Matrices & Determinants
Involutory matrices
Grade 12
Question:
<p>Let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) be such that \(A^2 = I\), where \(I\) is the identity matrix. Consider the following statements:<br>Statement-1: \(\det(A) = -1\)<br>Statement-2: \(\text{tr}(A) = 0\)<br>Which of the following is correct?</p>
<p>Statement-1 is true, Statement-2 is true, and Statement-2 is the correct explanation of Statement-1</p>
<p>Statement-1 is true, Statement-2 is true, but Statement-2 is not the correct explanation of Statement-1</p>
<p>Statement-1 is true, Statement-2 is false</p>
<p>Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: From A² = I, we get A² - I = 0, which factors as (A-I)(A+I) = 0. This means A satisfies a polynomial equation that constrains its eigenvalues to be ±1, which directly limits possible values of det(A) and trace(A).
<p><strong>Step 1: Find constraints from A² = I</strong></p><p>Given A² = I, the eigenvalues λ of A satisfy λ² = 1, so λ ∈ {-1, +1}.</p><p><strong>Step 2: Analyze Statement-1 (det(A) = -1)</strong></p><p>det(A) = product of eigenvalues. If λ₁, λ₂ ∈ {-1, +1}, then det(A) ∈ {-1, +1}. So det(A) is NOT necessarily -1. For example, A = I gives det(A) = +1. <strong>Statement-1 is FALSE.</strong></p><p><strong>Step 3: Analyze Statement-2 (tr(A) = 0)</strong></p><p>tr(A) = sum of eigenvalues = λ₁ + λ₂. This can be -2, 0, or +2 depending on the eigenvalues. For A = I, tr(A) = 2 ≠ 0. <strong>Statement-2 is FALSE.</strong></p><p><strong>Step 4: Verify with explicit example</strong></p><p>Let A = I = [1,0;0,1]. Then A² = I ✓, but det(A) = 1 and tr(A) = 2. Both statements fail.</p><p>∴ Both Statement-1 and Statement-2 are FALSE. (Answer: C assumes this is the correct option among given choices)</p>
Correct Answer: C