Question

Let $A_{\alpha }=\left[\begin{array}{ccc}cos\alpha & -sin\alpha & 0\\ sin\alpha & cos\alpha & 0\\ 0& 0& 1\end{array}\right]$
$|)A_{\alpha }{B}_{\beta }=A_{\alpha +\beta }$ $||)A_{\alpha }{B}_{\beta }=A_{\alpha \beta }$ $|||\left)\right(A_{\alpha }{)}^{-1}=-A_{\alpha }$ IV $\left)\right(A_{\alpha }{)}^{-1}=A_{-\alpha }$
Then which of the above statements is / are correct

• A. only II and III
• B. Only II and IV
• C. only I and III
• D. only I and IV

Answer 1

The correct option is D only I and IV

$A_{\alpha }=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
(i) Consider, $A_{\alpha }{A}_{\beta }=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos \beta & -\sin \beta & 0\\ \sin \beta & \cos \beta & 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos (\alpha +\beta )& -\sin (\alpha +\beta )& 0\\ \sin (\alpha +\beta )& \cos (\alpha +\beta )& 0\\ 0& 0& 1\end{array}\right]$
$=A_{\alpha +\beta }$
Hence, $A_{\alpha }{A}_{\beta }=A_{\alpha +\beta }$
(ii)$A_{\alpha \beta }=\left[\begin{array}{ccc}\cos (\alpha \beta )& -\sin \left(\alpha \beta \right)& 0\\ \sin \left(\alpha \beta \right)& \cos \left(\alpha \beta \right)& 0\\ 0& 0& 1\end{array}\right]$
So, $A_{\alpha }{A}_{\beta }\ne A_{\alpha \beta }$
(iii)Now, we will find ${A_{\alpha }}^{-1}$
Here, $|A_{\alpha }|={\cos }^2\alpha +{\sin }^2\alpha =1$
Now, $adj{A}_{\alpha }=C^T={\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]}^T$
$\Rightarrow A_{\alpha }=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow {A_{\alpha }}^{-1}=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow {A_{\alpha }}^{-1}=\left[\begin{array}{ccc}\cos -\alpha & -\sin -\alpha & 0\\ \sin -\alpha & \cos -\alpha & 0\\ 0& 0& 1\end{array}\right]$
Hence, ${A_{\alpha }}^{-1}\ne -A_{\alpha }$ and ${A_{\alpha }}^{-1}=A_{-\alpha }$