Page 54 - Matrix Analysis & Applied Linear Algebra

2.2 Reduced Row Echelon Form 47
2.2 REDUCED ROW ECHELON FORM

At each step of the Gauss–Jordan method, the pivot is forced to be a 1, and then
all entries above and below the pivotal 1 are annihilated. If A is the coefficient
matrix for a square system with a unique solution, then the end result of applying
the Gauss–Jordan method to A is a matrix with 1’s on the main diagonal and
0’s everywhere else. That is,

10 ··· 0
 
Gauss–Jordan  01 ··· 0 
A −−−−−−−−→  . . . .  .
 . . .
. . . . . 
00 ··· 1
But if the Gauss–Jordan technique is applied to a more general m × n matrix,
then the final result is not necessarily the same as described above. The following
example illustrates what typically happens in the rectangular case.
Example 2.2.1
Problem: Apply Gauss–Jordan elimination to the following 4 × 5 matrix and
circle the pivot positions. This is the same matrix used in Example 2.1.1:

12133
 
 .
 24044 
12355
A = 
24047
Solution:

2133 2 1 3 3 2 1 3 3
 1   1   1 
 2 4044   0 0 −2 −2   0 0 1 1 
1
-2
1 2355 0 0 2 2 2 0 0 2 2 2
 → → 
2 4047 0 0 −2 −2 1 0 0 −2 −21
 1   1   1 
2 0 2 2 2 0 2 2 2 0 2 2
 0 0 1 1   0 0 1 1   0 0 1 1 
1
1
1
0 0 0 0 0 0 0 0 0 0 0 0
1
→  → → 
0
3
0 0 0 0 3 0 0 0 0 0 0 0 0 0 0
2 0 2 0
 
1
 0 0 1 0 
1
0 0 0 0
→   .
1
0 0 0 0 0