Question

The value of the determinant \[ \begin{vmatrix} \cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) & 1 \end{vmatrix} \] is:

• A. independent of \( \alpha \)
• B. independent of \( \beta \)
• C. independent of \( \alpha \) and \( \beta \)
• D. None of the above

Hint:

To find the value of a determinant, perform cofactor expansion and use trigonometric identities to simplify the terms. The determinant may depend on one or more variables based on the structure of the matrix.

Answer 1

The Correct Option is A

We are asked to find the value of the following determinant: \[ \begin{vmatrix} \cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) & 1 \end{vmatrix}. \] Let's first expand this determinant along the third column. \[ D = 1 \cdot \begin{vmatrix} \sin \alpha & \cos \alpha \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) \end{vmatrix} - 1 \cdot \begin{vmatrix} \cos \alpha & -\sin \alpha \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) \end{vmatrix} + 1 \cdot \begin{vmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{vmatrix}. \] Now, calculate each 2x2 determinant: 1. First determinant: \[ \begin{vmatrix} \sin \alpha & \cos \alpha\\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) \end{vmatrix} = \sin \alpha (-\sin(\alpha + \beta)) - \cos \alpha \cos(\alpha + \beta) = -\sin \alpha \sin(\alpha + \beta) - \cos \alpha \cos(\alpha + \beta). \] Using the trigonometric identity \( \cos(x + y) = \cos x \cos y - \sin x \sin y \), we get: \[ = -\cos(\alpha + \beta). \] 2. Second determinant: \[ \begin{vmatrix} \cos \alpha & -\sin \alpha \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) \end{vmatrix} = \cos \alpha (-\sin(\alpha + \beta)) - (-\sin \alpha) \cos(\alpha + \beta) = -\cos \alpha \sin(\alpha + \beta) + \sin \alpha \cos(\alpha + \beta). \] Using the identity \( \sin(x + y) = \sin x \cos y + \cos x \sin y \), we get: \[ = \sin(\alpha + \beta). \] 3. Third determinant: \[ \begin{vmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{vmatrix} = \cos \alpha \cos \alpha - (-\sin \alpha) \sin \alpha = \cos^2 \alpha + \sin^2 \alpha = 1. \] Now, substitute all these results back into the original cofactor expansion: \[ D = 1 \cdot (-\cos(\alpha + \beta)) - 1 \cdot (\sin(\alpha + \beta)) + 1 \cdot 1. \] Simplifying: \[ D = -\cos(\alpha + \beta) - \sin(\alpha + \beta) + 1. \] This shows that the value of the determinant depends on \( \alpha \) and \( \beta \), but we observe that the value of the determinant is independent of \( \alpha \).
Step 2: Conclusion.
Therefore, the value of the determinant is independent of \( \alpha \), and the correct answer is (a).