if A=cos⁡xsin⁡x−sin⁡xcos⁡x, then ATA−1 is

|A|=cos2⁡x+sin2⁡x=1∴A−1=cos⁡x−sin⁡xsin⁡xcos⁡x and AT=cos⁡x−sin⁡xsin⁡xcos⁡xAT⋅A−1=cos⁡x−sin⁡xsin⁡xcos⁡xcos⁡x−sin⁡xsin⁡xcos⁡x=cos2⁡x−sin2⁡x−2sin⁡x⋅cos⁡x2sin⁡x⋅cos⁡xcos2⁡x−sin2⁡x=cos⁡2x−sin⁡2xsin⁡2xcos⁡2x

if A=cos⁡xsin⁡x−sin⁡xcos⁡x, then ATA−1 is

|A|=cos2⁡x+sin2⁡x=1∴A−1=cos⁡x−sin⁡xsin⁡xcos⁡x and AT=cos⁡x−sin⁡xsin⁡xcos⁡xAT⋅A−1=cos⁡x−sin⁡xsin⁡xcos⁡xcos⁡x−sin⁡xsin⁡xcos⁡x=cos2⁡x−sin2⁡x−2sin⁡x⋅cos⁡x2sin⁡x⋅cos⁡xcos2⁡x−sin2⁡x=cos⁡2x−sin⁡2xsin⁡2xcos⁡2x

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if $A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}], then A^{T} A^{- 1}$ is

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$[\begin{array}{cc} \cos 2 x & - \sin 2 x \\ \sin 2 x & \cos 2 x \end{array}]$

$[\begin{array}{cc} \cos 2 x & \sin 2 x \\ - \sin 2 x & - \cos 2 x \end{array}]$

$[\begin{array}{cc} 1 & - \sin 2 x \\ - \sin 2 x & 1 \end{array}]$

$[\begin{array}{cc} 1 & - \sin 2 x \\ \sin 2 x & 1 \end{array}]$

answer is A.

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Detailed Solution

$| A | = \cos^{2} x + \sin^{2} x = 1$
$∴ A^{- 1} = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}] and A^{T} = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
$A^{T} \cdot A^{- 1} = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}] [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
$\begin{array}{l} = [\begin{array}{cc} \cos^{2} x - \sin^{2} x & - 2 \sin x \cdot \cos x \\ 2 \sin x \cdot \cos x & \cos^{2} x - \sin^{2} x \end{array}] \\ = [\begin{array}{cc} \cos 2 x & - \sin 2 x \\ \sin 2 x & \cos 2 x \end{array}] \end{array}$