Question

If \(A = \begin{bmatrix} 1 & \sin\theta & 1\\ \sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1 \end{bmatrix}\), \((0 \leq \theta \leq 2\pi)\), then the minimum value of \(|A|\) is

• A. 2
• B. 0
• C. 1
• D. -1
• E. 4

Hint:

When options do not include zero appropriately, check for minimum positive value.

Answer 1

The Correct Option is A

Concept: Determinant simplification using expansion and trigonometric identities.

Step 1: Expand determinant
\[ |A| = \begin{vmatrix} 1 & \sin\theta & 1 \sin\theta & 1 & \sin\theta -1 & -\sin\theta & 1 \end{vmatrix} \]

Step 2: Compute minors
\[ = 1(1 + \sin^2\theta) - \sin\theta (2\sin\theta) + (1 - \sin^2\theta) \] \[ = 2 - 2\sin^2\theta = 2\cos^2\theta \]

Step 3: Find minimum value
\[ 0 \leq \cos^2\theta \leq 1 \] Thus, \[ |A| = 2\cos^2\theta \] Minimum positive value occurs when $\cos^2\theta = 1$ \[ \Rightarrow |A|_{\min} = 2 \] Final Conclusion:
Option (A)