Two parts to this question have answered the first part, stuck on this area on how to work out:
(Note: Matrix A corresponds to a rotation of the plane through theta radians, about the origin, anticlockwise.
ii) Find the matrix power A^3, and thus deduce identities for sin(3theta) and cos(3theta). Using matrix A = [cos theta -sin theta
sin theta cos theta]
iii) Find the least n >= 1 with A^n = I the identity matrix, where
A = [-1 0 0 0
0 0 0 1
0 1 0 0
0 0 1 0].
:o/

What did you get when you calculated A^3? I assume you know how to multiply a 2x2 matrix by itself a couple of times?

As for part (iii), notice that the matrix A is partitioned. The top left partition is simply [-1], while the bottom right is a 3x3 rotation matrix [0 0 1 ; 1 0 0 ; 0 1 0;]. This particular rotation matrix rotates through 120 degrees about the x1=x2=x3 axis. Thus, every three complete rotations (A^3, A^6, A^9, etc.) will result in the identity matrix for that partition. However, since the top left partition is -1, odd exponents will result in that partition being negative. In order to get a +1 in that partition, the exponent must be positive, so the minimum n to get the entire 4x4 matrix to be the identity matrix is 6.