A) \[f(-\theta )\]
B) \[f(-\theta )\]
C) \[f(2\theta )\]
D) None of these
Correct Answer: A
Solution :
Given, \[f(\theta )=\left[ \begin{matrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[|f(\theta )=1({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=1\] Now, \[adj\,\,\{f(\theta )\}=\left[ \begin{matrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{\{f(\theta )\}}^{-1}}=\frac{1}{|f(\theta )|}.adj\{f(\theta )\}\] \[=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=f(-\theta )\]
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