Up: Math 15 Selected Class 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.

+-               -+
|  0  2 -3  |  2  |       write the system as an augmented matrix
|  2  0  1  |  3  |
|  1 -1  3  |  1  |
+-               -+

+-               -+
|  1 -1  3  |  1  |       interchange first and third row
|  2  0  1  |  3  |       (to make top left entry non-zero)
|  0  2 -3  |  2  |
+-               -+

+-               -+
|  1 -1  3  |  1  |       Add -2 times first row to second row
|  0  2 -5  | -1  |       (to get 0 in first column of row 2)
|  0  2 -3  |  2  |
+-               -+

+-                   -+
|  1 -1  3    |  1    |   Divide second row by 2
|  0  1 -5/2  | -1/2  |   (to get a leading 1 in row 2)
|  0  2 -3    |  2    |
+-                   -+

+-                   -+
|  1  0  1/2  |  3/2  |   Add second row to first
|  0  1 -5/2  | -1/2  |   Add -2 times second row to third
|  0  0  2    |  1    |   (to get 0's in the second column)
+-                   -+

+-                   -+
|  1  0  1/2  |  3/2  |   Divide third row by 2
|  0  1 -5/2  | -1/2  |   (to get a leading 1)
|  0  0  1    |  1/2  |
+-                   -+

+-                 -+
|  1  0  0  |  5/4  |     Add -1/2 times third row to first row
|  0  1  0  |  7/4  |     Add 5/2 times third row to second row
|  0  0  1  |  1/2  |     (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.

Up: Math 15 Selected Class Notes