The relation indicated by this matrix can be described as
$$
(x,x) \quad x = 1,2,3\\
(2,3)\\
(3,1)
$$
In order for the relation to be transitively closed, you need to add $(2,1)$. So, the matrix we want is
$$
\pmatrix{
1&0&0\\
1&1&1\\
1&0&1
}
$$
Lets recall the definition of transitivity. A relation $R \subseteq A \times A$ on $A$ is called transitive, if we have
$$(a,b),(b,c) \in R \Rightarrow (a,c)$$
for all $a,b,c \in A$.
Your initial set is $R = \{(1,2),(2,3),(3,4),(4,1)\}$. The way you described your approach is basically the way to go. Just go through the set and if you find some $(a,b),(b,c)$ in it, add $(a,c)$.
The way the answer is given is a little bit confusing because it already tries to be explanatory :) The thing is, that they mean unions $\cup$ instead of compositions $\circ$. BUT they are writing it as a union to emphasize the steps taken in order to arrive at the solution:
- Step: Look at $R$. Since $(1,2),(2,3) \in R$, we need to add $(1,3)$ to the set. Analogously we have to add $(2,4),(3,1)$ and $(4,2)$. Lets call this new set $R_1$.
- Step: Now we have $R \subset R_1$, but $R_1$ is not yet transitive because $(1,3),(3,4) \in R_1$ but $(1,4)\notin R_1$. Hence, we have to add $(1,4)$ to $R_1$ and so on...
If we keep going we end up with the complete relation $R^+ = A \times A$ where $A = \{1,2,3,4\}$, i.e. $R^+$ contains ALL possible pairs of $1,2,3,4$.
By the way: I really like the idea to visualize the relation as a graph.
Best Answer
Your solution is correct
If our graph is small like this, we can simply draw it (loops at $a$ and $b$ and arrows $a\to c,\ b\to c,\ c\to a$), and then use that we have an arrow $x\to y$ in the transitive closure iff there is a path $x\leadsto y$ in the original graph.