and 
are two matrices such that the product AB is null matrix, then \[\alpha -\beta \] is
A) 0
B) multiple of \[\pi \]
C) an odd multiple of \[\frac{\pi }{2}\]
D) None of the above
Correct Answer: C
Solution :
Since, AB=0 \[\therefore \] \[\left[ \begin{matrix} {{\cos }^{2}}\alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & {{\sin }^{2}}\alpha \\ \end{matrix} \right]\] \[\times \left[ \begin{matrix} {{\cos }^{2}}\beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & {{\sin }^{2}}\beta \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\left[ \begin{matrix} \cos \alpha \cos \beta \cos (\alpha -\beta ) \\ \cos \beta \sin \alpha \cos (\alpha -\beta ) \\ \end{matrix} \right]\] \[\left. \begin{matrix} \cos \alpha \sin \beta \cos (\alpha -\beta ) \\ \sin \alpha \sin \beta \cos (\alpha -\beta ) \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\cos (\alpha -\beta )=0\] So, \[(\alpha -\beta )\]is an odd multiple of \[\frac{\pi }{2}.\]
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