Matematică, întrebare adresată de andreealuiza16, 8 ani în urmă

$A=\left(\begin{array}{cc}\sqrt{3} &-1\\1&\sqrt{3} \\\end{array}\right)\\A^{n} = ?$

smartest01: Am facut-o la repezeala pe o ciorna pana la un punct. Ridicand repetat la putere si tot dand factor comun ajungi undeva pe la puterea a 8-a la un factor comun * A. Deci dupa un inerval de ridicari la putere ajungi tot la A, doar mai adaugi in fata factorul comun, care din cate vad va fi o putere a lui 2 cu minus in fata.

Răspunsuri la întrebare

Răspuns de AsakuraHao

Răspuns:

Explicație pas cu pas:

$A = \begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} \dfrac{\sqrt{3}}{2} & \dfrac{-1}{2} \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{pmatrix} = \dfrac{1}{2}\begin{pmatrix} \cos \dfrac{\pi}{6} & -\sin\dfrac{\pi}{6} \\ \sin\dfrac{\pi}{6} & \cos\dfrac{\pi}{6}\end{pmatrix} \\\\$

$A^2 = \dfrac{1}{2^2}\begin{pmatrix}\cos \dfrac{\pi}{6} & -\sin\dfrac{\pi}{6} \\ \sin\dfrac{\pi}{6} & \cos\dfrac{\pi}{6}\end{pmatrix}\cdot \begin{pmatrix}\cos \dfrac{\pi}{6} & -\sin\dfrac{\pi}{6} \\ \sin\dfrac{\pi}{6} & \cos \dfrac{\pi}{6}\end{pmatrix} = \\$

$= \left(\dfrac{1}{2}\right)^2 \cdot \begin{pmatrix} \cos^2 \dfrac{\pi}{6} - \sin^2 \dfrac{\pi}{6} & -2 \cdot \sin\dfrac{\pi}{6}\cdot \cos\dfrac{\pi}{6}\\ 2\cdot \sin\dfrac{\pi}{6}\cdot \cos\dfrac{\pi}{6} & \cos^2\dfrac{\pi}{6}-\sin^2\dfrac{\pi}{6} \end{pmatrix} = \left(\dfrac{1}{2}\right)^2\cdot \begin{pmatrix}\cos\dfrac{2\pi}{6} & -\sin \dfrac{2\pi}{6} \\ \sin\dfrac{2\pi}{6} & \cos \dfrac{2\pi}{6}\end{pmatrix}\\\\\\\texttt{Prin inductie se demonstreaza: }\\$

$\boxed{A^n = \left(\dfrac{1}{2}\right)^n \begin{pmatrix} \cos\dfrac{n\cdot \pi}{6} & -\sin\dfrac{n\cdot \pi}{6} \\ \sin \dfrac{n\cdot \pi}{6} & \cos\dfrac{n\cdot \pi}{6}\end{pmatrix}}$