Question

If $\left[\begin{array}{cc}{\cos }^2\alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & {\sin }^2\alpha \end{array}\right]$ and $B=\left[\begin{array}{cc}{\cos }^2\beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & {\sin }^2\beta \end{array}\right]$ are two matrices such that the product $AB$ is null matirx, then $\alpha -\beta$ is

• A. $0$
• B. Multiple of $\pi$
• C. An odd number of $\frac{\pi }{2}$
• D. None of the above

Answer 1

Answer: (C)

An odd number of $\frac{\pi }{2}$

Given $AB=0$
$\therefore \left[\begin{array}{cc}{\cos }^2\alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & {\sin }^2\alpha \end{array}\right]\times \left[\begin{array}{cc}{\cos }^2\beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & {\sin }^2\beta \end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\cos \alpha \cos \beta \cos (\alpha -\beta )& \cos \alpha \sin \beta \cos (\alpha -\beta )\\ \cos \beta \sin \alpha \cos (\alpha -\beta )& \sin \alpha \sin \beta \cos (\alpha -\beta )\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \cos (\alpha -\beta )=0\Rightarrow \alpha -\beta$ is an odd multiple of $\pi /2$