I would like to represent the incidence matrix of the following graphs:
I know that each row of the incidence matrix represents a node and each column represents an arch. However, I'm having a hard time finding it. For example, why should we represent two incidence matrices? Also, what are exacly the passages we are doing in order to obtain them?
The followings are the solutions:
$$M_{R}=\left(\begin{array}{lllll}{1} & {1} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0}\end{array}\right), M_{R^{2}}=\left(\begin{array}{lllll}{1} & {1} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right)$$
It makes sense for you to be confused since the solution matrix $M_R$ you posted is the Adjacency matrix of the graph, not the incidence matrix.
You are correct that the incidence matrix relates vertices to edges, but I don’t see a way for you to meaningfully create an incidence matrix for a graph with unlabeled edges.
@David’s comment is correct, but the matrix $M_{R^2}$ should have twos instead of ones because that is what you’d get if you actually squared $M_R$. This makes sense since there are two paths from $a$ to itself, two from $a$ to $b$, etc.
Maybe the solution has ones instead of twos because the ones are meant to indicate there exists a path of length $2$ rather than there being exactly one path of length $2$.