JEE Main 2019 — Mathematics Algebra

$$A=[\begin{matrix}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix}]\Rightarrow |A|=|\begin{matrix}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix}|={cos}^{2}\theta +{sin}^{2}\theta =1$$A=[cosθsinθ​−sinθcosθ​]⇒∣A∣=∣cosθsinθ​−sinθcosθ​∣=cos2θ+sin2θ=1

And $adj(A)=[\begin{matrix}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix}]$adj(A)=[cosθ−sinθ​sinθcosθ​]

$$\Rightarrow {A}^{-1}=\frac{1}{|A|}adj(A)=[\begin{matrix}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix}]$$⇒A−1=∣A∣1​adj(A)=[cosθ−sinθ​sinθcosθ​]

$$\Rightarrow {A}^{-2}={({A}^{-1})}^{2}=({A}^{-1})({A}^{-1})$$⇒A−2=(A−1)2=(A−1)(A−1)

$$=[\begin{matrix}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix}][\begin{matrix}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix}]$$=[cosθ−sinθ​sinθcosθ​][cosθ−sinθ​sinθcosθ​]

$$=[\begin{matrix}{cos}^{2}\theta -{sin}^{2}\theta & 2sin\theta cos\theta \\ -2sin\theta cos\theta & {cos}^{2}\theta -sin2\theta \end{matrix}]$$=[cos2θ−sin2θ−2sinθcosθ​2sinθcosθcos2θ−sin2θ​]

$$=[\begin{matrix}cos2\theta & sin2\theta \\ -sin2\theta & cos2\theta \end{matrix}]$$=[cos2θ−sin2θ​sin2θcos2θ​]

It is visible that

$${({A}^{-1})}^{n}=[\begin{matrix}cosn\theta & sinn\theta \\ -sinn\theta & cosn\theta \end{matrix}]$$(A−1)n=[cosnθ−sinnθ​sinnθcosnθ​]

$$\Rightarrow {A}^{-50}={({A}^{-1})}^{50}=[\begin{matrix}cos50\theta & sin50\theta \\ -sin50\theta & cos50\theta \end{matrix}]$$⇒A−50=(A−1)50=[cos50θ−sin50θ​sin50θcos50θ​]

Now at $\theta =\frac{\pi }{12},sin50\theta =sin\frac{50\pi }{12}=sin\frac{\pi }{6}=\frac{1}{2}$θ=12π​,sin50θ=sin1250π​=sin6π​=21​

$$cos50\theta =cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}$$cos50θ=cos6π​=23​​

$$\Rightarrow {A}^{-50}=[\begin{matrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2}\end{matrix}]$$⇒A−50=[23​​2−1​​21​23​​​]