Limits using Cramer's Rule as determinant approaches 0 I'm in Linear Algebra 1, and having just covered Cramer's Rule, the prof showed this interesting case that I have a further question about the significance of.
Say we have a matrix containing a constant that can be adjusted, for instance, the system of equations
2cx+3y=6
4x+(c-1)y=4
giving
$\begin{pmatrix}2c&3\\4&c-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\4 \end{pmatrix}$
Since Cramer's rule only holds in cases where the determinant is nonzero, a typical question would be to find the values of c for which that is true. In this case, det(A)=0 when c=-2 or c=3. At c=-2 there are no solutions to the system, and at c=3 there are infinitely many solutions.
In the case of c=3, we cannot simply apply Cramer's rule, because the denominator of x=$\frac{detA(1)}{detA}$ and y=$\frac{detA(2)}{detA}$ are both detA=0.
However, what we can do is go back to the original system, leaving the variable c in the matrix, and calculate the values of detA, detA(1) and detA(2) in relation to c.
If I do that, and completely factor, I get:
detA=2(c+2)(c-3)
detA(1)=6(c-3)
detA(2)=8(c-3)
Now I can use limits to get an answer from the formulation of Cramer's rule in the case of c=3.
x= $\lim\limits_{c \to 3}\frac{6(c-3)}{2(c+2)(c-3)}$
y= $\lim\limits_{c \to 3}\frac{8(c-3)}{2(c+2)(c-3)}$
From which we can easily get the values of x=$\frac{3}{5}$ and y=$\frac{4}{5}$, which is a valid solution.
So Cramer's rule, despite its initial misgivings, has provided a solution to a system with a determinant of 0. My question (which my prof couldn't answer on the spot, which is why I'm bringing it here) is, which solution? What is special about these numbers, that they're the ones that happen to be spat out using this method? My first thought was that perhaps it's the particular solution, but it's not: the solution in parameterized form of the matrix when c=3 is
$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1\\0 \end{pmatrix}+s\begin{pmatrix}-1/2\\1 \end{pmatrix}$
So I am at a loss as to what these numbers "are," if they "are" anything in particular. Surely they're not just random?
 A: Maybe it helps to visualize the two equations as lines in the plane. As you vary $c,$ the lines move around in the plane in a certain way. Most of the time, they will have exactly one point of intersection. But if $c=-2,$ they are parallel and distinct, and if $c=3,$ they are identical and can both be described by the equation $2x+y=2$.
If you vary $c$ continuously, the point of intersection will also move around continuously in the plane. When $c$ approaches $3,$ the point of intersection will approach one particular point on the line described by the equation $2x+y=2,$ namely $\left(\frac 35,\frac 45\right).$
The way in which $c$ influences the matrix elements also determines the way the lines move in the plane when you vary $c$ and hence determines the point approached by the intersection of the lines. You can construct other systems of linear equations that also have infinitely many solutions of the form $2x+y=2$ for a particular value of a certain parameter, but the point, which makes the curve of the intersection points continuous, is somewhere else on that line.
As the location of that point is strongly coupled to the particular way your parameter modifies the system of equations, you cannot make more general statements about this point. This point is special only for your particular system of equations and your particular way of parameterizing it.
So all you can say is: If you want to get exactly one particular solution of the system of equations for each $c$ for which at least one solution exists, and if you want this solution to depend continuously on the parameter $c$ where continuity is possible, then you have to choose $(x,y) = \left(\frac 35,\frac 45\right)$ for $c=3.$
