Question

Let$f(\alpha) = \begin{bmatrix} \cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix},$ where $\alpha \in R$,then

$\lbrack f(\alpha)\rbrack^{- 1}$is equal to

• A. $f( - \alpha)$
• B. $f(\alpha^{- 1})$
• C. $f(2\alpha)$
• D. None

Answer 1

Answer: (A)

$f( - \alpha)$

Answer 2

$|f(\alpha)| = \left| \begin{matrix} \cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{matrix} \right| = 1$ , adj of

\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{matrix} \right|$$ $\lbrack f(\alpha)\rbrack^{- 1} = \left| \begin{matrix} \cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{matrix} \right|$ ......(i) and $f( - \alpha) = \left| \begin{matrix} \cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{matrix} \right|$......(ii) From (i) and (ii), $\left\lbrack f(\alpha) \right\rbrack^{- 1} = f\lbrack - \alpha\rbrack$