Question

Let $F\left(\alpha \right)=\left[\begin{array}{ccc}cos\alpha & -sin\alpha & 0\\ sin\alpha & cos\alpha & 0\\ 0& 0& 1\end{array}\right]$, where $\alpha ϵR$. Then $(F{\left(\alpha \right))}^{-1}$ is equal to

• A. $F{\left(\alpha \right)}^{-1}$
• B. $F(-\alpha )$
• C. $F\left(2\alpha \right)$
• D. none of these

Answer 1

Answer: (B)

$F(-\alpha )$

Given, $F\left(\alpha \right)=A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow |A|={\cos }^2\alpha +{\sin }^2\alpha =1$
$adjA=C^T={\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
Now, $A^{-1}=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
or $A^{-1}=\left[\begin{array}{ccc}\cos (-\alpha )& -\sin (-\alpha )& 0\\ \sin (-\alpha )& \cos (-\alpha )& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow A^{-1}=F(-\alpha )$