Solution 1: Denote by $P$ the transition matrix and by $P^n = (P^n(i,j))_{i,j=1,2,3}$ the $n$-th power of $P$. Let $(X_n)_{n \in \mathbb{N}}$ be a Markov chain with transition matrix $P$. It is known that
$$P^n(i,j) = \mathbb{P}^i (X_n=j), \tag{1}$$
i.e. $P^n(i,j)$ is the probability that the Markov chain starts at $x=i$ and moves to $x=j$ in $n$ steps. However, since $\mathbb{P}^2(X_1 = 2) = 1$, it follows (e.g. from the Markov property) that
$$\mathbb{P}^2(X_n = 2)=1$$
for all $n \in \mathbb{N}$. (Hint: Draw a graph of the chain!) This means that the Markov chain stays at $x=2$ whenever it starts at $x=2$. Therefore, $(1)$ is not satisfied for any $n \in \mathbb{N}$ for $i=2$, $j=1$.
Solution 2: Given two matrices $P,Q$ of the form
$$ P = \begin{pmatrix} p_{11} & 0 & p_{13} \\ 0 & 1 & 0 \\ p_{31} & 0 & p_{33} \end{pmatrix} \qquad \quad Q = \begin{pmatrix} q_{11} & 0 & q_{13} \\ 0 & 1 & 0 \\ q_{31} & 0 & q_{33} \end{pmatrix},$$
it is not difficult to see that $P \cdot Q$ is again of this form, i.e.
$$R := P \cdot Q = \begin{pmatrix} r_{11} & 0 & r_{13} \\ 0 & 1 & 0 \\ r_{31} & 0 & r_{33} \end{pmatrix}. \tag{2}$$
Since the transition matrix $P$ is of the form, we find that $P^2$ is also of this form and, by iteration, that $P^n$ is of the form $(2)$ for any $n \in \mathbb{N}$. Consequently, the matrix is not regular.
