Determinant of identity minus adjacency matrix

Question

Let $M$ be the adjacency matrix of a directed graph $G$. Is there any known relation between $\det(\textrm{id}-M)$ and the cycles of $G$?

It is easy to see that if $G$ is acyclic then this determinant is $1$ (because we can take $M$ to be strictly upper triangular). What does this measure in general?

Somewhat related: How to tell if a directed graph is acyclic from the adjacency matrix

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Using sage I computed these determinants for all the 9608 directed graphs on $5$ vertices and I found that we have $\det(I-M)=1$ iff $G$ is acyclic. Moreover this is the list of possible determinants together with the number of graphs with such determinant:

[(-48, 1), (-40, 1), (-36, 2), (-32, 6), (-30, 3), (-28, 9), (-27, 1),
(-26, 4), (-25, 4), (-24, 36), (-23, 4), (-22, 18), (-21, 9), (-20, 49),
(-19, 12), (-18, 75), (-17, 23), (-16, 144), (-15, 76), (-14, 124),
(-13, 69), (-12, 361), (-11, 116), (-10, 339), (-9, 290), (-8, 676),
(-7, 294), (-6, 917), (-5, 500), (-4, 1195), (-3, 889), (-2, 1144), (-1,
749), (0, 1166), (1, 302)]

Answer 1

According to my calculations in sage, there are 11 graphs (out 156) on six vertices such that $\det(I-M)=1$ and exactly two of these eleven are trees. This shows that is is going to be very difficult to determine information about cycles from the value of $\det(I-M)$. (There is nothing special about the value '1' here, for example example there are 35 graphs with $\det(I-M)=0$ and two of these are trees.)

Note that, trees aside, no useful combinatorial interpretation of $\det(M)$ is known, and there seems to be little reason to expect $\det(I-M)$ to work better.

Answer 2

If you have a graph with disjoint pairs of vertices such that you have just two edges that connect each pair of vertices in opposite directions so that you just have a bunch of disjoint cycles with 2 vertices each, then the determinant will be zero, regardless of how many cycles you have. Similarly, if you an arbitrary graph where two vertices are disconnected from all others and form a cycle among themselves, the determinant will similarly be zero regardless of the structure of the rest of the graph. However I'm not sure if the determinant will always be zero if you have arbitrary cycles, and if it's non-zero, then what it means in terms of the cycles.