Question
If $$AB = 0,$$ then for the matrices \[A = \left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta }&{\cos \theta \sin \theta }\\
{\cos \theta \sin \theta }&{{{\sin }^2}\theta }
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\phi }&{\cos \phi \sin \phi }\\
{\cos \phi \sin \phi }&{{{\sin }^2}\phi }
\end{array}} \right],\] $$\theta - \phi $$ is
A.
an odd number of $$\frac{\pi }{2}$$
B.
an odd multiple of $$\pi$$
C.
an even multiple of $$\frac{\pi }{2}$$
D.
$$0$$
Answer :
an odd number of $$\frac{\pi }{2}$$
Solution :
We have,
\[\begin{array}{l}
AB = \left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta }&{\cos \theta \sin \theta }\\
{\cos \theta \sin \theta }&{{{\sin }^2}\theta }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\phi }&{\cos \phi \sin \phi }\\
{\cos \phi \sin \phi }&{{{\sin }^2}\phi }
\end{array}} \right]\\
= \,\left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta \,{{\cos }^2}\phi + \cos \theta \,\cos \phi \,\sin \theta \,\sin \phi \,{{\cos }^2}\theta \,\cos \phi \,\sin \phi + \cos \theta \,\sin \theta \,{{\sin }^2}\phi }\\
{\cos \theta \,\sin \theta\, {{\cos }^2}\phi + {{\sin }^2}\theta \,\cos \phi \,\sin \phi \,\cos \theta \,\cos \phi \,\sin\theta \,\sin\phi + {\sin^2}\theta \,{{\sin }^2}\phi }
\end{array}} \right]\\
= \,\cos \left( {\theta - \phi } \right)\left[ {\begin{array}{*{20}{c}}
{\cos \theta \cos \phi }&{\cos \theta \sin \phi }\\
{\sin \theta \cos \phi }&{\sin \theta \sin \phi }
\end{array}} \right]
\end{array}\]
$$\eqalign{
& {\text{Since,}}\,AB = 0 \cr
& \therefore \,\cos \left( {\theta - \phi } \right) = 0 \cr} $$
$$\therefore \,\theta - \phi $$ is an odd multiple of $$\frac{\pi }{2}$$