Matrices & Determinants
Determinant and range of values
Grade None
Question:
<p>Given \(A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}\) where \(\theta \in \left(\dfrac{3\pi}{4}, \dfrac{5\pi}{4}\right)\), then \(\det(A)\) lies in the interval:</p>
<p>\(\left(\dfrac{3}{2}, 3\right)\)</p>
<p>\((1, 2)\)</p>
<p>\((2, 3)\)</p>
<p>\(\left(\dfrac{1}{2}, 2\right)\)</p>
Step-by-Step Solution
Key Concept: Expand the determinant along the first row, then recognize that the resulting expression simplifies to a form involving sin²θ and can be bounded using the constraint on θ to determine the range of det(A).
<p><strong>Step 1:</strong> Expand det(A) along the first row:</p><p>det(A) = 1·|<begin>matrix</begin>1 & sin θ \\ -sin θ & 1<end>matrix</end>| - sin θ·|<begin>matrix</begin>-sin θ & sin θ \\ -1 & 1<end>matrix</end>| + 1·|<begin>matrix</begin>-sin θ & 1 \\ -1 & -sin θ<end>matrix</end>|</p><p><strong>Step 2:</strong> Calculate the 2×2 determinants:</p><p>|<begin>matrix</begin>1 & sin θ \\ -sin θ & 1<end>matrix</end>| = 1 + sin²θ</p><p>|<begin>matrix</begin>-sin θ & sin θ \\ -1 & 1<end>matrix</end>| = -sin θ + sin θ = 0</p><p>|<begin>matrix</begin>-sin θ & 1 \\ -1 & -sin θ<end>matrix</end>| = sin²θ + 1</p><p><strong>Step 3:</strong> Substitute back:</p><p>det(A) = 1(1 + sin²θ) - sin θ(0) + 1(sin²θ + 1) = 1 + sin²θ + sin²θ + 1 = 2 + 2sin²θ</p><p><strong>Step 4:</strong> Analyze the range for θ ∈ (3π/4, 5π/4):</p><p>In this interval, sin θ ∈ (-1, -1/√2), so sin²θ ∈ (1/2, 1)</p><p>Therefore: det(A) = 2 + 2sin²θ ∈ (3, 4)</p><p>∴ Answer: C</p>
Correct Answer: C