00:03
We are given graphs, the graphs represent each graph with an adjacency matrix.
00:10
In part a, the graph is k4.
00:17
So recall that k4 is a simple graph with four vertices, and you can edge between every pair of vertices.
00:26
Because it has four vertices, it follows that the matrix, or the adjacency matrix, is a 4x4 matrix.
00:39
Also, the simple graph does not contain any loop.
00:43
There is never an edge from a vertex to itself, and therefore it follows that there are zeros on the main diagonal.
01:11
Also, we know that there is an edge between every pair of vertices, so all of their elements in the matrix should be one.
01:40
And so the adjacency matrix is matrix of rows 0111, 1, 1 ,011, and 1 ,011.
01:51
11, 1101, and 11110.
02:03
Okay.
02:04
In part b, we're asked to find the adjacency matrix for k -14.
02:17
Well, the definition, k -14 has two sets u and v, where u contains one vertex and v contains four vertices.
02:31
Therefore it has a total of five vertices and therefore the adjacency matrix is going to be a five by five matrix now you know that this does not contain any loops my definition this tells us there's never an edge from a vertex to itself so the matrix should contain zeros on the main diagonal we know that there are edges between every element in u and every element in v, so the matrix should contain only ones in the first column and in the first row.
04:31
There are, of course, no edges among any elements of u or among any elements of v, and therefore all other entries should be zero.
05:04
And so the adjacency matrix for this graph is going to be the matrix of rows 0 -1 -1 -1 -1 -1 -0 -0 -0.
05:19
0000, 1 ,000, 1 ,000 ,000, and 1 ,000000, and 1 ,000000.
06:06
Then, in part c, we're asked to find the adjacency matrix for the graph k23.
06:15
This is a bipartisan -type graph.
06:17
For definition, it has two sets of u and v.
06:20
For the magnitude of u is two vertices, and the magnitude of v is three vertices, and there or has a total of five vertices, and therefore the adjacency matrix is going to be a 5x5 matrix again.
06:44
Now, the bipartite graph, so it has no loops.
06:52
This means there's never an edge from a vertex to itself, and therefore the matrix should contain zeros on the main diagonal.
07:11
Now, there are edges between every element in u and every element in v.
07:32
This tells us the first two columns should contain ones in the last three rows.
07:58
So the two columns is for the two and the last three rows is for the three.
08:05
And the first two rows, these should contain ones in the last three columns.
08:36
This is because the matrix is symmetric.
08:38
Finally, there are no edges among any elements of you or any elements of v.
08:54
Therefore, the other entries of the matrix should be 0.
09:03
And therefore, the matrix is the matrix of the rows, 0 -1 -1 -1 -11 -11 -1 -1 -1 -1 -1 -0.
09:25
And 1 -1 -0.
09:38
The matrix looks something like this.
09:41
In part b, we're given the graph c4.
09:59
This is a cyclic graph on four vertices.
10:05
Well, this has four vertices, therefore the adjacency matrix is a 4 by 4 matrix.
10:28
Now, this is a simple graph, so it has no loops.
10:32
As we saw before, this implies that there are zeros on the main diagonal.
10:45
Now, the only edges are b1 to v2, b2 to b3, b3 to b4, and v4 to v1.
11:11
Now, of course, these are undirected edges.
11:14
This means that the matrix should only contain ones for the elements a12, a23, a34, and a41, as well as the elements which are the mears, a21, a32, a43, and a14.
11:51
And so the matrix is matrix of rows 0 -1 -1, 1 -0 -1, 1 -0 -1, and 1 -0 -1, and 1 -0 -1 -0 -1, and 1 -0 -0 -1 -1, and 1 -0 -0 -1 -0.
12:15
So we get this checkerboard pattern.
12:26
Then in part e, we're given the graph w4.
12:33
I'm going to have to delete some of the previous parts.
12:36
Make room for this.
12:39
I'll draw this in red to help distinguish.
13:12
This is the wheel, which is a cycle with four vertices with an additional vertex connected to all other vertices.
13:20
Therefore, it has 4 plus 1 or 5 verticals.
13:25
Vertices and so the adjacency matrix is going to be a 5 by 5 matrix...