Question

Let $F(\alpha) = \begin{bmatrix} \cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Then$F(\alpha).F(\alpha')$ is equal to

• A. $F(\alpha\alpha')$
• B. $F(\alpha/\alpha')$
• C. $F(\alpha + \alpha')$
• D. $F(\alpha - \alpha')$

Answer 1

Answer: (C)

$F(\alpha + \alpha')$

Answer 2

We have $F(\alpha) = \begin{bmatrix} \cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$,

{\cos\alpha}^{'} & - {\sin\alpha}^{'} & 0 \\ {\sin\alpha}^{'} & {\cos\alpha}^{'} & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$F(\alpha).F(\alpha^{'}) = \begin{bmatrix} \cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 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(\alpha + \alpha^{'}) & - \sin(\alpha + \alpha^{'}) & 0 \\ \sin(\alpha + \alpha^{'}) & \cos(\alpha + \alpha^{'}) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(\alpha + \alpha^{'})$$