Răspuns:
Explicație pas cu pas:
[tex]A = \begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} \dfrac{\sqrt{3}}{2} & \dfrac{-1}{2} \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{pmatrix} = \dfrac{1}{2}\begin{pmatrix} \cos \dfrac{\pi}{6} & -\sin\dfrac{\pi}{6} \\ \sin\dfrac{\pi}{6} & \cos\dfrac{\pi}{6}\end{pmatrix} \\\\[/tex]
[tex]A^2 = \dfrac{1}{2^2}\begin{pmatrix}\cos \dfrac{\pi}{6} & -\sin\dfrac{\pi}{6} \\ \sin\dfrac{\pi}{6} & \cos\dfrac{\pi}{6}\end{pmatrix}\cdot \begin{pmatrix}\cos \dfrac{\pi}{6} & -\sin\dfrac{\pi}{6} \\ \sin\dfrac{\pi}{6} & \cos \dfrac{\pi}{6}\end{pmatrix} = \\[/tex]
[tex]= \left(\dfrac{1}{2}\right)^2 \cdot \begin{pmatrix} \cos^2 \dfrac{\pi}{6} - \sin^2 \dfrac{\pi}{6} & -2 \cdot \sin\dfrac{\pi}{6}\cdot \cos\dfrac{\pi}{6}\\ 2\cdot \sin\dfrac{\pi}{6}\cdot \cos\dfrac{\pi}{6} & \cos^2\dfrac{\pi}{6}-\sin^2\dfrac{\pi}{6} \end{pmatrix} = \left(\dfrac{1}{2}\right)^2\cdot \begin{pmatrix}\cos\dfrac{2\pi}{6} & -\sin \dfrac{2\pi}{6} \\ \sin\dfrac{2\pi}{6} & \cos \dfrac{2\pi}{6}\end{pmatrix}\\\\\\\texttt{Prin inductie se demonstreaza: }\\[/tex]
[tex]\boxed{A^n = \left(\dfrac{1}{2}\right)^n \begin{pmatrix} \cos\dfrac{n\cdot \pi}{6} & -\sin\dfrac{n\cdot \pi}{6} \\ \sin \dfrac{n\cdot \pi}{6} & \cos\dfrac{n\cdot \pi}{6}\end{pmatrix}}[/tex]
