1.III.1 - Gauss-Jordan Reduction
MTH 214 - Linear Algebra
Dr. Lew
*
Depending on how Gauss reduction is applied, several different echelon
forms for the same matrix can occur. For example, the linear system and
its corresponding augmented matrix
2x + 2y = 0
4x + 3y = 0
=⇒
"
#
2 2 0
4 3 0
has several different echelon forms possible depending on the row
operations and order they are applied:
"
#
2 2 0
0 −1 0
"
#
2 0 0
0 −1 0
#
"
1 1 0
0 −1 0
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Since there are several different echelon forms, the natural question to
ask is whether or not this might lead to different free variables, which
creates different solutions sets for a given linear system.
There is an extension of Gauss reduction which answers the question. In
particular, we’ll see that the cost is additional elementary reduction
operations, but the advantage is that the general solution is expressed
quite easily.
The extension of Gauss reduction is called Gauss-Jordan reduction.
The idea is to take the linear system and first reduce it to echelon form.
From there, perform additional row reductions to satisfy some additional
conditions.
The result is a linear system that is in (row) reduced echelon form.
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Row Reduced Echelon Form (RREF)
An linear system is in reduced echelon form if it is in echelon form
AND satisfies the additional conditions:
The leading entry in each nonzero row is 1.
Each leading entry is the only nonzero entry in its corresponding
column.
The process of getting a linear system to reduced echelon form is called
Gauss-Jordan reduction.
This definition can be applied equally to the corresponding augmented
matrix representing the linear system.
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Example
Suppose we want to solve the linear system:



x + y − 2z = −2


y + 3z = 7



x
− z = −1
.
Setting up the augmented matrix using Gauss reduction gives




1 1 −2 −2
1 1 −2 −2

 −1ρ1 +ρ3 

−→
7
7
0 1 3
0 1 3
1ρ2 +ρ3
1 0 −1 −1
0 0 4
8
This is in echelon form. If we continue with Gauss-Jordan reduction,
we need a leading entry of 1 in the third row.
Then we can use the leading entries to zero out the entries above and
below them in each column:
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

1 1 −2 −2


7
0 1 3
0 0 4
8
2ρ3 +ρ1
−→
−3ρ3 +ρ2
(1/4)ρ3
−→


1 1 0 2


0 1 0 1
0 0 1 2


1 1 −2 −2


7
0 1 3
0 0 1
2
−1ρ2 +ρ1
−→


1 0 0 1


0 1 0 1
0 0 1 2
From the reduced echelon form, notice the solution is immediately apparent: There is exactly one solution given by x = 1, y = 1, and z = 2.
The process of using the leading entry to zero out the entries above and
below it is called pivoting.
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Before some morel examples, let’s make note of some basic definitions
and results regarding the connection between equivalent linear systems,
row operations, and reduced echelon form.
Each of the three elementary row reductions is “reversible”. In other
words, each row operation can be undone by a row operation of the
same type:
The row swap operation ρi ↔ ρj is reversed by ρi ↔ ρj .
The row scaling operation kρi is reversed by k1 ρi .
The row operation kρi + ρj is reversed by −kρi + ρj .
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If a linear system is obtained from another linear system through a sequence of elementary row reductions, then say the two systems are row
equivalent.
A linear system is row equivalent to any of its echelon forms.
Row equivalent systems have the same solution set.
Each linear system is row equivalent to a unique reduced echelon form
system.
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Exercise
Solve the following linear system by setting up the augmented matrix
and using Gauss-Jordan reduction and determine the general solution.
x
−z=4
2x + 2y
=1
Exercise
Solve the following linear system by setting up the augmented matrix
and using Gauss-Jordan reduction and determine the general solution.
− z
=1
y + 2z − w = 1
x + 2y + 3z − w = 7
x
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