Each node in K$10$ has $9$ edges incident to it while each node in the Petersen graph has $3$

edges incident to it$.$ So it is plausible that K$10$ can be covered perfectly $($the technical word

is tiled$)$ by $3$ Petersen graphs. This means that you can lay down three Petersens on K$10$

so that vertices go to vertices and each edge of K$10$ lies under an edge of exactly one of the

three Petersens. In the exercise below we will use several things we have learned so far to

argue that it is NOT possible to cover K$10$ with $3$ Petersens.

Fact: The adjacency matrix of t