= Solution
The <matrix> $r$ has distinct <eigenvalues> $\alpha,\alpha^2$. Any one-dimensional <invariant subspace> for the defining <representation> would therefore have to be one of its two coordinate <eigenspaces>. But $s$ exchanges those lines, so neither is invariant under the whole <group>. Hence $R_2$ is an <irreducible representation> over the <complex numbers>.
There are three complex <irreducible representations> up to <equivalence of group representations>, because there are three <conjugacy classes>. The <trivial representation> and the <sign representation> already give two inequivalent <one-dimensional representations>. Together with $R_2$, their <dimensions> satisfy the <sum of squares of irreducible representation dimensions>:
$$
\boxed{1^2+1^2+2^2=6=|G|,\qquad\text{dimensions }1,1,2.}
$$
Thus there are no further <irreducible representations>. Alternatively the <character> of $R_2$ is $(2,-1,0)$, and its <character inner product> with itself is $(4+2)/6=1$, independently confirming irreducibility.
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