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 $A({\alpha ,\beta )}^{-1}$ in terms of function of $A$ is

• A. $A\left(\alpha ,-\beta \right)$
• B. $A\left(-\alpha ,-\beta \right)$
• C. $A\left(-\alpha ,\beta \right)$
• D. none of these

Answer 1

The correct option is B $A\left(-\alpha ,-\beta \right)$

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

Finding det about $R_3$

$e^{\beta }({\cos }^2\alpha +{\sin }^2\alpha )=e^{\beta }$

$detA(\alpha ,\beta )=e^{\beta }$

Minor $A(\alpha ,\beta )=\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & e^{\beta }\sin \alpha & 0\\ e^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

Co factor$A(\alpha ,\beta )=\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & e^{\beta }\sin \alpha & 0\\ e^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$

$=\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & {-e}^{\beta }\sin \alpha & 0\\ {-e}^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

Adjoint=Transpose of co factor$=\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & {-e}^{\beta }\sin \alpha & 0\\ {-e}^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

Inverse$=A{(\alpha ,\beta )}^{-1}$

$=\frac{adjA}{detA}$

$=\frac{1}{e^{\beta }}\left[\begin{array}{ccc}{e}^{\beta }\cos \alpha & {-e}^{\beta }\sin \alpha & 0\\ {-e}^{\beta }\sin \alpha & e^{\beta }\cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

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

$\therefore$Option B is correct