# 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:
 éęë 021 20-1 -313 ďďď 231 ůúű
write the system as an augmented matrix
 éęë 120 -102 31-3 ďďď 132 ůúű
interchange first and third row
(to make top left entry non-zero)
 éęë 100 -122 3-5-3 ďďď 112 ůúű
Add -2 times first row to second row
(to get 0 in first column of row 2)
 éęë 100 -112 3-5/2-3 ďďď 11/22 ůúű
Divide second row by 2
(to get a leading 1 in row 2)
 éęë 100 -110 3-5/22 ďďď 11/21 ůúű

Add -2 times second row to third
(to get 0's in the second column)
 éęë 100 -110 3-5/21 ďďď 11/21/2 ůúű
Divide third row by 2
(to get a leading 1)
 éęë 100 -110 001 ďďď -1/27/41/2 ůúű
Add -3 times third row to first row
Add 5/2 times third row to second row
(to get 0's in third column)
 éęë 100 010 001 ďďď 5/47/41/2 ůúű
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. Math 15H (Fall 2002) web pages Created: 26 Mar 2002 Last modified: Nov 14, 2002 10:17:22 PM Comments to: `dpvc@union.edu`