Determinants Matrices Question 113
Question: If AB = O, then for the 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 \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{bmatrix} ,\theta -\phi $ is
Options:
A) An odd number of $ \frac{\pi }{2} $
B) An odd multiple of $ \pi $
C) An even multiple of $ \frac{\pi }{2} $
D) 0
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Answer:
Correct Answer: A
Solution:
- [a] We have, $ AB= \begin{bmatrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{bmatrix} $
$ \begin{bmatrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{bmatrix} $
$ = \begin{bmatrix} {{\cos }^{2}}\theta {{\cos }^{2}}\phi +\cos \theta \cos \phi \sin \theta \sin \phi \\ \cos \theta \sin \theta {{\cos }^{2}}\phi +{{\sin }^{2}}\theta \cos \phi \sin \phi \\ \end{bmatrix} . $
$ \begin{bmatrix} & {{\cos }^{2}}\theta \cos \phi \sin \phi +\cos \theta \sin \theta {{\sin }^{2}}\phi \\ & \cos \theta \cos \phi \sin \theta \sin \phi +{{\sin }^{2}}\theta {{\sin }^{2}}\phi \\ \end{bmatrix} $
$ =\cos (\theta -\phi ) \begin{bmatrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\ \sin \theta \cos \phi & \sin \theta \sin \phi \\ \end{bmatrix} $
Since $ AB=0,\therefore \cos (\theta -\phi )=0 $
$ \therefore \theta -\phi $ is an odd multiple of $ \frac{\pi }{2} $
BETA
