#### Abstract

A digraph is called oriented if there is at most one arc between two distinct vertices. An oriented graph is called nonsingular (singular) if its adjacency matrix is nonsingular (singular). In this paper, we study the singularity of the oriented outerplanar graph with a given number of inner edges.

#### 1. Introduction

A digraph consists of a nonempty finite set of elements called vertices and a finite set of ordered pairs of distinct vertices called arcs. Its adjacency matrix is the (0, 1)-matrix with if and only if is an arc in . The rank of is the rank of its adjacency matrix and is called singular (respectively, nonsingular) if is singular (respectively, nonsingular). A digraph can be obtained from an undirected graph by replacing each edge of with an arc or with a couple of arcs and . In this case, is called an orientation of and is called the underlying graph of . Two vertices are called adjacent if they are connected by an arc. If there is an arc from vertex to vertex , we indicate this by writing . Note that if whenever then becomes an undirected graph and is symmetric. The rank of is equal to the number of nonzero eigenvalues and is nonsingular if and only if has no zero eigenvalues. We deal with digraphs where loops are not permitted and there is at most one arc between two distinct vertices; these are usually called oriented graphs.

In fact, the singularity of undirected graphs is very important in chemistry. For a conjugated hydrocarbon molecule, its chemical stability depends on the singularity of the corresponding molecule graph [1]. In 1957, Collatz and Sinogowitz [2] posed the problem of characterising all singular (or nonsingular) graphs. The problem seems to be very hard and only a few particular results are known. For undirected graphs, it is easy to see that each complete graph is nonsingular [3]. Cvetković and Gutman [4, 5] showed that if is bipartite and has no cycles with length 0 (mod 4), then , where is the matching number of . This means that is nonsingular if and only if has a perfect matching. Guo et al. [6] determined all nonsingular unicyclic graphs. Li et al. [7] characterised the singularity of line graphs of unicyclic graphs with depth one. Gutman and Sciriha [8] presented a beautiful result showing that the line graph of a tree is either nonsingular or has exactly one zero eigenvalue. For bipartite graphs [9], bicyclic graphs [10], and tricyclic graphs [11], the rank sets of each type have been determined, but the characterisation of nonsingular graphs remains open. For further results on the singularity of undirected graphs, see [12].

There have been a few investigations on the singularity of oriented graphs. Monsalve and Rada [13] proposed the sink-source orientations, which preserve the rank for an oriented graph, and thus are very useful in singularity problems. Based on this, Zhang et al. [14] characterised the oriented graphs with a rank no larger than two. Chen et al. [15] determined the singularities of oriented graphs from the following three classes: graphs in which cycles are vertex disjoint, graphs in which all cycles share exactly one common vertex, and graphs formed by cycles sharing a common path. Ma and Jiang [16] give a necessary condition for an oriented bipartite graph to be nonsingular, characterise nonsingular oriented bipartite graphs when min . A planar graph is called an outerplanar graph if all of its vertices lie on the boundary of the same plane. The edge on the boundary is called the outer edge, the other edges are called the inner edge. In this paper, we aim to characterise the singularity of oriented outerplanar graphs with a given number of inner edges. By controlling the number of inner edges of the outerplanar graph, the singularity when the outer edge is directed and the existence of sink-source vertices are discussed, respectively.

#### 2. Preliminaries

In this section, we make preparations for the main results. An oriented graph is connected if its underlying graph is connected. Two vertices are adjacent if they are connected by an arc. We indicate an arc from to by writing . For a vertex of , we denote by (respectively, ) the set of vertices such that (respectively, ) is an arc of . The out-degree (respectively, in-degree) of is (respectively, ). The degree of in is . If , we call a pendant vertex. We call a sink (respectively, source) vertex if (respectively, ) and a sink-source vertex if . If there is a sink-source vertex there must be a sink vertex and a source vertex. Let be an oriented graph. An oriented graph is called a subgraph of if and . We denote by the subgraph of formed by deleting the arc from . When if and only if for each ordered pair of vertices and in , then is called an induced subgraph of . If is an induced subgraph of , then is the induced subgraph of with . If , we write as for brevity.

