= Solution
On the <normal subgroup> $N=\langle a,b\rangle$, the first diagonal entry is the <linear character> $\chi(a^i b^j)=\varepsilon^i\eta^j$. Its image has order at most three, although $|N|=9$, so its <kernel of a group homomorphism> has order at least three. The second entry is the inverse character and has the same <kernel of a group homomorphism>. Thus
$$
\boxed{\rho\text{ is never faithful}}.
$$
Elements of the other coset have off-diagonal image and cannot be the identity. If $(\varepsilon,\eta)\ne(1,1)$, the <kernel of a group homomorphism> has order exactly three; otherwise it is all of $N$.
When both roots equal one, only $C$ acts nontrivially and its eigenlines spanned by $(1,1)$ and $(1,-1)$ are invariant, so the representation is reducible. Otherwise one of $A,B$ has two distinct <eigenvalues>. Any invariant line would be one of its coordinate eigenlines, but $C$ interchanges these two lines. Hence no invariant line exists, and
$$
\boxed{\rho\text{ is irreducible}\iff(\varepsilon,\eta)\ne(1,1)}.
$$
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