How Does the Determinant of a Covariance Matrix Relate to System Entropy?

[trimota]

Hi,
I'm working on a project of calculating the complexity of a system which is represented by a matrix which shows the connective between each element. Normally this matrix is symmetry and the diagonal elements are 0. From the past paper, it's ln((2*pi*exp(n))*det(COV(X))) where X is the system and COV(X) means the covariance matrix which is calculated by:Q= the inverse of (1-CON(X)), COV(X)=(transpose of Q)*Q. Even though these formulays are given.
I'm still confused about how to relate the determinant of COV(X) with the entropy of the system. (The detail process)And what can we do about a matrix with a determinant of COV(X) is 0.

 

[Aaron Barnard]

The relationship between the determinant of the covariance matrix and the entropy of the system is that the determinant of the covariance matrix essentially represents the variance of the system. The higher the variance, the more complex the system. Therefore, the greater the value of the determinant of the covariance matrix, the greater the entropy of the system.

In the case of a matrix with a determinant of COV(X) equal to zero, it suggests that there is no variance in the system and thus the entropy is also 0. Furthermore, this could indicate that the matrix is singular, meaning that it is not invertible and cannot be used to determine the entropy of the system.