Non-square linear systems with exterior product

[JonnyMaddox]

Hi, how can I compute the general solution of a system of linear equations? Non-square systems for example. I have the book Linear Algebra via exterior products, but it is the worst book in the history of math books, I think I'll burn it somehow, whatever. I can calculate the solution with the exterior product easily with Cramer's rule. For non-square systems, or systems with infinite many solutions Winitzki talks gibberish about a maximal non-zero exterior product between the vectors of the matrix and then does some magic with it to calculate the homogeneous solution. Can someone explain this to me in a clear way? I think one should calculate it by first finding the non-zero exterior product between the vectors of the matrix (which should be a subspace) and then somehow express the remaining vectors in terms of the non-zero set? But that doesn't make sense. hm

 

[JonnyMaddox]

Can someone help me with this? I give an example of the book:

Let's say the vectors a,b,c are not a basis in the vectorspace V, then there exists a maximal nonzero exterior product (which should also tell you what the rank is) 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?