A) \[\left[ \begin{matrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
Given that \[A=\left[ \begin{matrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\] Here, cofactors are \[{{C}_{11}}=\cos 2\theta ,\,{{C}_{12}}=-\sin 2\theta \] \[{{C}_{21}}=\sin 2\theta ,\,{{C}_{22}}=\cos 2\theta \] \[\therefore \] \[|A|=|{{\cos }^{2}}2\theta +{{\sin }^{2}}2\theta |=1\] and inverse of A is \[{{A}^{-1}}\], i.e., \[{{A}^{-1}}=\frac{1}{|A|}\left[ \begin{matrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\] \[=\frac{1}{1}\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
