Question

If $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].J=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]andB=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right],thenB=$

• A. $I\cos \theta +Jsin\theta$
• B. $Isin\theta +Jcos\theta$
• C. $Icos\theta -Jsin\theta$
• D. $-Icos\theta -Jsin\theta$

Answer 1

The correct option is A $I\cos \theta +Jsin\theta$

Given,

$B=\left(\begin{array}{cc}\cos \left(\theta \right)& \sin \left(\theta \right)\\ -\sin \left(\theta \right)& \cos \left(\theta \right)\end{array}\right)$

$I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

$J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$

Now,

$I\cos \theta +J\sin \theta$

$=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\cos \left(\theta \right)+\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\sin \left(\theta \right)$

$=\left(\begin{array}{cc}\cos \left(\theta \right)& 0\\ 0& \cos \left(\theta \right)\end{array}\right)+\left(\begin{array}{cc}0& \sin \left(\theta \right)\\ -\sin \left(\theta \right)& 0\end{array}\right)$

$=\left(\begin{array}{cc}\cos \left(\theta \right)& \sin \left(\theta \right)\\ -\sin \left(\theta \right)& \cos \left(\theta \right)\end{array}\right)$

$\therefore B=I\cos \theta +J\sin \theta$