Question

If $A=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right];B=\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$ show that $AB$ is zero matrix if $\theta$ and $\varphi$ differ by an odd multiple of $\frac{\pi }{2}$

Answer 1

If $\theta =\frac{(2n+1)\pi }{2}+\varphi$, let $\varphi =x$

else if $\varphi =\frac{(2n+1)\pi }{2}+\theta$, let $\theta =x$

$\cos (\frac{(2n+1)\pi }{2}+x)=\cos (n\pi +\frac{\pi }{2}+x)=\{\begin{array}{c}-\sin x,whenniseven\\ \sin x,whennisodd\end{array}$

$\sin (\frac{(2n+1)\pi }{2}+x)=\sin (n\pi +\frac{\pi }{2}+x)=\{\begin{array}{c}\cos x,whenniseven\\ -\cos x,whennisodd\end{array}$

$\Rightarrow \cos (\frac{(2n+1)\pi }{2}+x)\sin (\frac{(2n+1)\pi }{2}+x)=-\cos x\sin x$

When $\theta =x$,

$A=\left[\begin{array}{cc}{\cos }^{2}x& \cos x\sin x\\ \cos x\sin x& {\sin }^{2}x\end{array}\right]$ and $B=\left[\begin{array}{cc}{\sin }^{2}x& -\cos x\sin x\\ -\cos x\sin x& {\cos }^{2}x\end{array}\right]$

$AB=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

When $\varphi =x$,

$B=\left[\begin{array}{cc}{\cos }^{2}x& \cos x\sin x\\ \cos x\sin x& {\sin }^{2}x\end{array}\right]$ and $A=\left[\begin{array}{cc}{\sin }^{2}x& -\cos x\sin x\\ -\cos x\sin x& {\cos }^{2}x\end{array}\right]$

$AB=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$