Matrices & Determinants
Evaluation of Determinants
Grade 12
Question:
<p>If \(\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\) and \(\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}\), \(x \neq 0\); then for all \(\theta \in \left(0, \dfrac{\pi}{2}\right)\):</p>
<p>\(\Delta_1 + \Delta_2 = -2(x^3 + x - 1)\)</p>
<p>\(\Delta_1 - \Delta_2 = -2x^3\)</p>
<p>\(\Delta_1 + \Delta_2 = -2x^3\)</p>
<p>\(\Delta_1 - \Delta_2 = x(\cos 2\theta - \cos 4\theta)\)</p>
Step-by-Step Solution
Key Concept: Recognize the determinant structure and use row/column operations combined with trigonometric identities (sin²θ + cos²θ = 1) to simplify both Δ₁ and Δ₂ to the same form, showing their relationship is independent of θ.
<p><strong>Step 1:</strong> Expand Δ₁ along the first row or apply row operations. Notice the pattern: the matrix has a special symmetric structure with trigonometric entries satisfying sin²θ + cos²θ = 1.</p><p><strong>Step 2:</strong> Compute Δ₁ by expanding:</p><p>Δ₁ = x[(-x)·x - 1·1] - sinθ[(-sinθ)·x - 1·cosθ] + cosθ[(-sinθ)·1 - (-x)·cosθ]</p><p>= x(-x² - 1) - sinθ(-x·sinθ - cosθ) + cosθ(-sinθ + x·cosθ)</p><p>= -x³ - x + x·sin²θ + sinθ·cosθ - sinθ·cosθ + x·cos²θ</p><p>= -x³ - x + x(sin²θ + cos²θ)</p><p>= -x³ - x + x = <strong>-x³</strong></p><p><strong>Step 3:</strong> Similarly expand Δ₂ using sin2θ and cos2θ where sin²2θ + cos²2θ = 1:</p><p>Δ₂ = x[(-x)·x - 1·1] - sin2θ[(-sin2θ)·x - 1·cos2θ] + cos2θ[(-sin2θ)·1 - (-x)·cos2θ]</p><p>= -x³ - x + x·sin²2θ + sin2θ·cos2θ - sin2θ·cos2θ + x·cos²2θ</p><p>= -x³ - x + x(sin²2θ + cos²2θ)</p><p>= <strong>-x³</strong></p><p><strong>Step 4:</strong> Since Δ₁ = Δ₂ = -x³ for all θ ∈ (0, π/2), we have <strong>Δ₁ = Δ₂</strong> (independent of θ).</p><p>∴ Answer: C</p>
Correct Answer: C