MCQ
Inverse of the matrix $\left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{ - \sin 2\theta }\\{\sin 2\theta }&{\cos 2\theta }\end{array}} \right]$ is
  • A
    $\left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{ - \sin 2\theta }\\{\sin 2\theta }&{\cos 2\theta }\end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{\sin 2\theta }\\{\sin 2\theta }&{ - \cos 2\theta }\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{\sin 2\theta }\\{\sin 2\theta }&{\cos 2\theta }\end{array}} \right]$
  • $\left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{\sin 2\theta }\\{ - \sin 2\theta }&{\cos 2\theta }\end{array}} \right]$

Answer

Correct option: D.
$\left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{\sin 2\theta }\\{ - \sin 2\theta }&{\cos 2\theta }\end{array}} \right]$
d
(d) Let $A = \left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{ - \sin 2\theta }\\{\sin 2\theta }&{\cos 2\theta }\end{array}} \right]$, $|A| = 1$

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

${A^{ - 1}} = \frac{{adj\,(A)}}{{|A|}} = \left[ {\begin{array}{*{20}{c}}{\cos 2\theta }&{\sin 2\theta }\\{ - \sin 2\theta }&{\cos 2\theta }\end{array}} \right]$.

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