Question

If $AB=O$ where $A=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$ and $B=\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$ then $|\theta -\varphi |$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\pi$

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$A=\left[\begin{array}{cc}{cos}^2\theta & cos\theta sin\theta \\ cos\theta sin\theta & {sin}^2\theta \end{array}\right]$ and
$B=\left[\begin{array}{cc}{cos}^2\varphi & cos\varphi sin\varphi \\ cos\varphi sin\varphi & {sin}^2\varphi \end{array}\right]$
given that $AB$ is a null matrix.
$AB=\left[\begin{array}{cc}{cos}^2\theta & cos\theta sin\theta \\ cos\theta sin\theta & {sin}^2\theta \end{array}\right]\left[\begin{array}{cc}{cos}^2\varphi & cos\varphi sin\varphi \\ cos\varphi sin\varphi & {sin}^2\varphi \end{array}\right]$
$\Rightarrow AB==\left[\begin{array}{cc}\cos \theta \cos \varphi \cdot \cos (\theta -\varphi )& \cos \theta \sin \varphi \cdot \cos (\theta -\varphi )\\ \sin \theta \cos \varphi \cdot \cos (\theta -\varphi )& \sin \theta \sin \varphi \cdot \cos (\theta -\varphi )\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow \cos (\theta -\varphi )=0$
$\therefore |\theta -\varphi |=\frac{\pi }{2}$
Hence, option B.