If $F(\propto)=\left[\begin{array}{ccc}\cos \propto & -\sin \propto & 0 \ \sin \propto & \cos \propto & 0 \ 0 & 0 & 1\end{array} ight]$, where $\propto \in R$, then $[F(\propto)]^{-1}$

Question

If $F(\propto)=\left[\begin{array}{ccc}\cos \propto & -\sin \propto & 0 \ \sin \propto & \cos \propto & 0 \ 0 & 0 & 1\end{array}ight]$, where $\propto \in R$, then $[F(\propto)]^{-1}$

Aicharya · Accepted Answer

$\begin{aligned} & F(\propto)=\left[\begin{array}{ccc} \cos \propto & -\sin \propto & 0 \ \sin \propto & \cos \propto & 0 \ 0 & 0 & 1 \end{array}ight] \ & 	herefore|F(\alpha)|=\cos \alpha(\cos \alpha)+\sin \alpha(\sin \alpha)=\cos ^2 \alpha+\sin ^2 \alpha=1 \ & 	herefore \operatorname{adj}[F(\alpha)]=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{array}ight]^{\mathrm{T}} \end{aligned}$ $\begin{aligned} & =\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{array}ight]=\left[\begin{array}{ccc} \cos (-\alpha) & -\sin (-\alpha) & 0 \ \sin (-\alpha) & \cos (-\alpha) & 0 \ 0 & 0 & 1 \end{array}ight] \ & {[F(\alpha)]^{-1}=\frac{\operatorname{adj}[F(\alpha)]}{|F(\alpha)|}=F(-\alpha)} \end{aligned}$

If $F(\propto)=\left[\begin{array}{ccc}\cos \propto & -\sin \propto & 0 \\ \sin \propto & \cos \propto & 0…

Solution