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

Question

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

Dr. Abubacker · Accepted Answer

(b)
We have,
$F(\alpha) = \begin{bmatrix}
\cos\alpha & - \sin\alpha & 0 \
\sin\alpha & \cos\alpha & 0 \
0 & 0 & 1
\end{bmatrix} \Rightarrow adj(F(\alpha)) = \left\lbrack - \begin{matrix}
\cos\alpha & \sin\alpha & 0 \
\sin\alpha & \cos\alpha & 0 \
0 & 0 & 1
\end{matrix} ightbrack$
Also,
det$\left( F(\alpha) ight) = 1$
$\Rightarrow \left\lbrack F(\alpha) ightbrack^{- 1} = \begin{bmatrix}
cos( - \alpha) & - sin( - \alpha) & 0 \
sin( - \alpha) & cos( - \alpha) & 0 \
0 & 0 & 1
\end{bmatrix} = F( - \alpha)$

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

Explanation

Reviewed by Dr. Abubacker

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