TITLE: Why is a singular matrix rare?
QUESTION [11 upvotes]: I am exploring patterns of integers in $n\times n$ matrices.  I have two matrices that have a determinant of $0$ and a circulant matrix that has positive determinants that differ depending on $n$.
I snipped this from Wikipedia and bolded the important part:

A square matrix that is not invertible is called singular or degenerate.
A square matrix is singular if and only if its determinant is 0. Singular
matrices are rare in the sense that if you pick a random square matrix
over a continuous uniform distribution on its entries, it will almost
surely not be singular.

The Good:
$$\left(
\begin{array}{ccccc}
 1 & 2 & 3 & 4 & 5 \\
 1 & 2 & 3 & 4 & 5 \\
 1 & 2 & 3 & 4 & 5 \\
 1 & 2 & 3 & 4 & 5 \\
 1 & 2 & 3 & 4 & 5 \\
\end{array}
\right)$$
The Bad:
$$\left(
\begin{array}{ccccc}
 1 & 2 & 3 & 4 & 5 \\
 5 & 1 & 2 & 3 & 4 \\
 4 & 5 & 1 & 2 & 3 \\
 3 & 4 & 5 & 1 & 2 \\
 2 & 3 & 4 & 5 & 1 \\
\end{array}
\right)$$
And the Ugly:
$$\left(
\begin{array}{ccccc}
 11 & 12 & 13 & 14 & 15 \\
 16 & 17 & 18 & 19 & 20 \\
 21 & 22 & 23 & 24 & 25 \\
 26 & 27 & 28 & 29 & 30 \\
 31 & 32 & 33 & 34 & 35 \\
\end{array}
\right)$$
The Good is same as Ugly mod n and both are singular.
The Bad is the circulant Good and has determinant $>0$.
Two questions:

What makes a singular matrix rare?
Has anyone documented the differences? (preferably, using $n$ or $n^2$)

REPLY [3 votes]: There's at least one very easy way to think about this. Imagine the $n$ column vectors as just points in $\mathbf{R}^n$. Now the matrix is singular if and only if these points all lie in a single $(n-1)$-dimensional subspace i.e hyperplane that goes through the origin. Typically if I just pick $n$ random points, there is basically no chance they will accidentally lie on a single hyperplane.