Question

If $I=I=\left[\begin{array}{c}1\\ 0\end{array}\begin{array}{c}0\\ 1\end{array}\right],j=\left[\begin{array}{c}0\\ -1\end{array}\begin{array}{c}1\\ 0\end{array}\right]andB=\left[\begin{array}{c}cos\theta \\ -sin\theta \end{array}\begin{array}{c}sin\theta \\ cos\theta \end{array}\right],$ then B =

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

Answer 1

The correct option is A $Icos\theta +Jsin\theta$

Given, $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],J=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

and $B=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

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

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