Can someone help find \(\begin{bmatrix}\sqrt3/2 & -1/2 \\1/2 & \sqrt3/2\end{bmatrix}^{2018}\)
+26367 Can someone help find \(\begin{bmatrix}\sqrt3/2 & -1/2 \\1/2 & \sqrt3/2\end{bmatrix}^{2018}\)
I assume the matrix is \(\begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018}\), then this matrix is a rotation matrix counterclockwise through an angle of \(30^\circ\)
\(\begin{array}{|rcll|} \hline \begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018} &=& \begin{bmatrix} \cos(30^\circ) & -\sin(30^\circ) \\\\ \sin(30^\circ) & \cos(30^\circ) \end{bmatrix}^{2018} \\ \hline \end{array} \) Rotation through an angle of \(30^\circ \times 2018\)
\(\begin{array}{|rcll|} \hline 30^\circ \times 2018 &=& 60540^\circ \\ &=& 60540^\circ-168\times 360^\circ \\ &=& \color{red}\mathbf{60^\circ} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \begin{bmatrix} \cos(30^\circ) & -\sin(30^\circ) \\\\ \sin(30^\circ) & \cos(30^\circ) \end{bmatrix}^{2018} &=& \begin{bmatrix} \cos( {\color{red}\mathbf{60^\circ}} ) & -\sin( {\color{red}\mathbf{60^\circ}} ) \\\\ \sin( {\color{red}\mathbf{60^\circ}} ) & \cos( {\color{red}\mathbf{60^\circ}} ) \end{bmatrix} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \begin{bmatrix} \cos( {\color{red}\mathbf{60^\circ}} ) & -\sin( {\color{red}\mathbf{60^\circ}} ) \\\\ \sin( {\color{red}\mathbf{60^\circ}} ) & \cos( {\color{red}\mathbf{60^\circ}} ) \end{bmatrix} &=& \begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt3}{2} \\\\ \dfrac{\sqrt3}{2} & \dfrac{1}{2} \end{bmatrix} \\\\ \begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018} &=& \begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt3}{2} \\\\ \dfrac{\sqrt3}{2} & \dfrac{1}{2} \end{bmatrix} \\ \hline \end{array} \)
