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# If $F(\alpha)=\begin{bmatrix}\cos \alpha&-\sin \alpha&0\\\sin \alpha&\cos \alpha&0\\0&0&1\end{bmatrix}$ $\alpha \in R$ then $[F(\alpha)]^{-1}$ is equal to

$\begin{array}{1 1}(A)\;F(-\alpha)&(B)\;F(\alpha^{-1})\\(C)\;F(2\alpha)&(D)\;\text{None of these}\end{array}$

Can you answer this question?

$F(\alpha)F(-\alpha)=\begin{bmatrix}\cos \alpha&-\sin \alpha&0\\\sin \alpha&\cos \alpha&0\\0&0&1\end{bmatrix}\times =\begin{bmatrix}\cos(- \alpha)&-\sin (-\alpha)&0\\\sin (-\alpha)&\cos \alpha&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\cos \alpha&-\sin \alpha&0\\\sin \alpha&\cos \alpha&0\\0&0&1\end{bmatrix}\begin{bmatrix}\cos \alpha&\sin \alpha&0\\-\sin \alpha&\cos \alpha&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\cos^ \alpha+\sin^2\alpha&\sin \alpha\cos \alpha&0\\\cos\alpha\sin \alpha-\cos\alpha\sin \alpha&\sin^2\alpha+\cos^2\alpha&0\\0&0&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I$
$F(-\alpha)=[F(\alpha)]^{-1}$
Hence (A) is the correct answer.
answered Apr 22, 2014