A) \[A\,(\theta )\]
B) \[A\,\left( \frac{\theta }{2} \right)\]
C) \[A\,\left( -\,\theta \right)\]
D) \[A\,\left( \frac{-\,\theta }{2} \right)\]
Correct Answer: C
Solution :
\[\because \]\[A(\theta )=\left[ \begin{matrix} 1 & \tan \theta \\ -\tan \theta & 1 \\ \end{matrix} \right]\] Also, \[AB=I.\] \[\Rightarrow \] \[B={{A}^{-1}}\] \[=\frac{1}{1+{{\tan }^{2}}\theta }\left[ \begin{matrix} 1 & -\tan \theta \\ \tan \theta & 1 \\ \end{matrix} \right]\] \[=\frac{1}{{{\sec }^{2}}\theta }\left[ \begin{matrix} 1 & -\tan \theta \\ \tan \theta & 1 \\ \end{matrix} \right]\] \[\Rightarrow ({{\sec }^{2}}\theta )B=\left[ \begin{matrix} 1 & -\tan \theta \\ \tan \theta & 1 \\ \end{matrix} \right]\] \[=A\,(-\theta ).\]
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