Solve systems by Gauss elimination:

To illustrate what is going on here I wrote down on the left-hand side the system of equations, and on the right side the corresponding augmented Matrix . Normally we wouldn't do this but would work with the augmented matrix only!

Alright, here we go!

Equations Augmented Matrix

̃ Subtract: 1 times the pivot from second equation

20 times the pivot from fourth equation

Our goal is, of course, to make the augmented matrix upper triangular ….

Ooops, our first row operation made the second row disappear! The reason is that row 1 and row 2 in the initial augmented matrix were multiples (-1) of each other. This can happen! Let's just move this empty row to the bottom and continue …

Sort & identify a pivot to eliminate

And again ….

Subtract 3 times the pivot from third equation

Voila! Our matrix is upper triangular. And now row number 3 reads as -

We found an equation for i_{3}!

̃

…and for i_{2} ….

solved

…and for i_{1}

Problem solved!

It is not always this nice and smooth. Sometimes the system we are going to solve is underdetermined….

Example: (underdetermined system)

or short

(we should get used to work with the augmented matrix only!)

Anyway, we have three equations to find 4 unknowns - how will this work out?

A: It shouldn't worry us - we just follow procedure and try to make the augmented matrix upper triangular ….

Subtract: 0.2 times the first row from the second

0.4 times the first row from the third

Subtract -1 times the second row from the third

This is all we can do - It's even worse than we expected!

Resubstitute:

or

- and,

This is our solution! Or better "solutions"

We found

- a relation between x
_{1}and x_{4}, and - another between x
_{2}and x_{3}, x_{4}, but the - values for x
_{3}and x_{4}can arbitrarily been chosen.

So we have an infinite number of solutions here!

But - it is not always that bad! Here is an example were you find just one solution:

Example: Unique solution

How does the equivalent system of linear equations look like?)

Right, is the same as

Row 2 + 3Row 1

Row 3 - Row 1

Row 3 - Row 2

Here we are: The augmented matrix is in upper triangular shape …

….and we are ready to retrieve the final solution:

This is a "unique" solution: , , and

This were three example on how Gauss Elimination helps us to solve system of linear equations. In the next module we will continue the discussion with the question how to determine if a system of linear equations has solutions at all, etc .

Copyrights 1999, 2000 by Peter Dragovitsch and Ben A. Fusaro