Question

If $A=\begin{bmatrix}\cos ^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin ^2 \alpha\end{bmatrix}$ and $B=\begin{bmatrix}\cos ^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin ^2 \beta\end{bmatrix}$ are two matrices such that $A B=0$, then $\alpha-\beta$ is equal to

• A. 0
• B. $\pi$
• C. $\frac{\pi}{2}$
• D. None of these

Answer 1

Answer: (C)

The given matrices are
$A=\begin{bmatrix}\cos ^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin ^2 \alpha\end{bmatrix}$
and $B=\begin{bmatrix}\cos ^2 \beta & \cos \beta \sin \beta \\\cos \beta \sin \beta & \sin ^2 \beta\end{bmatrix}$
It is given that $A B=0$
$\Rightarrow\begin{bmatrix}\cos ^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin ^2 \alpha\end{bmatrix}$
$\begin{bmatrix}\cos ^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin ^2 \beta\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\cos ^2 \alpha \cos ^2 \beta+\cos \alpha \sin \alpha \cos \beta \sin \beta \cos ^2 \alpha \cos \beta \sin \beta+\cos \alpha \sin \alpha \sin ^2 \beta \\ \cos \alpha \sin \alpha \cos ^2 \beta+\sin ^2 \alpha \cos \beta \sin \beta \cos \alpha \sin \alpha \cos \beta \sin \beta+\sin ^2 \alpha \sin ^2 \beta\end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\cos \alpha \cos \beta \cdot \cos (\alpha-\beta) & \cos \alpha \sin \beta \cos (\alpha-\beta) \\ \cos \beta \sin \alpha \cos (\alpha-\beta) & \sin \alpha \sin \beta \cos (\alpha-\beta)\end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
On equating the corresponding parts, we have
$\cos \alpha \cdot \cos \beta \cos (\alpha-\beta)=0$
$\cos \beta \sin \alpha \cos (\alpha-\beta)=0$
$\cos \alpha \sin \beta \cos (\alpha-\beta)=0$
$\sin \alpha \sin \beta \cos (\alpha-\beta)=0$
$\Rightarrow \cos (\alpha-\beta)=0$
$\therefore \alpha-\beta=\frac{\pi}{2}$