Matrices & Determinants
Special Matrices and Determinants
Grade 12
Question:
<p>If \(\alpha, \beta, \gamma\) are three real numbers and</p><p>\(A = \begin{bmatrix} 1 & \cos(\alpha-\beta) & \cos(\alpha-\gamma) \\ \cos(\beta-\alpha) & 1 & \cos(\beta-\gamma) \\ \cos(\gamma-\alpha) & \cos(\gamma-\beta) & 1 \end{bmatrix}\),</p><p>then which of following is/are true?</p>
<p>\(A\) is singular</p>
<p>\(A\) is symmetric</p>
<p>\(A\) is orthogonal</p>
<p>\(A\) is not invertible</p>
Step-by-Step Solution
Key Concept: Recognize that A is a Gram matrix formed by unit vectors in ℝ³ with dot products as entries; such matrices are always positive semi-definite, making their determinant ≥ 0 and all eigenvalues ≥ 0. Use the symmetry property cos(x-y) = cos(y-x) and apply determinant expansion or the Cauchy-Schwarz inequality in its matrix form.
<p><strong>Step 1: Recognize the structure</strong> – Matrix A is symmetric with 1's on diagonal and cos(αᵢ-αⱼ) as off-diagonal entries. This is a Gram matrix formed by three unit vectors <strong>u₁, u₂, u₃</strong> where uᵢ·uⱼ = cos(αᵢ-αⱼ).</p><p><strong>Step 2: Apply Gram matrix properties</strong> – For any Gram matrix:</p><ul><li>det(A) ≥ 0 (positive semi-definite)</li><li>det(A) = 0 iff vectors are linearly dependent</li><li>det(A) ≤ 1 with equality iff α = β = γ</li></ul><p><strong>Step 3: Verify eigenvalues</strong> – Since A is positive semi-definite (Gram matrix), all eigenvalues λ ≥ 0. The trace = 3, so Σλᵢ = 3.</p><p><strong>Step 4: Check typical claims</strong>:</p><ul><li><strong>A:</strong> det(A) ≥ 0 ✓ (Gram matrix property)</li><li><strong>B:</strong> All eigenvalues ≥ 0 ✓ (positive semi-definite)</li><li><strong>C:</strong> det(A) = 1 ✗ (only when α = β = γ)</li><li><strong>D:</strong> A is always invertible when α,β,γ distinct ✓ (vectors linearly independent)</li></ul><p>∴ <strong>Answer: A, B, D</strong></p>
Correct Answer: A,B,D