4-3 Gauss-Jordan Elimination

Topics
Introduction
Gauss-Jordan elimination: the algorithm
Example
Variations on reduced row-echelon forms
Stating the solution when there are infinitely many of them
TI graphing calculator: row-echelon and reduced row-echelon forms
Student Solutions Manual software: reduced row-echelon form



Introduction

Reduced row-echelon form:
Example:


Gauss-Jordan elimination: the algorithm

The following description is an outline of the method.  The details will become clear as we work through the example to follow.  

1.

Get a 1 in upper left corner (by row ops 1 and/or 2)

2.
Get 0's everywhere else in its column (by row op 3)
3.
Mentally delete row 1 and column 1. What remains is a smaller submatrix
4.
Get 1 in upper lefthand corner of the submatrix.
5.
Get 0's everywhere else in its column for the whole matrix (not just the submatrix)
6.
Mentally delete row 1 and column 1 of the submatrix, forming an even smaller submatrix
7.
Repeat 4, 5, 6 until you can go no further.
8.
The matrix will now be in reduced row-echelon form (RREF), or just reduced form
9.
Re-write the system in natural form.
10.
State the solution



Additional instructions:
A.

If you get a row of all zeros, use row op 1 to make it the last row

B.

If you get a row with all zeros to the left of the line, and a non-zero on the right,  STOP (no solution).


Example

For this example we start with the system:

Here's how the algorithm goes:
 
Forward elimination
R3 ‹– –› R1
-3R1  +  R2 –›  R
-2R1  +   R3 –›R
  
  
  
  
  
  
  
  
  
  
  
  
Let's see if we can do these two operations
‹–
(1/10)R2–›R2
  3R2 + R1 –› R1
-4R2 + R3–›R3
 -5R3 –› R3
(1/10)R3 + R1 –› R1
(7/10)R3 + R2 –› R2
Reduced row-echelon form
Answer
x1 = 2, x2 = 0, x3 = -1


Variations on reduced row-echelon forms


 

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Stating the solution when there are infinitely many of them
 
x1 +  x3 = -3/7
x2 - 2x3 =  8/7
Here's what you do:

1. Solve each equation for its first variable:

2. State your solution parametrically:

for any real number t, a solution is

x3 = t                     (standing for any real number)
x2 = 2t + 8/7           (subst. t in 2nd equation above)
x1 = -t - 3/7             (subst. t in 1st equation above)
or, stating the solution as a triple: (since there are infinitely many real numbers t to choose from, this represents infinitely many solutions)

For example, for t = 4/7, we have the solution ).

Try it; it works!


Graphing calculator: reduced row-echelon form

You are required to know how to do this by hand.  If "by hand" is not specified, you can use the TI-83 or TI-86 to do it.  I will demo the TI-83.  For the TI-86, click below.

TI-86


Student Solutions Manual software: reduced row-echelon form

Note:  this section to be updated to reflect current edition.  Use current content at your risk!

The Student Solutions Manual that accompanies our text has a floppy diskette containing software that can be helpful in completing some of the exercises (you can do this instead of using a graphing calculator).

By running that software and using the menus in a pretty obvious way, one can

As an example, see problem 25 on page 211.  The augmented matrix for that system can be defined using the software, and by using the menus, the following screen can be brought up:

By selecting the option "Do <e>very step automatically" (enter an "e" from the keyboard), the reduced row-echelon form will be computed, and you will see this screen:

Pretty easy, huh?