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.