<s>[INST] Solutions to $X^3 = I_2$ Show that there are an infinite number of solutions to $X^3 = I_2$ in $M_2(\mathbb{Q})$.
$\operatorname{det} (X) = 1$ because $X^3 = I_2$ and $\operatorname{det} (X) \in \mathbb{Q}$
$X^2 = X^{-1}$ and, using Cayley-Hamilton $X^2 = \operatorname{tr}(X) X - I_2$ so $$ X^{-1} = \operatorname{tr}(X)X - I_2 $$ but when i try to solve this equation the solution it gives do not verify $X^3 = I_2$. What am I doing wrong? 
 [/INST] Cayley-Hamilton gives us $X^2-\text{tr}(X)X+I = 0$, from which we find $X(X-\text{tr}(X)I) = -I$, and hence $X^{-1} = \text{tr}(X)I-X$. Letting
\begin{align*}
X = \left( \begin{matrix}
a & b \\
c & d
\end{matrix} \right)
\end{align*}
we get the following system of equations
\begin{align*}
a^2+bc =& d \\
b(a+d) =& -b \\
c(a+d) =& -c \\
d^2+bc =& a
\end{align*}
Which yields either $X = I$ or 
\begin{align*}
X = \left( \begin{matrix}
a & b \\
c & -1-a
\end{matrix} \right)
\end{align*}, where $bc = -1-a-a^2$
</s>