Matrices & Determinants
Evaluation of Determinants
Grade 12
Question:
<p>If \(A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}\); then for all \(\theta \in \left(\dfrac{3\pi}{4}, \dfrac{5\pi}{4}\right)\), \(\det(A)\) lies in the interval:</p>
<p>\(\left[1, \dfrac{5}{2}\right]\)</p>
<p>\(\left[\dfrac{5}{2}, 4\right]\)</p>
<p>\(\left(0, \dfrac{3}{2}\right]\)</p>
<p>\(\left(\dfrac{3}{2}, 3\right]\)</p>
Step-by-Step Solution
Key Concept: Expand the determinant using the first row, recognize that the resulting expression is a function of sin(θ), and determine its range over the given interval where sin(θ) ∈ (-1, -√2/2).
<p><strong>Step 1:</strong> Expand det(A) along the first row:</p><p>det(A) = 1·(1 - (-sin²θ)) - sin(θ)·(-sin(θ) - (-sin(θ))) + 1·(sin²(θ) + 1)</p><p>= 1 + sin²(θ) - sin(θ)·0 + sin²(θ) + 1</p><p>= 2 + 2sin²(θ)</p><p><strong>Step 2:</strong> For θ ∈ (3π/4, 5π/4), determine the range of sin(θ):</p><p>At θ = 3π/4: sin(θ) = √2/2</p><p>At θ = π: sin(θ) = 0</p><p>At θ = 5π/4: sin(θ) = -√2/2</p><p>Therefore sin(θ) ∈ (-1, -√2/2) in the given interval (the minimum is -1 at θ = 3π/2, but this is outside our interval).</p><p><strong>Step 3:</strong> Find the range of sin²(θ):</p><p>Since sin(θ) ∈ (-1, -√2/2), we have sin²(θ) ∈ (1/2, 1)</p><p><strong>Step 4:</strong> Substitute into det(A) = 2 + 2sin²(θ):</p><p>det(A) ∈ (2 + 2(1/2), 2 + 2(1)) = (3, 4)</p><p>∴ Answer: D</p>
Correct Answer: D