Question

If $\alpha -\beta =\left(2n+1\right)\frac{\pi }{2};n\in z$ then mod of $\left[\begin{array}{cc}{\cos }^2\alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & {\sin }^2\alpha \end{array}\right]\left[\begin{array}{cc}{\cos }^2\beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & {\sin }^2\beta \end{array}\right]$

• A. $0$
• B. $I$
• C. $2I$
• D. $-I$

Answer 1

Answer: (A)

$0$

$\left[\begin{array}{cc}{\cos }^2\alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & {\sin }^2\alpha \end{array}\right]\left[\begin{array}{cc}{\cos }^2\beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & {\sin }^2\beta \end{array}\right]$

$=\left[\begin{array}{cc}{\cos }^2\alpha {\cos }^2\beta +\cos \alpha \sin \alpha \cos \beta \sin \beta & {\cos }^2\alpha \cos \beta \sin \beta +\cos \alpha \sin \alpha {\sin }^2\beta \\ \cos \alpha \sin \alpha {\cos }^2\beta +{\sin }^2\alpha \cos \beta \sin \beta & \cos \alpha \sin \alpha \cos \beta \sin \beta +{\sin }^2\alpha {\sin }^2\beta \end{array}\right]$

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

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

Given $\alpha -\beta =(2n+1)\frac{\pi }{2}$ so, $\cos (\alpha -\beta )=0$

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