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