Lemma 1 (see [15]). Let be an oriented graph, If has a sink-source vertex, then is singular. In particular, if has a pendant vertex, then is singular.

Lemma 2 (see [14]). Let be an oriented graph and let , be two distinct vertices of .(i)If , then , where is the subgraph of obtained by deleting all for .(ii)If , then , where is the subgraph of obtained by deleting all for .(iii)If and , then , where is the subgraph of obtained by deleting all for and all for .

Lemma 3 (see [15]). Let be an oriented cycle. Then is nonsingular if and only if is directed.

Lemma 4 (see [15]). Let D be an oriented connected graph in which cycles are disjoint. Then D is nonsingular if and only if D satisfies all the following conditions:(i)D has no pendant vertices(ii)Each cycle in D is directed(iii)Two cycles are adjacent in D if and only if they are connected by an arc

#### 3. Oriented Outerplanar Graph with A Given Inner Edge

In this section, we characterise the singularity of the oriented outerplanar graph with a given inner edge.

Theorem 1. Let be an oriented outerplanar graph. Then is singular if and only if there is a pair sink-source vertex on the outer edge of without considering the inner edge, see Figure 1.

Proof. Let and be two neighbours of in . Assume that . Since , so the rows that correspond to the adjacency matrix are the same, the case is singular according to the properties of the determinant. In the case of and , they have a sink vertex or source vertex on the outer edge and are not connected to the inner edge. Then and all of them are singular by Lemma 1.
Let be an oriented outerplanar graph and it is directed on the outer edge. Let and neighbour of and . Assume that . Since , it follows that by Lemma 2. Since is a directed cycle, it is nonsingular by Lemma 3. NowAnd is nonsingular.
Theorem 1 has thus been proved.

Example 1. (1)The adjacency matrix of isSet E- = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is nonsingular.(2)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is nonsingular.(3)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is singular.(4)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is singular.

#### 4. Oriented Outerplanar Graph with Given Two Inner Edges

In this section, we characterise the singularity of the oriented outerplanar graph with given two inner edges. Here are two main cases. First, discuss the singularity of the graph where the outer edge is directed. Then discuss the singularity of the graph where the outer edge has a sink-source vertex. The sink-source vertex mentioned here refers to the sink or source vertex that appears on the outer edge without considering the existence of the inner edge.

Theorem 2. Let be an outerplanar graph and the outer edge is directed. Then is singular if and only if can be obtained from the graph shown in Figure 2.

Proof. In order to prove that Theorem 2 is true, it is proved that the case in Figure 2 is singular, and the other cases are nonsingular. Let be an outerplanar graph with the outer edge is directed. Let the four vertices of the two inner edges. Assume that . Since , so the rows that correspond to the adjacency matrix are the same, by subtracting two rows to get a zero row. Thus and are singular.
Next, we will prove the other cases are nonsingular. When there are no inner edges, each vertex has an out- and in-degree of 1. We want to develop from the vertex outer degree to think and solve problems.Case 1. One vertex has an out-degree of 3 and the others have an out-degree of 1.This case is when two inner edges are assumed to be are connected to the same vertex , and at the same time the vertex is an out-degree of 3 and the others have an out-degree of 1. Therefore, by Lemma 2, two inner edges can be deleted, . Since is a directed cycle, it is nonsingular by Lemma 3. NowAnd is nonsingular.Case 2. Two vertexes have an out-degree of 2 and the others have an out-degree of 1. There are roughly two different cases for the connection of the two inner edges.(1)Two inner edges have terminal vertices connected at the same vertex.In this case, after the inner edge is joined, we will denote the vertices with out-degree 2 as and . And the terminal vertex shared by these two inner edges as . Since only the vertices connecting the initial vertices of the two inner edges are out-degree of 2, the remaining vertices are out-degree of 1. There must be some vertex that has an initial vertex out-degree of 1 and a terminal vertex is . Therefore, an inner edge can be deleted by Lemma 2. Another inner edge can be deleted in the same way according to Lemma 2. So by deleting the inner edge you get a directed cycle, it is nonsingular by Lemma 3.(2)The terminal vertices of the two inner edges are not connected to the same vertex.The proof is the same as (1), after the inner edge is joined, the out-degree of the vertex is 2, and the vertex adjacent to the other terminal vertices will always have an out-degree of 1 to that vertex. Therefore, an inner edge can be deleted by Lemma 2. Another inner edge can be deleted in the same way according to Lemma 2. So by deleting the inner edge, we will get a directed cycle, it is nonsingular by Lemma 3.Theorem 2 has thus been proved.

