How many $3 \times 3$ matrices are singluar? How many $3 \times 3$ matrices are singluar? Describe the methodology used to achieve the result.
 A: For example the integers are made up of $50 \%$ evens and $50\%$ odds.  The evens are in the equivalence class $2\Bbb{Z}$ and the odds are in $2\Bbb{Z} + 1$.  Where $2\Bbb{Z}$ is the obvious additive subgroup of $\Bbb{Z}$.  So asking how many evens there were was equivalent to taking the number classes of interest ($2\Bbb{Z}$, so $1$). And dividing by the total number of classes present as elements of $\Bbb{Z} / 2 \Bbb{Z}$.  Thus you might want to take your ring of $3\times 3$ matrices $R$ and forming the quotient ring $R / GL(3,k)$, where $GL(3,k)$ is the group of invertible matrices that are $3\times 3$ and have elements in the algebraic structure $k$ ($k$ would be $\Bbb{R}$ for example).  So you want to count the number of elements of $R / GL(3,k)$, however it seems infinite (when $k = \Bbb{R}$).  So a better way might be to let $k$ be finite of order $n$, find a formula for such a count over this finite structure, then take the limit as $n \to \infty$.  Probably won't work though.
A: Let's try to compare the size of $G = GL(3, k)$ to $k^{3\times 3}$.  A matrix $A \in G \iff \det(A) \neq 0$.  Well if $A$ has a column or a row of all zeros then $\det(A) = 0$ by the expansion formula along a column or row.
Let's look at the $2\times 2$ case.  We want all $A = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$ such that $ad - bc \neq 0$.  Thus your first choice is for the RHS of $\det(A) = e$.  One of $ad$ or $bc$ has to be nonzero, so wlog let $ad \neq 0$, then $a = \frac{e - bc}{d}$.  Thus we have a total of $|\Bbb{R}|^2 (|\Bbb{R}| - 1)^2$ when $d \neq 0$ and if $d = 0$ then we have $b = \frac{e}{c}$ with a total of $(|\Bbb{R}| - 1)^2$ choices with a total of $(|\Bbb{R}|^2 + 1)(|\Bbb{R}| - 1)^2$ choices.  Thus the probability of selecting an invertible matrix from $k^{2\times 2}, k = \Bbb{R}$ is $\lim f(n) = \lim_{n \to |\Bbb{R}|} \frac{(n^2 +1)(n-1)^2}{n^4}$, which I think is $1$??  That's weird.  Good luck on your problem.... lols
Maybe as you increase the size of matrix you'll get something different, but in $\Bbb{R}^{2\times 2}$ the above says that they're practically all non-singular, but that doesn't seem right so I probably messed up somewhere.
