2.2. GAUSS ELIMINATION 41

It has exactly the same information as the original system but here the columns correspondto the variables and the rows correspond to the equations in the system.

To solve the system, we can use Gauss elimination in the usual way. The solution set isnot changed by using the row operations. Take 3 = −2 times the top equation and add tothe second.  1 3 1 0

0 0 1 30 2 1 4

Now switch the bottom two rows.  1 3 1 0

0 2 1 40 0 1 3

Then take 4 times the bottom row and add to the top two. 1 3 0 2

0 2 0 10 0 1 3

Next multiply the second row by 3 1 3 0 2

0 1 0 30 0 1 3

Now take 2 times the second row and add to the top. 1 0 0 3

0 1 0 30 0 1 3

Therefore, the solution is x = y = z = 3. How do you know when to stop? You certainlyshould stop doing row operations if you have gotten a matrix in row reduced echelon formdescribed next. The leading entry of a row is the first nonzero row encountered whenstarting at the left entry and moving from left to right along the row.

Definition 2.2.1 An augmented matrix is in row reduced echelon form if

1. All nonzero rows are above any rows of zeros.

2. Each leading entry of a row is in a column to the right of the leading entries of anyrows above it.

3. All entries in a column above and below a leading entry are zero.

4. Each leading entry is a 1, the only nonzero entry in its column.

Echelon form means that the leading entries of successive rows fall from upper left tolower right.