Example 2. (1)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is singular.(2)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is singular.(3)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is nonsingular.(4)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is nonsingular.

Theorem 3. Let be an outerplanar graph and outer edge with sink-source vertices. Then is nonsingular if and only if can be obtained from the graph shown in Figure 3.

Proof. In order to prove that Theorem 2 is true, it is proved that the case in Figure 3 is nonsingular, and the other cases are singular. For in the case of Figure 3 assume that . Since , it follows that by Lemma 2. Since follow the two cycles are adjacent and connected by an arc. Thus it is nonsingular by Lemma 4. NowAnd is nonsingular.
For in the case of Figure 3 assume that . Since , it follows that by Lemma 2. Since follow the two cycles are adjacent and connected by an arc. Thus it is nonsingular by Lemma 4. NowAnd is nonsingular.
For in the case of Figure 3 assume that . Since , it follows that by Lemma 4. Since follow the two cycles are adjacent and connected by an arc. Thus it is nonsingular by Lemma 4. NowAnd is nonsingular.
Next, we will prove the other cases are singular. We are going to split it up into two cases to prove that the rest of the cases are singular. These are cases where the outer edge has a pair of sink-source vertices and two pairs of sink-source vertices.Case 1. The outer edge has a pair of sink-source vertices.(1)When the two inner edges are connected at the sink vertex and source vertex, respectively, see Figure 4 and . For , let and be two neighbours of , . Since , so the rows that correspond to the adjacency matrix are the same, by subtracting two rows to get a zero row. Thus and are singular. For , let and be two neighbours of , . Since , so the rows that correspond to the adjacency matrix are the same, by subtracting two rows to get a zero row. Thus and are singular.(2)When there is an inner edge connected at the sink-source vertex. When the out-degree of the two adjacent vertices of the sink vertex is 1, then it is singular. When the vertex adjacent to the sink vertex has an out-degree of 2, the other terminal vertices of the other inner edge can be connected to the source vertex or the vertex separated by two or more vertex from the source vertex. At this point, except for the source vertex and one of the adjacent vertices of sink vertices has an out-degree of 2, all other vertices have an out-degree of 1. Therefore, an inner edge can be deleted according to Lemma 2 to obtain graph . Thus graph is singular.Case 2. The outer edge has two pairs of sink-source vertices.When there are two pairs of sink-source vertices on the outside, then the two inner edges need to be connected at the sink-source vertex. Otherwise, the graph has sink-source vertices that are singular, by Lemma 1. When the terminal vertices of the two inner edges are separated by several vertices, the out-degree of the two adjacent vertices at the sink vertex is 1, then the graph is singular.Theorem 3 has thus been proved.

Example 4. (1)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is nonsingular.(2)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is nonsingular.(3)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is singular.(4)The adjacency matrix of isSet E −  = 0, all the eigenvalues of can be obtained by calculation. The results are as follows:The value of is equal to the product of all its eigenvalues, therefore . Thus, is singular.

#### Data Availability

The data used to support the findings of this study are included within the article.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

The research was partially supported by the National Science Foundation of China (Grant No. 12171190) and the Natural Science Foundation of Anhui Province (Grant No. 2008085MA01).