If A=cos⁡θ−sin⁡θsin⁡θcos⁡θ, then the matrix A−50 when θ=π12, is equal to

We have __A=cos⁡θ−sin⁡θsin⁡θcos⁡θ∴ |A|=cos2θ+sin2θ=1 And adj A=cos⁡θsin⁡θ−sin⁡θcos⁡θ∵ If A=a bc d, then adj A=d−b−ca⇒ A−1=cosθsinθ−sinθcosθ ∵A−1=adj A|A| Note that, A−50=A−150 Now, A−2=A−1A−1⇒A−2=cos⁡θsin⁡θ−sin⁡θcos⁡θcos⁡θsin⁡θ−sin⁡θcos⁡θ=cos2⁡θ−sin2⁡θcos⁡θsin⁡θ+sin⁡θcos⁡θ−cos⁡θsin⁡θ−cos⁡θsin⁡θ−sin2⁡θ+cos2⁡θ=cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ Also, A−3=A−2A−1A−3=cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θ=cos⁡3θsin⁡3θ−sin⁡3θcos⁡3θ Similarly, A−50=cos⁡50θsin⁡50θ−sin⁡50θcos⁡50θ=cos⁡256πsin⁡256π−sin⁡256πcos⁡256π when θ=π12=cos⁡π6sin⁡π6−sin⁡π6cos⁡π6∵cos⁡25π6=cos⁡4π+π6=cos⁡π6 and sin⁡25π6=sin⁡4π+π6=sin⁡π6=3212−1232

If A=cos⁡θ−sin⁡θsin⁡θcos⁡θ, then the matrix A−50 when θ=π12, is equal to

We have __A=cos⁡θ−sin⁡θsin⁡θcos⁡θ∴ |A|=cos2θ+sin2θ=1 And adj A=cos⁡θsin⁡θ−sin⁡θcos⁡θ∵ If A=a bc d, then adj A=d−b−ca⇒ A−1=cosθsinθ−sinθcosθ ∵A−1=adj A|A| Note that, A−50=A−150 Now, A−2=A−1A−1⇒A−2=cos⁡θsin⁡θ−sin⁡θcos⁡θcos⁡θsin⁡θ−sin⁡θcos⁡θ=cos2⁡θ−sin2⁡θcos⁡θsin⁡θ+sin⁡θcos⁡θ−cos⁡θsin⁡θ−cos⁡θsin⁡θ−sin2⁡θ+cos2⁡θ=cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ Also, A−3=A−2A−1A−3=cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θ=cos⁡3θsin⁡3θ−sin⁡3θcos⁡3θ Similarly, A−50=cos⁡50θsin⁡50θ−sin⁡50θcos⁡50θ=cos⁡256πsin⁡256π−sin⁡256πcos⁡256π when θ=π12=cos⁡π6sin⁡π6−sin⁡π6cos⁡π6∵cos⁡25π6=cos⁡4π+π6=cos⁡π6 and sin⁡25π6=sin⁡4π+π6=sin⁡π6=3212−1232

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$If A = [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}], then the matrix A^{- 50} when θ = \frac{π}{12}, is equal to$

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$[\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

$[\begin{matrix} \frac{\sqrt{3}}{2} & - \frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$

$[\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$

$[\begin{matrix} \frac{1}{2} & - \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

answer is C.

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Detailed Solution

$We \underline{\underline{have}} A = [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}]$

$∴ | A | = \cos^{2} θ + \sin^{2} θ = 1$

$And adj A = [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}]$

$(∵ If A = [\begin{array}{ll} a b \\ c d \end{array}], then adj A = [\begin{array}{cc} d & - b \\ - c & a \end{array}])$

$\Rightarrow A^{- 1} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] (∵ A^{- 1} = \frac{a d j A}{| A |})$

$Note that, A^{- 50} = {(A^{- 1})}^{50}$

$Now, A^{- 2} = (A^{- 1}) (A^{- 1})$

$\begin{array}{l} \Rightarrow A^{- 2} = [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] \\ = [\begin{array}{cc} \cos^{2} θ - \sin^{2} θ & \cos θ \sin θ + \sin θ \cos θ \\ - \cos θ \sin θ - \cos θ \sin θ & - \sin^{2} θ + \cos^{2} θ \end{array}] \\ = [\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array}] \end{array}$

$Also, A^{- 3} = (A^{- 2}) (A^{- 1})$

$\begin{array}{l} A^{- 3} = [\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array}] [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] \\ = [\begin{array}{cc} \cos 3 θ & \sin 3 θ \\ - \sin 3 θ & \cos 3 θ \end{array}] \end{array}$

$Similarly, A^{- 50} = [\begin{array}{cc} \cos 50 θ & \sin 50 θ \\ - \sin 50 θ & \cos 50 θ \end{array}]$

$\begin{array}{l} = [\begin{array}{cc} \cos \frac{25}{6} π & \sin \frac{25}{6} π \\ - \sin \frac{25}{6} π & \cos \frac{25}{6} π \end{array}] (when θ = \frac{π}{12}) \\ = [\begin{array}{cc} \cos \frac{π}{6} & \sin \frac{π}{6} \\ - \sin \frac{π}{6} & \cos \frac{π}{6} \end{array}] [\begin{array}{ll} ∵ \cos (\frac{25 π}{6}) = \cos (4 π + \frac{π}{6}) = \cos \frac{π}{6} \\ and \sin (\frac{25 π}{6}) = \sin (4 π + \frac{π}{6}) = \sin \frac{π}{6} \end{array}] \\ = [\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{array}] \end{array}$