradiction)</p><p>• If eigenvalues are 1 and -1: det A = (1)(-1) = -1 and tr(A) = 1 + (-1) = 0</p><p><strong>Step 4: Verify the statements</strong></p><p>Statement-1: When A ≠ ±I, eigenvalues must be {1, -1}, giving det A = -1. <strong>TRUE</strong></p><p>Statement-2: When A ≠ ±I, we have tr(A) = 0, so "tr(A) ≠ 0" is <strong>FALSE</strong></p><p><strong>Conclusion:</strong> Only Statement-1 is true.</p><p>∴ Answer: A</p>
Correct Answer: A