I'm reading the book Linear algebra via exterior products by Sergei Winitzki (which is the worst book, ever) and he shows that you can solve linear systems with a general solution with Cramer's rule and the exterior product. Let's say the vectors a,b,c are not a basis in the vectorspace V, then there exists a maximal nonzero exterior product of a linear independent subset of a,b,c Take an example
$2x+y=1$
$2x+2y+z=4$
$y+z=3$
Now $a=(2,2,0)$, $b=(1,2,1)$, $c=(0,1,1)$, $p=(1,4,3)$
We see that $a \wedge b \wedge c=0$. And the maximal nonzero exterior product can be written as $\omega =a \wedge b$ (which is not equal to zero. Now we can check if $p$ is a subset of the span ${a,b}$ with $\omega \wedge p=0$ so $p$ can be expressed with $a,b$. We can find the coefficients with Cramer's rule
$\alpha = \frac{p\wedge b}{a\wedge b}=-1$
$\beta = \frac{a\wedge p}{a\wedge b}= 3$
Therefore $p=-a+3b$ so the inhomogeneous solution is $x^{1}= (-1,3,0)$ Now to determine the space of homogeneous solutions, the vector $c$ get's decomposed into a linear combination of $a$ and $b$ again by Cramer's rule. This gives $c=-\frac{1}{2}a+b$ And the space of homogeneous solutions is given by the span of $x_{i}^{(0)(1)}=(-\frac{1}{2},1,-1)$. So the general solution is
$x_{i}=x_{i}^{1}+ \beta x_{i}^{(0)(1)}= (-1-\frac{1}{2}\beta, 3+\beta, -\beta)$
Then he gives another example with a non-square system
$x+y=1$
$y+z=1$
And the general solution is $x_{i}=(1,0,1)+\alpha (1,-1,1)$ (with no explanation).
I don't see a pattern here. What is going on?
OK. I'll try to explain the first example for you.
Our task is to solve the system of linear equations: $\begin{cases}2x+y=1 \\ 2x+2y+z=4 \\ y+z=3 \end{cases}$
Said another way, we're trying to find the space of vectors $(x,y,z)$ that get mapped to $\vec p = (1,4,3)$ under the given transformation $T$. That is: $T(x,y,z) = x(2,2,0) + y(1,2,1) + z(0,1,1) = (1,4,3) = \vec p$.
First, for reference, let's see how we'd do it using matrix algebra.
So let's use Gauss-Jordan elimination: $$ \left[\begin{array}{ccc|c} 2 & 1 & 0 & 1 \\ 2 & 2& 1 & 4 \\ 0 & 1 &1 & 3 \end{array} \right] \sim \left[\begin{array}{ccc|c} 2 & 1 & 0 & 1 \\ 0 & 1 & 1 & 3 \\ 0 & 1 & 1 & 3 \end{array}\right] \sim \left[\begin{array}{ccc|c} 2 & 1 & 0 & 1 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]$$
Let's then set $z=t$ for some arbitrary value $t \in \Bbb R$. Then $y=3-t$ and $x=\frac 12(1-y) = \frac 12 (1-3+t) = -1+\frac 12 t$. So our general solution is $\vec x = \begin{bmatrix} -1 + \frac 12 t \\ 3-t \\ t\end{bmatrix}$. Which we can see is equivalent to the answer Dr. Winitzki got by the change of variable $t \mapsto -\beta$.
Now, let's try to do this with exterior algebra and Cramer's rule.
We need to first note that there is a theorem in linear algebra which says that the solution to a system of linear equations is the sum of any particular solution with the general solution to the corresponding homogeneous set of equations.
That is, if you have some transformation $T(x,y,z)=(1,4,3)$, then knowing some particular vector $(x_0,y_0,z_0)$ which transforms into $(1,4,3)$ via this transformation and the general solution to the related problem $T(t,u,v)=(0,0,0)$, your solution set will be $(x_0+t,y_0+u,z_0 + v)$.
So our first task is to find a vector $(x_0,y_0,z_0)$ that solves this system.
We first note that $\vec p \in \vec a \wedge \vec b$ because $x\vec a + y \vec b + z \vec c = \vec p \iff \vec p \in \text{span}(\vec a, \vec b, \vec c)$, but we know that $\text{span}(\vec a, \vec b, \vec c)=\text{span}(\vec a, \vec b)$ because $\vec a \wedge \vec b \wedge \vec c = 0$ and $\vec a \wedge \vec b \ne 0$. We also know that $\text{span}(\vec a, \vec b)$ is representable by $\vec a \wedge \vec b$.
Thus $\vec p = \alpha \vec a + \beta \vec b$ for some scalars $\alpha$ and $\beta$.
You will see why Cramer's rule works by simply taking the exterior product of that last equation with $\vec a$ and $\vec b$: $\ \vec p\wedge \vec b = (\alpha \vec a + \beta \vec b)\wedge \vec b = \alpha \vec a \wedge \vec b + \beta \vec b \wedge \vec b= \alpha \vec a \wedge \vec b + 0=\alpha \vec a \wedge \vec b$. Then dividing by $\vec a \wedge \vec b$ on the far left and far right sides gives $\alpha = \frac{\vec p\wedge \vec b}{\vec a \wedge \vec b}$. $\ \beta$ can be found similarly. Thus $\vec p=-\vec a+3\vec b=x_0\vec a + y_0\vec b + z_0\vec c$.
So now we've found $\vec x_p=(-1,3,0)$ which is a particular solution to our system. Note here that Cramer's rule is not powerful enough to get us EVERY solution to our system, but it can find us $1$ (assuming there is at least $1$).
Now we need the homogeneous solution. This means we need to find the solution to : $\begin{cases}2x+y=0 \\ 2x+2y+z=0 \\ y+z=0 \end{cases}\ \ \ \ $ Equivalently, we need to solve $x\vec a + y\vec b + z\vec c =0$.
But we know that $\vec c$ is a linear combination of vectors $\vec a$ and $\vec b$. We can then use the exact same method to determine that $\vec c = -\frac 12 \vec a + \vec b$.
Adding $-\vec c$ to both sides we get $-\frac 12 \vec a + \vec b -\vec c =0$. So apparently one solution is $(x,y,z)=(-\frac 12, 1, -1)$. From the rank-nullity theorem, we know that our homogeneous solution (AKA nullspace) is $1$-dimensional, so now that we've found $1$ solution, we know that all others will be a scalar multiple of that one.
So our general homogeneous solution is $\beta (-\frac 12, 1, -1)$, for some arbitrary $\beta \in \Bbb R$.
Putting this together with the particular solution we already found, we find that the general solution to the system of equations: $\begin{cases}2x+y=1 \\ 2x+2y+z=4 \\ y+z=3 \end{cases}\ \ \ $ is $(x,y,z) = (-1,3,0) + \beta(-\frac 12, 1, -1)$, which is just what we found from our matrix methods. Yay!