Solutions to $X^3 = I_2$

Question

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?

Answer 1

Any matrices with characteristic polynomial $\lambda^2+\lambda+1$ will do: both eigenvalues satisfy $\lambda^3=1$. For example, for any integer $n$ one can take
$$ X=\left[ \begin{array}{rrr} n, & -1-n-n^2\\ 1, & -1-n \end{array}\right]. $$

Answer 2

The general solution, if that is what you want, is (apart from $I$ itself)

$\begin{pmatrix}a&b\\c&-1-a\\\end{pmatrix}$, where $bc=-(a^2+a+1)$.

Answer 3

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 $bc = -1-a-a^2$