= Solution
The <irreducible representations> can all be specified on the generators:
$$
\boxed{\begin{array}{c|cc}
&R(r)&R(s)\\\hline
\mathbf1&1&1\\
\varepsilon&1&-1\\
R_2&\begin{pmatrix}\alpha&0\\0&\alpha^2\end{pmatrix}&\begin{pmatrix}0&1\\1&0\end{pmatrix}
\end{array}}
$$
The first is the <trivial representation> and the second is the <sign representation>. To see directly that these exhaust the <one-dimensional representations>, their scalar images commute, so the relation $srs=r^{-1}$ forces $r=r^{-1}$. Combining $r^2=1$ with $r^3=1$ gives $r=1$, and then $s$ may be either sign.
Taking <traces> gives the <character table>, in the <conjugacy class> order of <2/a/solution>:
$$
\boxed{\begin{array}{c|rrr}
&I_2&r&s\\\hline
\text{class size}&1&2&3\\\hline
\mathbf1&1&1&1\\
\varepsilon&1&1&-1\\
R_2&2&-1&0
\end{array}}
$$
Here $\alpha+\alpha^2=-1$ gives the middle entry in the last row. The weighted <character inner products> of distinct rows vanish, and the norm of each row is one. Together with <2/b/solution>, this proves completeness of the table.
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