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Gauss-Jordon Elimination Notes:

The procedure for Guass-Jordan Elimination is as follows:
  1. Find the leftmost column that is not all zeros and swap its row with the top row.

  2. Make the leading entry in the first row a "1" (If the top entry is a, then multiply the top row by 1/a).

  3. Use the top row to make all the entries in the column below the leading one be zero.

  4. Ignoring the top row, repeat steps 1 to 4 until there are no more leading ones.

  5. Finally, use each leading 1 to make all entries on the column above the zeros.


Example:

Solve for x, y, and z in:
2y - 3z=2
2x +   z=3
x - y + 3z=1.
Solution:
é
ê
ë
0
2
1
2
0
-1
-3
1
3
ï
ï
ï
2
3
1
ù
ú
û
write the system as an augmented matrix
é
ê
ë
1
2
0
-1
0
2
3
1
-3
ï
ï
ï
1
3
2
ù
ú
û
interchange first and third row
(to make top left entry non-zero)
é
ê
ë
1
0
0
-1
2
2
3
-5
-3
ï
ï
ï
1
1
2
ù
ú
û
Add -2 times first row to second row
(to get 0 in first column of row 2)
é
ê
ë
1
0
0
-1
1
2
 3
-5/2
-3
ï
ï
ï
1
1/2
2
ù
ú
û
Divide second row by 2
(to get a leading 1 in row 2)
é
ê
ë
1
0
0
-1
1
0
3
-5/2
2
ï
ï
ï
1
1/2
1
ù
ú
û

Add -2 times second row to third
(to get 0's in the second column)
é
ê
ë
1
0
0
-1
1
0
3
-5/2
1
ï
ï
ï
1
1/2
1/2
ù
ú
û
Divide third row by 2
(to get a leading 1)
é
ê
ë
1
0
0
-1
1
0
0
0
1
ï
ï
ï
-1/2
7/4
1/2
ù
ú
û
Add -3 times third row to first row
Add 5/2 times third row to second row
(to get 0's in third column)
é
ê
ë
1
0
0
0
1
0
0
0
1
ï
ï
ï
5/4
7/4
1/2
ù
ú
û
Add second row to first row

This is now in reduced row-echelon form, so we can read off the answer: x = 5/4, y = 7/4, and z = 1/2.

Check that the answer satisfies the initial equations (in case we made arithmatic errors):

2y - 3z = 2(7/4) - 3(1/2) = 7/2 - 3/2 = 4/2 = 2
2x + z = 2(5/4) + 1/2 = 5/2 + 1/2 = 6/2 = 3
x - y + 3z = (5/4) - (7/4) + (1/2) = -(2/4) + 3/2 = -(1/2) + 3/2 = 2/2 = 1
All of these check out, so our solution is correct.


[HOME] Math 15 (Winter 2001) web pages
Created: 06 Feb 1998
Last modified: 11 Mar 2001 23:59:59
Comments to: dpvc@union.edu
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