Q. The product of matrices $A = \begin{bmatrix} \cos^{2}\theta & {\cos\theta\sin\theta} \\ \cos\theta\sin\theta & \sin^{2}\theta \end{bmatrix}$and $B = \begin{bmatrix} \cos^{2}\phi & \cos\varnothing\sin\varnothing \\ \cos\varnothing\sin\varnothing & \sin^{2}\varnothing \end{bmatrix}$ is a null matrix if $\theta - \phi =$

A) $2n\pi,\ n \in Z$
B) $n\frac{\pi}{2}$, $n \in Z$
C) $(2n + 1)\frac{\pi}{2}$, $n\ \in Z$
D) $n\pi,\ n \in Z$
Explanation

(c)
$AB = \begin{bmatrix}
\cos^{2}\theta & \cos\theta\sin\theta \\
\cos\theta\sin\theta & \sin^{2}\theta
\end{bmatrix}\begin{bmatrix}
\cos^{2}\varnothing & \cos\varnothing\sin\varnothing \\
\cos\varnothing\sin\varnothing & \sin^{2}\varnothing
\end{bmatrix}$
$= \begin{bmatrix}
\cos^{2}\theta\cos^{2}\varnothing + \cos\theta\cos\varnothing\sin\theta\sin\varnothing & \cos^{2}{\theta\cos\varnothing\sin\varnothing + \cos\theta\sin{\theta\sin^{2}\varnothing}} \\
\cos\theta\sin\theta\cos^{2}\varnothing + \cos\theta\sin\theta\cos\varnothing\sin\varnothing & \cos\theta\cos\varnothing\sin\theta\sin\varnothing + \sin^{2}\theta\sin^{2}\varnothing
\end{bmatrix}$
$= \begin{bmatrix}
\cos\theta\cos\varnothing(\cos{(\theta - \varnothing)}) & \cos\theta\sin\varnothing\ (\cos{(\theta - \varnothing)}) \\
\sin\theta\cos\varnothing(\cos{(\theta - \varnothing)}) & \sin\theta\sin\varnothing(\cos{(\theta - \varnothing)})
\end{bmatrix}$
$= (\cos{(\theta - \varnothing)})\begin{bmatrix}
\cos{\theta\cos\varnothing} & \cos{\theta\sin\varnothing} \\
\sin\theta\cos\varnothing & \sin{\theta\sin\varnothing}
\end{bmatrix}$
Now, $AB = O \Rightarrow \cos{(\theta - \varnothing)} = 0 \Rightarrow \theta - \varnothing = (2n + 1)\frac{\pi}{2,\ n \in Z}.$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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