I came across this question, asked in a competitive exam. It is as follows.
Given a matrix M = \begin{bmatrix}2&1\\1&2\end{bmatrix} what is the value of cos(πM/6)?
I've tried series expansion but I think there is an alternative way doing it, any help is appreciated.
Options given are
\begin{bmatrix}1/2&1\\1&1/2\end{bmatrix} \begin{bmatrix}\sqrt3/4&-\sqrt3/4\\-\sqrt3/4&\sqrt3/4\end{bmatrix} \begin{bmatrix}\sqrt3/4&\sqrt3/4\\\sqrt3/4&\sqrt3/4\end{bmatrix} \begin{bmatrix}1/2&\sqrt3/2\\\sqrt3/2&1/2\end{bmatrix}
It is not clear from the question whether you are looking for the series expansion solution or an alternative.
For the series expansion, remember that
$$\cos(M) = I - \frac{M^2}{2!} + \frac{M^4}{4!} - ...$$
and that $M^n$ can be found by diagonalising.
An alternative solution has been given in part by user akhmeteli on the Physics SE.
It is easy to show that $3$ is an eigenvalue of $M$ with eigenvector $\left( \begin{matrix} 1 \\ 1 \end{matrix} \right)$. By virtue of the series expansion, we know that this is an eigenvector of $\cos(\frac{\pi M}{6})$ and so the corresponding eigenvalue is $\cos(\frac{\pi 3}{6}) = 0$. Thus we are looking for a matrix with determinant $0$ and with an eigenvalue (corresponding to our eigenvector) of $0$. Only one of the solutions satisfies this condition.