Question

The determinant \( \begin{vmatrix} \lambda & \sin\theta & \cos\theta \\ -\sin\theta & -\lambda & 1 \cos\theta & 1 & \lambda \end{vmatrix} \) is equal to:

• A. \( -\lambda^3 \)
• B. \( \lambda^3 \)
• C. 1
• D. 0

Hint:

When trigonometric functions are involved, look for identities like \(\sin^2\theta + \cos^2\theta = 1\) to simplify the expansion.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the value of the $3 \times 3$ determinant, we expand it along the first row. The formula is \( a(ei - fh) - b(di - fg) + c(dh - eg) \).

Step 2: Key Formula or Approach:
Expand the determinant: \[ D = \lambda \begin{vmatrix} -\lambda & 1 \\ 1 & \lambda \end{vmatrix} - \sin\theta \begin{vmatrix} -\sin\theta & 1 \\ \cos\theta & \lambda \end{vmatrix} + \cos\theta \begin{vmatrix} -\sin\theta & -\lambda \\ \cos\theta & 1 \end{vmatrix} \]

Step 3: Detailed Explanation:
1. First term: \( \lambda(-\lambda^2 - 1) = -\lambda^3 - \lambda \).
2. Second term: \( -\sin\theta(-\lambda \sin\theta - \cos\theta) = \lambda \sin^2\theta + \sin\theta \cos\theta \).
3. Third term: \( \cos\theta(-\sin\theta + \lambda \cos\theta) = -\sin\theta \cos\theta + \lambda \cos^2\theta \).
Summing these: \[ D = -\lambda^3 - \lambda + \lambda \sin^2\theta + \sin\theta \cos\theta - \sin\theta \cos\theta + \lambda \cos^2\theta \] \[ D = -\lambda^3 - \lambda + \lambda (\sin^2\theta + \cos^2\theta) \] Since \( \sin^2\theta + \cos^2\theta = 1 \): \[ D = -\lambda^3 - \lambda + \lambda(1) = -\lambda^3 \]

Step 4: Final Answer:
The determinant is equal to \( -\lambda^3 \).