A less cumbersome way to represent a linear system is to write it as an augmented matrix. For example the linear system, 18.4 can be written as

It has exactly the same information as the original system but here it is understood there is an x column,

Now when you replace an equation with a multiple of another equation added to itself, you are just taking a row of this augmented matrix and replacing it with a multiple of another row added to it. Thus the first step in solving 18.4 would be to take

Note how this corresponds to 18.5. Next take

This augmented matrix corresponds to the system

which is the same as 18.6. By back substitution you obtain the solution x = 1,y = 6, and z = 3.
In general a linear system is of the form
 (18.7) 
where the x_{i} are variables and the a_{ij} and b_{i} are constants. This system can be represented by the augmented matrix
 (18.8) 
Changes to the system of equations in 18.7 as a result of an elementary operations translate into changes of the augmented matrix resulting from a row operation. Note that Theorem 18.1.4 implies that the row operations deliver an augmented matrix for a system of equations which has the same solution set as the original system. To summarize, the following gives the row operations.
Gauss elimination is a systematic procedure to simplify an augmented matrix to a reduced form. In the following definition, the term “leading entry” refers to the first nonzero entry of a row when scanning the row from left to right.
Definition 18.1.7 An augmented matrix is in echelon form if
How do you know when to stop doing row operations? You might stop when you have obtained an echelon form as described above, but you certainly should stop doing row operations if you have gotten a matrix in row reduced echelon form described next.
Definition 18.1.8 An augmented matrix is in row reduced echelon form if
Example 18.1.9 Here are some matrices which are in row reduced echelon form.

Example 18.1.10 Here are matrices in echelon form which are not in row reduced echelon form but which are in echelon form.

Example 18.1.11 Here are some matrices which are not in echelon form.

Definition 18.1.12 A pivot position in a matrix is the location of a leading entry in an echelon form resulting from the application of row operations to the matrix. A pivot column is a column that contains a pivot position.
For example consider the following.
Replace the second row by −3 times the first added to the second. This yields

This is not in reduced echelon form so replace the bottom row by −4 times the top row added to the bottom. This yields

This is still not in reduced echelon form. Replace the bottom row by −1 times the middle row added to the bottom. This yields

which is in echelon form, although not in reduced echelon form. Therefore, the pivot positions in the original matrix are the locations corresponding to the first row and first column and the second row and second columns as shown in the following:

Thus the pivot columns in the matrix are the first two columns.
The following is the algorithm for obtaining a matrix which is in row reduced echelon form.
This algorithm tells how to start with a matrix and do row operations on it in such a way as to end up with a matrix in row reduced echelon form.
This row reduction procedure applies to both augmented matrices and non augmented matrices. There is nothing special about the augmented column with respect to the row reduction procedure.
Example 18.1.15 Here is a matrix.

Do row reductions till you obtain a matrix in echelon form. Then complete the process by producing one in row reduced echelon form.
The pivot column is the second. Hence the pivot position is the one in the first row and second column. Switch the first two rows to obtain a nonzero entry in this pivot position.

Step two is not necessary because all the entries below the first pivot position in the resulting matrix are zero. Now ignore the top row and the columns to the left of this first pivot position. Thus you apply the same operations to the smaller matrix

The next pivot column is the third corresponding to the first in this smaller matrix and the second pivot position is therefore, the one which is in the second row and third column. In this case it is not necessary to switch any rows to place a nonzero entry in this position because there is already a nonzero entry there. Multiply the third row of the original matrix by −2 and then add the second row to it. This yields

The next matrix the steps in the algorithm are applied to is

The first pivot column is the first column in this case and no switching of rows is necessary because there is a nonzero entry in the first pivot position. Therefore, the algorithm yields for the next step

Now the algorithm will be applied to the matrix

There is only one column and it is nonzero so this single column is the pivot column. Therefore, the algorithm yields the following matrix for the echelon form.

