# Non-square linear systems with exterior product

1. Sep 11, 2014

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

Last edited by a moderator: Sep 29, 2014
2. Sep 12, 2014

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?