If F(alpha) = begin{bmatrix} cos alpha & -sin | vedclass

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(A) Given $F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$.First,we find the d

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$F(-\alpha)$

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(A) Given $F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$.First,we find the determinant $|F(\alpha)|$:$|F(\alpha)| = \cos \alpha(\cos \alpha - 0) - (-\sin \alpha)(\sin \alpha - 0) + 0 = \cos^2 \alpha + \sin^2 \alpha = 1$.Since $|F(\alpha)| = 1 
eq 0$,the inverse exists.The inverse is given by $[F(\alpha)]^{-1} = \frac{1}{|F(\alpha)|} 	ext{adj}(F(\alpha))$.The cofactor matrix $C$ is calculated by finding the cofactor of each element.The adjoint is the transpose of the cofactor matrix:$	ext{adj}(F(\alpha)) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$.We know that $\cos(-\alpha) = \cos \alpha$ and $\sin(-\alpha) = -\sin \alpha$.Substituting these,we get $	ext{adj}(F(\alpha)) = \begin{bmatrix} \cos(-\alpha) & -\sin(-\alpha) & 0 \ \sin(-\alpha) & \cos(-\alpha) & 0 \ 0 & 0 & 1 \end{bmatrix} = F(-\alpha)$.Thus,$[F(\alpha)]^{-1} = F(-\alpha)$.

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

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