= Solution
A <group representation> on a vector space $W$ is a <group homomorphism> $\rho:G\to GL(W)$: $\rho(gh)=\rho(g)\rho(h)$ and $\rho(e)=I$. An <invariant subspace> $U$ obeys $\rho(g)U\subseteq U$ for every $g$. A nonzero representation is <irreducible> if its only invariant subspaces are $0$ and $W$. It is a <faithful representation> if it is injective, equivalently if its <group kernel> is $\{e\}$. Its <character> is $\chi_\rho(g)=\operatorname{tr}\rho(g)$. For inequivalent complex <irreducible representations> of a finite group, <character orthogonality> is
$$
\boxed{\frac1{|G|}\sum_{g\in G}\overline{\chi_\alpha(g)}\chi_\beta(g)=\delta_{\alpha\beta}.}
$$
Use right-to-left composition of <permutations> and assign the matrix labels consistently by
$$
I=e,\quad A=(123),\quad B=(132),\quad C=(13),\quad D=(12),\quad E=(23).
$$
The resulting <Cayley table>, with row multiplying column, is
$$
\begin{array}{c|cccccc}
&I&A&B&C&D&E\\\hline
I&I&A&B&C&D&E\\
A&A&B&I&E&C&D\\
B&B&I&A&D&E&C\\
C&C&D&E&I&A&B\\
D&D&E&C&B&I&A\\
E&E&C&D&A&B&I
\end{array}.
$$
Conjugation relabels the entries of a cycle, preserving its length. Thus the <conjugacy classes> are
$$
\boxed{\{I\},\qquad\{A,B\},\qquad\{C,D,E\}.}
$$
Define the <permutation representation> by $P_\sigma e_j=e_{\sigma(j)}$, or $(P_\sigma)_{ij}=\delta_{i,\sigma(j)}$. Applying it to a basis vector proves $P_\sigma P_\tau=P_{\sigma\tau}$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_\sigma=I$ forces $\sigma(j)=j$ for every $j$, so the representation is faithful.
Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional <invariant subspace> carrying the <trivial representation>. Its <orthogonal complement> is
$$
V_0^\perp=\{(v_1,v_2,v_3):v_1+v_2+v_3=0\},
$$
since the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decomposition
$$
v=\frac{v_1+v_2+v_3}{3}(1,1,1)+\left[v-\frac{v_1+v_2+v_3}{3}(1,1,1)\right].
$$
Thus the <permutation representation> is the <direct sum> of <group representations> on the line and the plane.
To find the <eigenvectors> of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta=e^{2\pi i/3}$, its three <eigenvalues> and eigenspaces are
$$
\boxed{\begin{array}{c|c}
1&\operatorname{span}(1,1,1)\\
\zeta&\operatorname{span}(1,\zeta^2,\zeta)\\
\zeta^2&\operatorname{span}(1,\zeta,\zeta^2)
\end{array}.}
$$
The only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real <eigenvector> in $V_0^\perp$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb R$.
For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb C$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and $\zeta^2$ eigenlines. But $C$ exchanges them:
$$
C(1,\zeta^2,\zeta)=\zeta(1,\zeta,\zeta^2),\qquad
C(1,\zeta,\zeta^2)=\zeta^2(1,\zeta^2,\zeta).
$$
Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is therefore
$$
\boxed{P\cong\mathbf1\oplus W,\qquad \dim W=2,}
$$
where $W$ is the <standard representation of the symmetric group> for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.
Finally, the <sign representation> gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm1$, the trivial and sign choices. Together with the two-dimensional $W$, these give
$$
1^2+1^2+2^2=6=|\Sigma_3|.
$$
The <sum of squares of irreducible representation dimensions> leaves room for no other irreducible. Thus there are \b[exactly two one-dimensional and one two-dimensional complex irreducible representations].
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