If $A=\begin{bmatrix}
\cos^{2}\alpha & \cos\alpha\sin\alpha\\
\cos\alpha\sin\alpha & \sin^{2}\alpha\\
\end{bmatrix}$ and $B=\begin{bmatrix}
\cos^{2}\beta & \cos\beta\sin\beta\\
\cos\beta\sin\beta & \sin^{2}\beta\\
\end{bmatrix}$ then if $AB=0$, $\alpha-\beta$ is a multiple of
Answer & explanation
Correct answer: option 4
: $AB=\cos(\alpha-\beta)\begin{bmatrix}
\cos\alpha\cos\beta & \cos\alpha\sin\beta\\
\sin\alpha\cos\beta & \sin\alpha\sin\beta\\
\end{bmatrix}$. Hence $AB=0$ gives $\cos(\alpha-\beta)=0$. So $\alpha-\beta$ is a multiple of $\pi/2$.