Question

Let \( F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \), where \( \alpha \in \mathbb{R} \). Then \( [F(\alpha)]^{-1} \) is equal to:

• A. \( F(-\alpha) \)
• B. \( [F(\alpha)]^{-1} \)
• C. \( F(2\alpha) \)
• D. None of these

Hint:

Rotation matrices are orthogonal, meaning their inverses are identical to their transposes.

Answer 1

The Correct Option is A

The matrix \( F(\alpha) \) is a rotation matrix. For a rotation matrix, the inverse is equal to its transpose, i.e., \[ [F(\alpha)]^{-1} = [F(\alpha)]^T, \]

which can also be written as \( F(-\alpha) \), as shown by the following calculation:

\[ F(\alpha) \cdot F(-\alpha) = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = I, \]

where \( I \) is the identity matrix, and \( \cos^2 \alpha + \sin^2 \alpha = 1 \).