If \[ f(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \] then find \(f(\theta) \cdot f(\alpha)\):
Step-by-step Solution:
Let us multiply the two matrices: \[ f(\theta) f(\alpha) = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} \cos (\theta + \alpha) & -\sin (\theta + \alpha) & 0 \\ \sin (\theta + \alpha) & \cos (\theta + \alpha) & 0 \\ 0 & 0 & 1 \end{bmatrix} \] \[ = f(\theta + \alpha) \]