A) \[A{{(\theta )}^{-1}}=A(\pi -\theta )\]
B) \[A(\theta )+A(\pi +\theta )\]is a null matrix
C) \[A(\theta )\]is invertible for all\[\theta \in R\]
D) \[A{{(\theta )}^{-1}}=A(-\theta )\]
Correct Answer: D
Solution :
Now,\[|A(\theta )|={{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,\]it is invertible \[A(\pi +\theta )=\left[ \begin{matrix} \sin (\pi +\theta ) & i\cos (\pi +\theta ) \\ i\cos (\pi +\theta ) & \sin (\pi +\theta ) \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -\sin \theta & -i\cos \theta \\ -i\cos \theta & -\sin \theta \\ \end{matrix} \right]=-A(\theta )\] \[adj(A(\theta ))=\left[ \begin{matrix} \sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \\ \end{matrix} \right]\] \[\Rightarrow \]\[A{{(\theta )}^{-1}}=\left[ \begin{matrix} \sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \\ \end{matrix} \right]=A(\pi -\theta )\]
You need to login to perform this action.
You will be redirected in
3 sec
