# 9.2.2 When a System of Linear Equations has No Solution

Dr. Robert van de Geijn:
So let’s have a look at how we can show that a linear
system does not have any solutions and let’s use the example from
the last unit, the example that we claimed to have no
solution, to illustrate this. In this particular case, we’re
starting with a linear system that has a square matrix
associated with it. Notice that this here has as
many rows as it has columns but we could have had more rows than
columns or fewer rows than columns. It doesn’t really matter. Let’s apply a step of
Gaussian elimination to this. Let me get this and then let’s see. It looks like we’re
actually doing Gauss-Jordan because we’re introducing zeroes
above and below the diagonal. But notice that, in the process,
we also introduced a 0 right here. And notice that by swapping
rows, we can’t fix that. And in matter of fact, in this case,
there is no row below it to swap with. Now, if we take this and we translate
it back into the linear system that it represents, notice
that the last equation here says that 0 is equal to 1. Well, for what choices of
chi 0, chi 1, and chi 2 is it the case that we
can make 0 equal to 1? Well, the fact is you
can’t make 0 equal to 1. And therefore, there
is no choice for chi 0, chi 1, and chi 2 that will
make that last equation true. And hence, the system of linear
equations has no solution. If Gaussian elimination or
Gauss-Jordan elimination, introduces an inconsistent
equation along the way, then the system of linear
equations has no solution. 