<s>[INST] Why doesn't $A^2=I$ imply $A=\pm I$? Im having trouble believing this T/F Question:
if $\mathrm A^2=I$ then $\mathrm A = \pm \mathrm I$
The answer is False but why?
If the matrix is $\mathrm A = \mathrm I,$ say 
\begin{bmatrix}1& 0\\ 
    0 & 1 \end{bmatrix}
then $\mathrm A^2$ is also that. And if $\mathrm A = -\mathrm I,$ then it is 
\begin{bmatrix}-1 & 0\\
                   0 &-1 
\end{bmatrix}
and that squared is also the same? Where am i going wrong?
 [/INST] for matrices in $\mathbb{M}_2(\mathbb{R})$
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = 
\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix} = 
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} 
$$
requires either $b=c=0, a,d= \pm1$ or $a=-d, bc=1-a^2$
the first type of solution gives $4$ solutions which form a multiplicative group isomorphic to $V_4$ (Klein's Viergruppe).
the second type of solution gives a two-parameter family of matrices:
$$
\begin{pmatrix} k & r \\ \frac{1-k^2}r & -k \end{pmatrix}
$$
in case  $|k| \le 1$ we may also (after re-scaling $r$) write these latter matrices as:
$$
\begin{pmatrix} \cos \theta & r \sin \theta \\ \frac{\sin \theta}r & -\cos \theta \end{pmatrix}
$$
looking at the subfamily with $r=1$ we obtain the solutions
$$
\begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}
$$
</s>