V = {A, B, C, D, E, F, G, H, I} E = {e1 = (A, B), e2 = (A, C), e3 = (A, D), e4 = (B, C), e5 = (B, F), e6 = (B, H), e7 = (C, D), e8 = (C, E), e9 = (C, F), e10 = (D, E), e11 = (D, I), e12 = (E, F), e13 = (E, G), e14 = (F, G), e15 = (F, H), e16 = (G, H), e17 = (G, I), e18 = (H, I)} W = {22, 9, 12, 35, 36, 34, 4, 33, 30, 18, 23, 24, 39, 25, 21, 19} The order of node (vertex)-edge incidence matrix for the undirected graph is .................. .
9×18
The node (vertex)–edge incidence matrix of a graph has one row for every vertex and one column for every edge. The entry in row v and column e is 1 when edge e is incident on vertex v and 0 otherwise, so in an undirected graph every column holds exactly two 1s. Its order is therefore (number of vertices) × (number of edges); the edge weights in W play no part in the size.
Step 1 — Count the vertices: V = {A, B, C, D, E, F, G, H, I} gives \(|V| = 9\).
Step 2 — Count the edges: E lists e1 to e18, so \(|E| = 18\).
Step 3 — The incidence matrix is \(|V| \times |E| = 9 \times 18\).
A 9×9 matrix is the size of the vertex adjacency matrix, not the incidence matrix. 18×9 is the transpose (edges as rows), which is the edge–vertex arrangement rather than the node–edge one asked for. 18×18 would relate edges to edges, as in the adjacency matrix of the line graph.
9×18 option is correct.
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