<s>[INST] A $2 \times 2$ matrix $A$ such that $A^n$ is the identity matrix So basically determine a $2 \times 2$ matrix $A$ such that $A^n$ is an identity matrix, but none of $A^1, A^2,..., A^{n-1}$ are the identity matrix. (Hint: Think geometric mappings) 
I don't understand this question at all, can someone help please?
 [/INST] Here's a cute way of looking at it:  let
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$
then
$J^2 = -I. \tag{2}$
From (1) and (2) it follows that
$((\cos \theta)I + (\sin \theta)J)^n = ((\cos n \theta)I + (\sin n \theta)J); \tag{3}$
the proof of (3) is virtually identical to that of de Moivre's formula (see http://en.m.wikipedia.org/wiki/De_Moivre's_formula); indeed, the algebraic maneuvers are essentially the same in either case.   Noting that
$(\cos \theta)I + (\sin \theta)J =  \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end {bmatrix}, \tag{4}$
it follows from (3) that
$\begin{bmatrix} \cos \theta & -\sin \theta \\  \sin \theta & \cos \theta \end{bmatrix}^n = \begin{bmatrix} \cos n \theta & -\sin n \theta \\ \sin n \theta & \cos n \theta \end{bmatrix}. \tag{5}$
Now setting $\theta = \frac{2 \pi}{n}$, we let
$A = \begin{bmatrix} \cos \frac{2  \pi}{n} & -\sin \frac{2\pi}{n} \\  \sin \frac{2\pi}{n} & \cos \frac{2\pi}{n} \end{bmatrix}, \tag{6}$
and we see from (5) that
$A^n = \begin{bmatrix} \cos 2\pi & -\sin 2\pi \\ \sin 2\pi & \cos 2\pi \end{bmatrix} = I, \tag{7}$
whereas for $k$, $1 \le k \le n -1$,
$A^k = \begin{bmatrix} \cos \frac{2\pi k}{n} & -\sin \frac{2\pi k}{n}  \\ \sin \frac{2\pi k}{n} & \cos \frac{2 \pi k}{n} \end{bmatrix} \ne I, \tag{8}$
establishing the desired result.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
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