The result of raising the enclosed matrix to the 120th matrix power is?

Compute the determinant of the matrix and simplify. It is 1. This does not select any of the answers. It does indicate that it is unitary a.k.a. rotation matrix.

Since a rotation matrix is of the general form $\left( \begin{array}{cc} \cos (\alpha ) & -\sin (\alpha ) \\ \sin (\alpha ) & \cos (\alpha ) \\ \end{array} \right)$ , using a calculator, compute the numeric value of the matrix's upper right entry ( $-0.052335956*10^{-2}$ ).

Compute the arcsin of ( $-0.052335956*10^{-2}$ ) to get $-0.0523598775598*10^{-2}$ . Let us divide that in $2\pi$ and get $-120.$ or $-3^{\circ}$ .

One could compute the ``sine of 3 degrees in radicals'' using half-angle, sum and difference trigonometric identities; but, let us just search the Internet for that phrase instead. The English Wikipedia page supplies that as $\frac{1}{16} \left(\left(\sqrt{3}+1\right) \left(\sqrt{10}-\sqrt{2}\right)+2 \left(1-\sqrt{3}\right) \sqrt{\sqrt{5}+5}\right)$ .

Is the Wikipedia value the negative of the upper right value in the given matrix? Yes.

Therefore the $120^{th}$ matrix power is a rotation matrix for a $-360^{\circ}$ . Since the rotation angle can be reduced to the remainder after division by $360^{\circ}$ , that is a rotation matrix with no rotation or the identity matrix $\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ .

A identity matrix is the most specific correct answer. The only answer not correct at all is ``very messy matrix.''