# Explaining this solution regarding finding the transitive closure.

Graph Theory: The question is to find the transitive closure.

Let $G$ be a graph. A directed path $v_1 \rightarrow v_3 \rightarrow v_4$ connects the vertex $v_1$ to $v_4$. $G$ has these additional directed edges: $v_2 \rightarrow v_1$, $v_3 \rightarrow v_2$, and $v_2 \rightarrow v_3$.

\begin{bmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\0&0&0&0\end{bmatrix}

Can you explain how this can be? From what I understand there is a path from $v \rightarrow v_3 \rightarrow v_1$ thus the 1st column first row should be $1$.
Also there is a path from $v_2$ to $v_2$ as well so that is $1$.
And finally there is a path from $v_3$ to $v_3$ so that is $1$.