MHT CET · Maths · Matrices
\(A(\propto)\left[\begin{array}{cc}\cos \propto & \sin \propto \\ -\sin \propto & \cos \propto\end{array}\right]\), then \(\left[A^2(\propto)\right]^{-1}=\)
- A \(A(\propto)\)
- B \(\mathrm{A}^2(\propto)\)
- C \(A(-2 \propto)\)
- D \(A(2 \propto c)\)
Answer & Solution
Correct Answer
(C) \(A(-2 \propto)\)
Step-by-step Solution
Detailed explanation
\(A(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \Rightarrow| A (\alpha)|=\cos ^2 \alpha\) \(+\sin ^2 \alpha=1\)
\(A ^2(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\)
\(\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & 2 \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha & \cos ^2 \alpha-\sin ^2 \alpha\end{array}\right]=\) \(\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\)
\(\therefore \operatorname{adj}\left[ A ^2(\alpha)\right]\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]\)
\(\therefore\left[ A ^2(\alpha)\right]^{-1}=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right] \quad \ldots\left[\because\left| A ^2\right|=| A |^2=1\right]\)
\(\therefore\left[ A ^2(\alpha)\right]^{-1}=\left[\begin{array}{cc}\cos (-2 \alpha) & \sin (-2 \alpha) \\ -\sin (-2 \alpha) & \cos (-2 \alpha)\end{array}\right]= A (-2 \alpha)\)
\(A ^2(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\)
\(\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & 2 \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha & \cos ^2 \alpha-\sin ^2 \alpha\end{array}\right]=\) \(\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\)
\(\therefore \operatorname{adj}\left[ A ^2(\alpha)\right]\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]\)
\(\therefore\left[ A ^2(\alpha)\right]^{-1}=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right] \quad \ldots\left[\because\left| A ^2\right|=| A |^2=1\right]\)
\(\therefore\left[ A ^2(\alpha)\right]^{-1}=\left[\begin{array}{cc}\cos (-2 \alpha) & \sin (-2 \alpha) \\ -\sin (-2 \alpha) & \cos (-2 \alpha)\end{array}\right]= A (-2 \alpha)\)
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