Math 15H (Notes)

# Gauss-Jordon Elimination Notes:

The procedure for Guass-Jordan Elimination is as follows:
1. Find the leftmost column that is not all zeros.

2. Interchange the top row (if necessary) with another row to bring a non-zero entry to the top of the column.

3. If the top entry is a, then multiply the top row by 1/a to form a leading 1 in that row.

4. Add multiples of this row to the other rows so that all other rows have a 0 in this column.

5. Cover up the top row and go back to step 1, considering only the rows bewlo this one (until step 4). Continue until the matrix is in reduced row-echelon form.

## Example:

Solve for x, y, and z in:
 2y - 3z = 2 2x +   z = 3 x - y + 3z = 1
Solution:
 éêë 021 20-1 -313 ïïï 231 ùúû
write the system as an augmented matrix
 éêë 120 -102 31-3 ïïï 132 ùúû
interchange first and third row
(to make top left entry non-zero)
 éêë 100 -122 3-5-3 ïïï 112 ùúû
Add -2 times first row to second row
(to get 0 in first column of row 2)
 éêë 100 -112 3-5/2-3 ïïï 11/22 ùúû
Divide second row by 2
(to get a leading 1 in row 2)
 éêë 100 010 1/2-5/22 ïïï 3/21/21 ùúû
Add -2 times second row to third
(to get 0's in the second column)
 éêë 100 010 1/2-5/21 ïïï 3/21/21/2 ùúû
Divide third row by 2
 éêë 100 010 001 ïïï 5/47/41/2 ùúû
Add -1/2 times third row to first row
Add 5/2 times third row to second row
(to get 0's in third column)
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 2000) web pages Created: 06 Feb 1998 Last modified: 10 Nov 2000 11:07:09 Comments to: `dpvc@union.edu`