= Solution
There are three complex <irreducible representations>, because there are three <conjugacy classes>. Two are the <trivial representation> $\mathbf1$ and the <sign representation> $\varepsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the <one-dimensional representations>, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.
The defining two-dimensional <matrix representation> $D$ has <character>
$$
\chi_D(I)=2,\qquad \chi_D(r)=\chi_D(r^2)=\alpha+\alpha^2=-1,\qquad \chi_D(r^ks)=0.
$$
Its <character inner product> with itself is
$$
\langle\chi_D,\chi_D\rangle=\frac{1}{6}(1\cdot4+2\cdot1+3\cdot0)=1,
$$
so it is an <irreducible representation>. Directly, the two coordinate lines are the distinct <eigenspaces> of $r$, and $s$ exchanges them, leaving no common invariant line. The <sum of squares of irreducible representation dimensions> is $1^2+1^2+2^2=6$, proving completeness. In the <conjugacy class> order of <2/a/solution>, the <character table> is
$$
\boxed{\begin{array}{c|rrr}
&I&r&s\\\hline
\text{class size}&1&2&3\\\hline
\mathbf1&1&1&1\\
\varepsilon&1&1&-1\\
D&2&-1&0
\end{array}}
$$
Representatives $r,s$ in this table stand for their full <conjugacy classes>. This lists all <irreducible representations> up to <equivalence of group representations>.
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