MHT CET · Maths · Matrices
If \(F(\propto)=\left[\begin{array}{ccc}\cos \propto & -\sin \propto & 0 \\ \sin \propto & \cos \propto & 0 \\ 0 & 0 & 1\end{array}\right]\), where \(\propto \in R\), then \([F(\propto)]^{-1}\)
- A \(\mathrm{F}(-\propto)\)
- B \(F(2 \propto)\)
- C \(F(\propto)\)
- D \(\mathrm{F}(3 \propto)\)
Answer & Solution
Correct Answer
(A) \(\mathrm{F}(-\propto)\)
Step-by-step Solution
Detailed explanation
\(F(\propto)=\left[\begin{array}{ccc}\cos \propto & -\sin \propto & 0 \\ \sin \propto & \cos \propto & 0 \\ 0 & 0 & 1\end{array}\right]\)
\(\therefore|F(\alpha)|=\cos \alpha(\cos \alpha)+\sin \alpha(\sin \alpha)=\cos ^2 \alpha\) \(+\sin ^2 \alpha=1\)
\(\therefore \operatorname{adj}[F(\alpha)]=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]^{ T }\)
\(=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]=\) \(\left[\begin{array}{ccc}\cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1\end{array}\right]\)
\([F(\alpha)]^{-1}=\frac{\operatorname{adj}[F(\alpha)]}{|F(\alpha)|}=F(-\alpha)\)
\(\therefore|F(\alpha)|=\cos \alpha(\cos \alpha)+\sin \alpha(\sin \alpha)=\cos ^2 \alpha\) \(+\sin ^2 \alpha=1\)
\(\therefore \operatorname{adj}[F(\alpha)]=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]^{ T }\)
\(=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]=\) \(\left[\begin{array}{ccc}\cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1\end{array}\right]\)
\([F(\alpha)]^{-1}=\frac{\operatorname{adj}[F(\alpha)]}{|F(\alpha)|}=F(-\alpha)\)
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