Why is a singular matrix rare? 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 $n^2$)

 A: Here is an extension of Gerry's argument.  There are $q^{n^2}$ matrices in $M_{n\times n}(\mathbb{F}_q)$ and $\prod_{k=0}^{n-1}(q^n-q^k)$ elements in $GL_n(\mathbb{F}_q)$.
$$\lim_{q\to\infty}\frac{\prod_{k=0}^{n-1}(q^n-q^k)}{q^{n^2}}=\lim_{q\to\infty}(q^{-n};q)_n=\lim_{q\to\infty}\prod_{k=0}^{n-1}\left(1-q^{-n+k}\right)=1$$ where $(q^{-n};q)_n$ is the $q$-Pochammer symbol.
Note that the fact that $\mathbb{F}_q$ has nonzero characteristic doesn't affect this argument, since the formula $\prod_{k=0}^{n-1}(q^n-q^k)$ is based on picking $n$ linearly independent vectors combinatorially from $(\mathbb{F}_q)^n$, a process which extends to the infinite case independently of characteristic.
A: If you think about the interpretation of a matrix as a system of equations, and take the 2-variable case as an easy to visualize example, most pairs of equations are not parallel lines.  If the 2nd column is a multiple of the first, then they are parallel. Inb this case, the 2nd column will be a multiple of the first.
Of course, for the 3 dimensional case, there are linear combinations, so this no longer holds in the same sense, but you still have the nth column as a "multiple" of one or both of the other columns.
Does that clarify, or is there something more complex you are noticing?
A: 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.
A: Thinking in terms of probability helps. If you have a continuous probability distribution defined on some space of matrices, then typically the singular matrices will have probability zero. Thinking in terms of the determinant: The determinant is a polynomial in the entries of the matrix. Setting it to zero gives a polynomial equation, which are defining (implicitely) some surface in the matrix space. This surface will have a reduced dimension , so its (Lebesgue) measure will be zero. 
A: The number of $2\times2$ matrices over a field of $q$ elements is $q^4$. 
The number of non-singular $2\times2$ matrices over a field of $q$ elements is $$(q^2-1)(q^2-q)=q^4-q^3-q^2+q$$ which means only $q^3+q^2-q$ out of $q^4$ are singular. 
