Let F(alpha ) = begin{bmatrix} cos alpha & -s | vedclass

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(A) We are given the matrix $F(\alpha ) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$.

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$F(-\alpha )$

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(A) We are given the matrix $F(\alpha ) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$.To find the inverse $[F(\alpha )]^{-1}$,we check the product $F(\alpha ) \cdot F(-\alpha )$:$F(\alpha ) \cdot F(-\alpha ) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos(-\alpha ) & -\sin(-\alpha ) & 0 \ \sin(-\alpha ) & \cos(-\alpha ) & 0 \ 0 & 0 & 1 \end{bmatrix}$Since $\cos(-\alpha ) = \cos \alpha$ and $\sin(-\alpha ) = -\sin \alpha$,we have:$F(\alpha ) \cdot F(-\alpha ) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$Multiplying the matrices:$= \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha & 0 \ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$Since $F(\alpha ) \cdot F(-\alpha ) = I$,it follows that $[F(\alpha )]^{-1} = F(-\alpha )$.

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

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