If A = [ * 20 c , , , & - & ], then A^2 = - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}{\,\,\,\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$, then ${A^2} = $

A
$\left[ {\begin{array}{*{20}{c}}{\cos 2\alpha }&{\sin 2\alpha }\\{\sin 2\alpha }&{\cos 2\alpha }\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}{\cos 2\alpha }&{ - \sin 2\alpha }\\{\sin 2\alpha }&{\cos 2\alpha }\end{array}} \right]$
✓
$\left[ {\begin{array}{*{20}{c}}{\,\,\,\cos 2\alpha }&{\sin 2\alpha }\\{ - \sin 2\alpha }&{\cos 2\alpha }\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}{ - \cos 2\alpha }&{\sin 2\alpha }\\{ - \sin 2\alpha }&{ - \cos 2\alpha }\end{array}} \right]$

✓

Answer

Correct option: C.

$\left[ {\begin{array}{*{20}{c}}{\,\,\,\cos 2\alpha }&{\sin 2\alpha }\\{ - \sin 2\alpha }&{\cos 2\alpha }\end{array}} \right]$

c
(c) Since ${A^2} = A\,.\,A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$
= $\left[ {\begin{array}{*{20}{c}}{\cos 2\alpha }&{\sin 2\alpha }\\{ - \sin 2\alpha }&{\cos 2\alpha }\end{array}} \right]$.

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