How is determinant defined by variable elimination? I've been chasing origins of determinants for quite a while now, and having depleted all literature I have access to, I'm trying to find some hints here.
Many books describe determinants as solutions to systems of linear equation obtained by elimination of variables. I can't get the expression for determinant to fall out of systems of linear equations larger that 2. For a $2 \times 2$ system, e.g.
$$ ax + by = 0 \\ cx + dy = 0$$
multiplying the first equation by $c$ and subtracting it from the second equation multiplied by $a$ gives us
$$(ac - ca)x + (ad - bc)y = 0 \rightarrow (ad - bc)y = 0$$
Therefore giving us the determinant.One one hand, this is a nice result since it shows that if the determinant is 0, y can be anything. On the other hand, if either $a$ or $c$ is 0, we would simply get the system of equations
$$ 0=0 \\ 0=0 $$
which tells us nothing. Also, the same technique does not work for systems larger that $2 \times 2$
My question is is how exactly does determinant manifest itself when solving systems of linear equations the straightforward way?
One may want to say that the determinant is the product of all diagonal entries of a matrix after reducing it to a diagonal form. That's true, but how is this related to obtaining solutions to systems of linear equations?
I hope what I'm looking for is clear, but is not, please let me know and I'll provide more clarification.
 A: Let's take a closer look at your $2 \times 2$ example. You start by assuming you have numbers $x$ and $y$ that satisfy the system
$$\begin{align*}
ax + by & = 0 \\
cx + dy & = 0.
\end{align*}$$
From there, you deduce that
$$\begin{align*}
(ad - bc)x & = 0 \\
(ad - bc)y & = 0.
\end{align*}$$
If $ad - bc$ is nonzero, then $x$ and $y$ have to be zero, so the system has only that trivial solution. If $ad - bc$ is zero, on the other hand, the equations you deduced don't force $x$ and $y$ to be zero. In this case, it turns out that the system always has nontrivial solutions—solutions where not all the free variables are zero. Now we have a guess about how the determinant is related to solving linear systems:

A homogeneous linear system has nontrivial solutions if and only if the associated determinant is zero.

To fill in more detail, we need to learn a little more about determinats.

Our $2 \times 2$ linear system can be rewritten as the equation
$$x\left[ \begin{array}{cc} a \\ c \end{array} \right]
+ y\left[ \begin{array}{cc} b \\ d \end{array} \right]
= \left[ \begin{array}{cc} 0 \\ 0 \end{array} \right].$$
To help us focus on the big picture, rather than worrying about details like the individual coordinates of the vectors, let's define
$$\begin{align*}
\mathbf{v} & := \left[ \begin{array}{cc} a \\ c \end{array} \right]
& \mathbf{w} & := \left[ \begin{array}{cc} b \\ d \end{array} \right],
\end{align*}$$
so the equation becomes
$$x\,\mathbf{v} + y\,\mathbf{w} = \mathbf{0}.$$
The determinant of the matrix with columns $\mathbf{v}$ and $\mathbf{w}$ can be defined as the signed area of the parallelogram with sides $\mathbf{v}$ and $\mathbf{w}$; that's just the ordinary area with a $+$ or $-$ sign that flips if you reverse the order of the sides. If you draw some pictures, representing the vectors as arrows and keeping in mind that multiplying a vector by a number $s$ means scaling it by a factor of $s$, you should be able to convince yourself that the equation has a solution if and only if the vectors $\mathbf{v}$ and $\mathbf{w}$ lie on the same line through $\mathbf{0}$ in $\mathbb{R}^2$, which in turn happens if and only if the parallelogram with sides $\mathbf{v}$ and $\mathbf{w}$ has zero area.
To emphasize the point of view that determinants are areas, I like to write the determinat of the matrix with columns $\mathbf{v}$ and $\mathbf{w}$ as $D(\mathbf{v}, \mathbf{w})$. Using that notation, let's summarize what we've learned about the relationship between determinants and linear systems.

Given vectors $\mathbf{v}, \mathbf{w} \in \mathbb{R}^2$, the equation
  $$x\,\mathbf{v} + y\,\mathbf{w} = \mathbf{0}$$
  has nontrivial solutions if and only if the vectors $\mathbf{v}$ and $\mathbf{w}$ lie on the same line through $\mathbf{0}$ in $\mathbb{R}^2$, which happens if and only if the area $D(\mathbf{v}, \mathbf{w})$ is zero.


In $n$-dimensional space, the determinant of the matrix with columns $\mathbf{v}_1, \ldots, \mathbf{v}_n$ is the signed volume of the parallelopiped with edges $\mathbf{v}_1, \ldots, \mathbf{v}_n$; once again, that's just the ordinary volume with a sign that flips if you exchange the order of two sides. Now we can generalize our statement above to any finite number of dimensions.

Given vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n \in \mathbb{R}^n$, the equation
  $$x_1\,\mathbf{v}_1 + \ldots + x_n\,\mathbf{v}_n = \mathbf{0}$$
  has nontrivial solutions if and only if the vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$ lie on the same hyperplane through $\mathbf{0}$ in $\mathbb{R}^n$, which happens if and only if the volume $D(\mathbf{v}_1, \ldots, \mathbf{v}_n)$ is zero.


When you're dealing with an inhomogeneous linear system, the determinant can do more than just tell you whether or not there are solutions. If the system has a unique solution, the determinant will give it to you! I'll sketch how this works in two dimensions.
Let's say $\mathbf{v}, \mathbf{w} \in \mathbb{R}^2$ don't lie on the same line through $\mathbf{0}$. This turns out to imply that, for any $\mathbf{b} \in \mathbb{R}^2$, the equation
$$x\,\mathbf{v} + y\,\mathbf{w} = \mathbf{b}$$
has a unique solution. You can find it by using the geometry of signed areas to deduce that
$$\begin{align*}
D(\mathbf{v}, \mathbf{b})
& = D(\mathbf{v}, x\,\mathbf{v} + y\,\mathbf{w}) \\
& = D(\mathbf{v}, x\,\mathbf{v}) + D(\mathbf{v}, y\,\mathbf{w}) \\
& = 0 + y\,D(\mathbf{v}, \mathbf{w}),
\end{align*}$$
so
$$y = \frac{D(\mathbf{v}, \mathbf{b})}{D(\mathbf{v}, \mathbf{w})}.$$
A similar calculation reveals that
$$x = \frac{D(\mathbf{w}, \mathbf{b})}{D(\mathbf{w}, \mathbf{v})}.$$
You can justify all the algebraic manipulations I just did by drawing pictures. Try it!
This method of solving inhomogeneous linear systems generalizes to any finite number of dimensions. It's called Cramer's rule.
