Sometimes it is necessary to solve systems of equations. For example the problem could be to find x and y such that
 (1.2) 
The set of ordered pairs,
 (1.3) 
The second equation was replaced by −2 times the first equation added to the second. Thus the solution is y = 2, from −3y = −6 and now, knowing y = 2, it follows from the other equation that x + 2 = 7 and so x = 5.
Why exactly does the replacement of one equation with a multiple of another added to it not change the solution set? The two equations of 1.2 are of the form
 (1.4) 
where E_{1} and E_{2} are expressions involving the variables. The claim is that if a is a number, then 1.4 has the same solution set as
 (1.5) 
Why is this?
If
The other thing which does not change the solution set of a system of equations consists of listing the equations in a different order. Here is another example.
To solve this system replace the second equation by
 (1.7) 
Now take
 (1.8) 
At this point, you can tell what the solution is. This system has the same solution as the original system and in the above, z = 3. Then using this in the second equation, it follows y + 6 = 8 and so y = 2. Now using this in the top equation yields x + 6 + 18 = 25 and so x = 1.
This process is not really much different from what you have always done in solving a single equation. For example, suppose you wanted to solve 2x + 5 = 3x − 6. You did the same thing to both sides of the equation thus preserving the solution set until you obtained an equation which was simple enough to give the answer. In this case, you would add −2x to both sides and then add 6 to both sides. This yields x = 11.
In 1.8 you could have continued as follows. Add

Now add

a system which has the same solution set as the original system.
It is foolish to write the variables every time you do these operations. It is easier to write the system 1.6 as the following “augmented matrix”

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 1.6 would be to take

Note how this corresponds to 1.7. Next take

which is the same as 1.8. You get the idea I hope. Write the system as an augmented matrix and follow the procedure of either switching rows, multiplying a row by a non zero number, or replacing a row by a multiple of another row added to it. Each of these operations leaves the solution set unchanged. These operations are called row operations.
Definition 1.11.2 The row operations consist of the following
It is important to observe that any row operation can be “undone” by another inverse row operation. For example, if r_{1},r_{2} are two rows, and r_{2} is replaced with r_{2}^{′} = αr_{1} + r_{2} using row operation 3, then you could get back to where you started by replacing the row r_{2}^{′} with −α times r_{1} and adding to r_{2}^{′}. In the case of operation 2, you would simply multiply the row that was changed by the inverse of the scalar which multiplied it in the first place, and in the case of row operation 1, you would just make the same switch again and you would be back to where you started. In each case, the row operation which undoes what was done is called the inverse row operation.
Example 1.11.3 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

Multiply the second row by 2, the first row by 5, and then take

Now, combining some row operations, take

Putting in the variables, the last two rows say z = 1 and z = 21. This is impossible so the last system of equations determined by the above augmented matrix has no solution. However, it has the same solution set as the first system of equations. This shows there is no solution to the three given equations. When this happens, the system is called inconsistent.
This should not be surprising that something like this can take place. It can even happen for one equation in one variable. Consider for example, x = x + 1. There is clearly no solution to this.
Example 1.11.4 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

Replace the last row with 2 times the top row added to 3 times the bottom row. This gives

Next take −1 times the middle row and add to the bottom.

Take the middle row and add to the top and then divide the top row which results by 3.

This says y = 10z and x = 3 + 5z. Apparently z can equal any number. 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 and this is a good description of the solutions. This is what it is all about, finding the solutions to the system.
The phenomenon of an infinite solution set occurs in equations having only one variable also. For example, consider the equation x = x. It doesn’t matter what x equals.
Definition 1.11.6 A system of linear equations is a list of equations,

where a_{ij} are numbers, f_{j} is a number, and 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. It turns out these are the only three cases which can occur for linear systems. 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. 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. These things are all the same.
Example 1.11.7 Give the complete solution to the system of equations,−41x+15y = 168, 109x − 40y = −447, −3x + y = 12, and 2x + z = −1.
The augmented matrix is

To solve this multiply the top row by 109, the second row by 41, add the top row to the second row, and multiply the top row by 1/109. Note how this process combined several row operations. This yields

Next take 2 times the third row and replace the fourth row by this added to 3 times the fourth row. Then take

Take −1∕2 times the third row and add to the bottom row. Then take 5 times the third row and add to four times the second. Finally take 41 times the third row and add to 4 times the top row. This yields

It follows x =
You should practice solving systems of equations. Here are some exercises.