4-3 Gauss-Jordan Elimination
We will do row operations :
1. Interchange two rows
Ri  Rj
2. Multiply a row by a constant
cRi  Ri
3. Add a multiple of one row
cRi + Rj  Rj
to another
 on an augmented matrix to solve a system
 using a method known as
GAUSS-JORDAN ELIMINATION:
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 all rows in the
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.
6. Re-write the system in natural form.
7. State the solution.
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).
4-3
p. 1
Example:
 2  2 1 3


 3 1 1 7 


1  3 2 0


R3  R1
1  3 2 0


 3 1 1 7 


 2  2 1 3


- 3R 1 + R 2  R 2
- 2R 1 + R 3  R 3
1  3 2 0


 0 10  7 7 


0 4  3 3


1
10R2  R2
2
0
1  3


7
7
0 1 


10 10 
0 4
 3 3 


 1

 0


 0

4-3
0
1
0
1
10
7

10
1

5

3R2 + R1  R1
-4R2 + R3  R3
21
10
7
10
1
5








-5R3  R3
p. 2

 1

 0


 0

1

0

0

0
1
0
1
10
7

10

1
0
0
1
0
0
1






1 

21
10
7
10
1
R3 + R1  R1
10
7
R3 + R2  R2
10
2

0

 1
Answer: x1 = 2, x2 = 0, x3 = -1
4-3
p. 3
Variations on reduced forms
1
0
0

2
1
0
3
2  no solution
4 
 1
0
0
1
1
2
 3/ 7
8 / 7 
infinitely many solutions
 when there are infinitely many solutions
 we state "the solution" parametrically.
 above reduced form corresponds to the system:
x1
x2
+ x3
- 2x3
=
=
-3/7
8/7
1. Solve each equation for its first variable:
x1 = -x3 x2 = 2x3 +
3/7
8/7
The variables x1 and x2 on the left appear nowhere else.
The variable (or variables) on the right can be chosen
arbitrarily to be any real numbers (we use “t” for this)
2. State your solution parametrically:
for any real number t, a solution is
x3 = t
x2 = 2t + 8/7
or, stated as a triple:
x1 = -t - 3/7
(-t - 3/7, 2t + 8/7, t)
(since there are infinitely many real numbers t to choose
from, this represents infinitely many solutions)
4-3
p. 4