If A = [ * 20 c ^2 & & ^2 ], B = [ * 20 c ^2 & & ^2 ] and and dif… - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\sin \theta \cos \theta }\\{\sin \theta \cos \theta }&{{{\sin }^2}\theta }\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\phi }&{\sin \phi \cos \phi }\\{\sin \phi \cos \phi }&{{{\sin }^2}\phi }\end{array}} \right]$ and $\theta $ and $\phi $ differs by $\frac{\pi }{2},$ then $AB = $

A
$I$
✓
$O$
C
$-I$
D
None of these

✓

Answer

Correct option: B.

$O$

b
(b) $AB = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\sin \theta \cos \theta }\\{\sin \theta \cos \theta }&{{{\sin }^2}\theta }\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\phi }&{\sin \phi \cos \phi }\\{\sin \phi \cos \phi }&{{{\sin }^2}\phi }\end{array}} \right]$
= $\left[ {\begin{array}{*{20}{c}}{\cos \theta \cos \phi \cos (\theta - \phi )}&{\cos \theta \sin \phi \cos (\theta - \phi )}\\{\cos \theta \sin \phi \cos (\theta - \phi )}&{\sin \theta \sin \phi \cos (\theta - \phi )}\end{array}} \right]$
= $\cos (\theta - \phi )\,\left[ {\begin{array}{*{20}{c}}{\cos \theta \cos \phi }&{\cos \theta \sin \phi }\\{\cos \theta \sin \phi }&{\sin \theta \sin \phi }\end{array}} \right]$
= $O$, $\left( {\because \theta - \phi = \frac{\pi }{2}} \right)$

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