To complete placing the matrix in reduced echelon form, multiply the third row by 3 and add −2 times the fourth row to it. This yields

Next multiply the second row by 3 and take 2 times the fourth row and add to it. Then add the fourth row to the first.

Next work on the fourth column in the same way.

Take −1∕2 times the second row and add to the first.

Finally, divide by the value of the leading entries in the nonzero rows.

The above algorithm is the way a computer would obtain a reduced echelon form for a given matrix. It is not necessary for you to pretend you are a computer but if you like to do so, the algorithm described above will work. The main idea is to do row operations in such a way as to end up with a matrix in echelon form or row reduced echelon form because when this has been done, the resulting augmented matrix will allow you to describe the solutions to the linear system of equations in a meaningful way. When you do row operations until you obtain row reduced echelon form, the process is called the Gauss Jordan method. Otherwise, it is called Gauss elimination.
Example 18.1.16 Give the complete solution to the system of equations, 5x + 10y − 7z = −2, 2x + 4y − 3z = −1, and 3x + 6y + 5z = 9.
The augmented matrix for this system is

If you follow the algorithm, you end up with the row reduced echelon form is

Thus there is no solution to the system of equations because you would have to satisfy an equation of the form 0x + 0y + 0z = 1. When this happens, the system is called inconsistent. In this case it is very easy to describe the solution set. The system has no solution.
Here is another example based on the use of row operations.
Example 18.1.17 Give the complete solution to the system of equations, 3x−y− 5z = 9, y − 10z = 0, and −2x + y = −6.
The augmented matrix of this system is

The row reduced echelon form is

The equations corresponding to this reduced echelon form are y = 10z and x = 3 + 5z. Apparently z can equal any number. Lets call this number t. ^{1}Therefore, the solution set of this system is x = 3 + 5t,y = 10t, and z = t where t is completely arbitrary. The system has an infinite set of solutions which are given in the above simple way. This is what it is all about, finding the solutions to the system.
There is some terminology connected to this which is useful. Recall how each column corresponds to a variable in the original system of equations. The variables corresponding to a pivot column are called basic variables. The other variables are called free variables. In Example 18.1.17 there was one free variable, z, and two basic variables, x and y. In describing the solution to the system of equations, the free variables are assigned a parameter. In Example 18.1.17 this parameter was t. Sometimes there are many free variables and in these cases, you need to use many parameters. Here is another example.
Example 18.1.18 Find the solution to the system

The augmented matrix is

The row reduced echelon form is

This is another example of a system which has an infinite solution set but this time the solution set depends on two parameters, not one. Most people find it less confusing in the case of an infinite solution set to first place the augmented matrix in row reduced echelon form rather than just echelon form before seeking to write down the description of the solution. Then the solution is y = 2 from the second row and x = −1 + z − w from the first. Thus letting z = s and w = t, the solution is given as follows.

As indicated above, you are probably less likely to become confused if you place the augmented matrix in row reduced echelon form rather than just echelon form.
In summary,
Definition 18.1.19 A system of linear equations is a list of equations,

where a_{ij} are numbers, and b_{j} is a number. The above is a system of m equations in the n variables, x_{1},x_{2}

It is desired to find
As illustrated above, such a system of linear equations may have a unique solution, no solution, or infinitely many solutions and these are the only three cases which can occur for any linear system. Furthermore, you do exactly the same things to solve any linear system. You write the augmented matrix and do row operations until you get a simpler system in which it is possible to see the solution, usually obtaining a matrix in echelon or reduced echelon form. All is based on the observation that the row operations do not change the solution set. You can have more equations than variables, fewer equations than variables, etc. It doesn’t matter. You always set up the augmented matrix and go to work on it.
Definition 18.1.20 A system of linear equations is called consistent if there exists a solution. It is called inconsistent if there is no solution.
These are reasonable words to describe the situations of having or not having a solution. If you think of each equation as a condition which must be satisfied by the variables, consistent would mean there is some choice of variables which can satisfy all the conditions. Inconsistent would mean there is no choice of the variables which can satisfy each of the conditions.