1. Tardigrade
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  3. Mathematics
  4. Let A=[ beginmatrix cos 2θ sin θ cos θ cos θ sin θ sin 2θ endmatrix ] and B=[ beginmatrix cos 2φ sin φ cos φ cos φ sin φ sin 2φ endmatrix ] then AB=0 if:

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Q. Let $ A=\left[ \begin{matrix} {{\cos }^{2}}\theta & \sin \theta \cos \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right] $ and $ B=\left[ \begin{matrix} {{\cos }^{2}}\phi & \sin \phi \cos \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right] $ then $ AB=0 $ if:

KEAMKEAM 2003

Solution:

$ \therefore $ $ AB=\left[ \begin{matrix} {{\cos }^{2}}\theta & \sin \theta \cos \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right] $ $ \left[ \begin{matrix} {{\cos }^{2}}\phi & \sin \phi \cos \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right] $ $ =\left[ \begin{matrix} {{\cos }^{2}}\theta {{\cos }^{2}}\phi +\sin \theta \cos \theta \cos \phi \sin \phi \\ {{\cos }^{2}}\phi \cos \theta \sin \theta +{{\sin }^{2}}\theta \sin \phi \cos \phi \\ \end{matrix} \right. $ $ \left. \begin{matrix} {{\cos }^{2}}\theta \sin \phi \cos \phi +{{\sin }^{2}}\phi \sin \theta \cos \theta \\ \cos \theta \sin \theta \sin \phi \cos \phi +{{\sin }^{2}}\theta {{\sin }^{2}}\phi \\ \end{matrix} \right] $ $ =\left[ \begin{matrix} \cos \theta \cos \phi \cos (\theta -\phi ) \\ \sin \theta \cos \phi \cos (\theta -\phi ) \\ \end{matrix} \right. $ $ \left. \begin{matrix} \sin \phi \cos \theta \cos (\theta -\phi ) \\ \sin \theta \sin \phi \cos (\theta -\phi ) \\ \end{matrix} \right] $ $ \because $ $ AB=0 $ $ \Rightarrow $ $ \cos (\theta -\phi )=0 $ $ \Rightarrow $ $ \cos (\theta -\phi )=\cos (2n+1)\frac{\pi }{2} $ $ \Rightarrow $ $ \theta =(2n+1)\frac{\pi }{2}+\phi $ where, $ n=0,1,\text{ }2,... $