MA 265 LECTURE NOTES: MONDAY, JANUARY 28
GaussJordan Reduction
Augmented Matrices.
Consider a system of linear equations in the form
a
11
x
1
+
a
12
x
2
+
· · ·
+
a
1
n
x
n
=
b
1
a
21
x
1
+
a
22
x
2
+
· · ·
+
a
2
n
x
n
=
b
2
.
.
.
.
.
.
.
.
.
.
.
.
a
m
1
x
1
+
a
m
2
x
2
+
· · ·
+
a
mn
x
n
=
b
m
Recall that we can express this system as a product of matrices:
A
=
a
11
a
12
...
a
1
n
a
21
a
22
...
a
2
n
.
.
.
.
.
.
.
.
.
.
.
.
a
m
1
a
m
2
...
a
mn
,
x
=
x
1
x
2
.
.
.
x
n
and
b
=
x
1
x
2
.
.
.
x
n
=
⇒
A
x
=
b
.
Consider the
augmented matrix
that keeps track of the coefficients to the right of the equal sign:
bracketleftbig
A
b
bracketrightbig
=
a
11
a
12
...
a
1
n
b
1
a
21
a
22
...
a
2
n
b
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
m
1
a
m
2
...
a
mn
b
m
The method where we place this matrix in (row) echelon form is called
Gaussian elimination
. The method
where we place this matrix in reduced (row) echelon form, is called
GaussJordan reduction
.
For example, if the matrix can be placed in the form
bracketleftbig
I
n
d
bracketrightbig
, then the unique solution to the system
is
x
=
d
.
Example.
Consider the following system of equations
x
+
2
y
+
3
z
=
6
3
x
+
y

z
=

2
2
x

3
y
+
2
z
=
14
The augmented matrix for this system is
bracketleftbig
A
b
bracketrightbig
=
1
2
3
6
3
1

1

2
2

3
2
14
.
We perform Gaussian elimination to find the reduced (row) echelon form. The leading one in the first row
exists, so we subtract to eliminate the entries in the first column:

3
r
1
+
r
2
→
r
2

2
r
1
+
r
3
→
r
3
=
⇒
1
2
3
6
0

5

10

20
0

7

4
2
.
Now divide the second row by

5 to find the “leading one” for that row:
(

1
/
5)
r
2
→
r
2
=
⇒
1
2
3
6
0
1
2
4
0

7

4
2
.
1
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MA 265 LECTURE NOTES: MONDAY, JANUARY 28
Let’s eliminate the other entries in the second column:

2
r
2
+
r
1
→
r
1
7
r
2
+
r
3
→
r
3
=
⇒
1
0

1

2
0
1
2
4
0
0
10
30
.
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 Spring '10
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 Linear Algebra, Invertible matrix, Row, 1 j, Elementary matrix

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