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 i3!
…and for i2 ….
…and for i1
It is not always this nice and smooth. Sometimes the system we are going to solve is underdetermined….
Example: (underdetermined system)
(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!
This is our solution! Or better "solutions"
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