Question

If $A(\alpha ,\beta )=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& e^{\beta }\end{array}\right]$,
then $A(\alpha ,\beta {)}^{-1}$ is equal to

• A. $A(-\alpha ,\beta )$
• B. $A(-\alpha ,-\beta )$
• C. $A(\alpha ,-\beta )$
• D. $A(\alpha ,\beta )$

Answer 1

Answer: (B)

$A(-\alpha ,-\beta )$

$A(\alpha ,\beta )=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& e^{\beta }\end{array}\right]$
$|A|=e^{\beta }$
cofactor matrix is $\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & e^{\beta }\sin \alpha & 0\\ -e^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
$adjA=\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & -e^{\beta }\sin \alpha & 0\\ e^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
$\therefore A(\alpha ,\beta {)}^{-1}=\frac{1}{e^{\beta }}\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & -e^{\beta }\sin \alpha & 0\\ e^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]=A(-\alpha ,-\beta )$
Hence, option B.