Question

nu gasesc nici o metoda de rezolvare !

Answer 1

[tex]A= \left[\begin{array}{ccc}\sqrt{3}&-1\\1&\sqrt 3\end{array}\right]=2 \left[\begin{array}{ccc}\dfrac{\sqrt{3}}{2}&\dfrac{-1}{2}\\\dfrac{1}{2}&\dfrac{\sqrt 3}{2}\end{array}\right]=2 \left[\begin{array}{ccc}\cos \dfrac{\pi}{6}&-\sin \dfrac{\pi}{6}\\\sin\dfrac{\pi}{6}&\cos \dfrac{\pi}{6}\end{array}\right]\\ \text{Se demonstreaza usor ca }B^n= \left[\begin{array}{ccc}\cos (na)&-\sin(na)\\\sin(na)&\cos(na)\end{array}\right],\text{unde}  [/tex][tex]B= \left[\begin{array}{ccc}\cos a&-\sin a\\\sin a&cos a\end{array}\right]\text{(eventual prin inductie)}.\\ \text{Atunci } A^{36}=2^{36} \left[\begin{array}{ccc}\cos 6\pi&-\sin 6\pi\\\sin 6\pi&\cos 6\pi\end{array}\right]=2^{36}\cdot I_2[/tex]