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ONE WAY TO DO GAUSSIAN-JORDAN ELIMINATION
by Jim Sears
(This is not the only way to accomplish Gaussian-Jordan elimination.)
To solve a system of equation using
matrices and Gaussian-Jordan Elimination,
organize the equations so the like variables,
equal signs and constants are aligned in
columns.
x  2 y  z  3
2 x  4 y  2 z  7
 2 x  2 y  3z  4
Create an augmented matrix from the
coefficients and constants.
1
2  1  3


 2  4 2  7
 2 2  3 4
Use proper matrix operations to reduce the
matrix to reduced row echelon form.
Or the ordered-triplet:
 13 13 11 
 , , 
 4 16 8 
Proper matrix operations include:
 Interchange two rows
 Replace a row with a non-zero
multiple of itself
 Replace a row with the sum of it and
another row
 Replace a row with the sum of nonzero multiples of itself and another
row
Work on getting the zero elements of the
matrix in the following order.
 1 62 5 1  3

1
42  7
2

4


2 2 32  3 4
th
1 0 0  13 4


0 1 0 13 16
0 0 1 11 8
The solution for the system is:
X = -13/4
Y = 13/16
Z = 11/8
st
nd
th
th
rd
Then, once you have the “two corners” filled
with zeros, put the ones on the diagonal.
In short follow the steps outlined in this
table for the seven steps to Gaussian-Jordan
elimination.
Zero’s
Position
1st
2nd
3rd
4th
5th
6th
7th
Work with
R1 & R2
R1 & R3
R2 & R3
R3 & R2
R3 & R1
R2 & R1
To replace
R2
R3
R3
R2
R1
R1
Put the ones on the
diagonal
Translation:
 To put a zero in the first position, work
with row 1 and row 2 to replace row 2.
 To put a zero in the second position, work
with row 1 and row 3 to replace row 3.
 Etc.
Example, solve the following system:
x  2 y  z  3
2 x  4 y  2 z  7
 2 x  2 y  3z  4
Create an augmented matrix:
1
2  1  3


 2  4 2  7
 2 2  3 4
Work with rows one and two to place a zero
in the first position of row two. To do this,
multiply row one by -2, add the result to row
two and replace row two with the sum.
1
2  1  3


 0  8 4  1
 2 2  3 4
Scratch work
-2 -4 2 | 6
2 -4 2 | -7
0 -8 4 | -1
Work with rows one and three to place a
zero in the first position of row three. To do
this, multiply row one by 2, add the result to
row three and replace row three with the
sum.
1

0
 0
2  1  3

 8 4  1
6  5  2
Scratch work
2 4 -2 | -6
-2 2 -3 | 4
0 6 -5 | -2
Work with rows two and three to place a
zero in the second position of row three. To
do this, multiply row two by 3 and row three
by 4, add the new products, and replace row
three with the sum.
1

0
 0
 1  3

 8 4  1
0  8  11
2
Scratch work
0 -24 12 | -3
0 24 -20 | -8
0 0 -8 |-11
Work with rows three and two to place a
zero in the third position of row two. To do
this, multiply row two by 2, add it to row
three and replace row two with the sum.
 1

 0
 0
2
 16
0
 1  3

0  13
 8  11
Scratch work
0 -16 8 | -2
0 0 -8 |-11
0 -16 0 |-13
Work with rows three and one to place a
zero in the third position of row one. To do
this, multiply row one by -8, add it to row
three and replace row one with the sum.
 8

 0
 0
 16
 16
0
13

0  13
 8  11
0
Scratch work
-8 -16 8 | 24
0 0 -8 | -11
-8 -16 0 | 13
Work with rows two and one to place a zero
in the second position of row one. To do
this, multiply row one by -1, add it to row
two, and replace row one with the sum.
 8

 0
 0
0
 16
0
0  26

0  13
 8  11
Scratch work
8 16 0 |-13
0 -16 0 |-13
8 0 0 |-26
Once the lower left and upper right corners
of the coefficient matrix are filled with
zeros, work on placing ones on the diagonal.
To do this, divide each row by the leading
non-zero coefficient of the row. For row
one in this example, divide by 8, row two
divide by -16, and row three divide by -8.
Reduce all fractions.
 8
 8
 0

  16
 0
 8

0
8
 16
 16
0
8
0
8
0
 16
8
8
 26  
 13 



8 
1 0 0 4 

 13
 = 0 1 0 13 
16 
 16  
 11 0 0 1  11
8 
 8  
  13 13  11 
, ,


 4 16 8 