Elementary Row Operations

The three elementary row operations are

1. Interchange two rows.

2. Multiply a row by a nonzero value.

3. Add a multiple of one row to another row.

Although you may not recognize these, you've previously used all of these in solving systems of equations. If you replace the word row by equation, you will realize that you can switch the order of the equations without changing the solution. Yes, you've multiplied an equation by a nonzero value. It's what we do in going from 3x = 38 to x = 38/3. You have even used the third elementary row operation many times although you may not realize it.

Before we go any further with the applications of the elementary row operations it is important to realize that the three elementary row operations , when applied to an augmented matrix, do not change the solution to the corresponding system.

In the rest of this little section we will demonstrate this graphically for systems of linear equations in two variables.

Example 1 (unique solution)

Consider the following system

( i ) 4x + 3y = 24

( ii ) -4x + y = -8

The pair (3, 4) is a solution to the system since 4(3) +3(4) = 24 and -4(3) + 4 = -8. As we perform elementary row operations on this system we will keep checking this graphically to confirm that the operations do not change the solution.

First convert the system to an augmented matrix.

Here is a graph of this system. Note that the lines intersect at (3, 4).

Now let's experiment with the first row operation by switching the two rows in the matrix.

Switch row 1 and row 2

-4x + y = -8 is now the first equation and 4x + 3y = 24 is now the second one. We now graph these. The solution is still (3, 4). The only difference you'll note is in the colors.

Now let's illustrate the second row operation by multiplying row 1 by -1/4.

Multiply row 1 by -1/4

Here is the graph corresponding to this system. Nothing has changed.

We will now demonstrate the third type of row operation and again use a graph to show that the solution has not changed. Before doing so recall that the last matrix we had was the following

The third type of row operation is usually used to eliminate a variable from an equation. We could eliminate x from the second equation if we could get its coefficient to be zero. Row operation type 3 can be used to do this. We need to add to row 2, -4 times row 1.

Add to row 2 -4 times row 1

The graph of this system is given below. Note that the red line has changed into a horizontal line. However, the two lines still go through (3, 4).

Note that because the red line is horizontal, it is very easy to see that y is 4 because of where the red line intersects the y-axis.

We've illustrated the three row operations and seen that they do not change the solution set. We will continue with this to see how we get a matrix which has a form in which it is very easy to find the values of our solution. Recall the last matrix we had.

The last line of the matrix corresponds to 4y = 16. To get y we must multiply both side by 1/4. This will be accomplished in our next row operation

Multiply row 2 by 1/4

We can see from this last row that y = 4. Since we saw before that row operation type 2 does not change the graph, we will skip the graph.

To eliminate y from the first equation ( so it will just have x in it) we need to add (1/4) times row 2 to row 1.

Add to row 1 1/4 times row 2

This matrix says that x = 3 and y = 4. The graph of this last system now has a vertical line and a horizontal line which makes it easy to see the solution.

In the next example we will look at what happens to the augmented matrix when we have a system that has no solution.

Example 2 (no solution)

First change the system

( i ) -x + y = 2

( ii ) -x + y = 4

into an augmented matrix

When we graph this system, we will see what appear to be parallel lines.

Multiply row 1 by -1

Add to row 2 1 times row 1

What does this last matrix tell you?

Putting the last row into equation form gives us 0x+0y = 2 or 0 =2. This is a contradiciton which means that the system has no solution. The lines are indeed parallel in this case.

In the next example we will see what happens to the augmented matrix when there are an infinite number of solutions.

Example 3 (infinite number of solutions)

First put the following system into an augmented matrix.

( i ) 4x - 3y = 12

( ii ) -8x + 6y = -24

Look at the graph of this system of equations.

What has happened to the second line? Since the second equation is -2 times the first equation, the two equations represent the same line. Let's see what happens to the augmented matrix that will tell us that we have this situation.

Here is the orginal augmented matrix.

If we start as we have in the other problems we would first divide the first row by 4 so that the first entry would be a 1 (corresponding to one times x).

Multiply row 1 by 1/4

To eliminate x from the second equation we'd have to use row operation type three. Specifically, we need to add to row two 8 times row one.

Add to row 2 8 times row 1

Notice the row of zeros in the augmented matrix. Look at the graph of the current system.

The graph is the same as it was at the start. The row of zeros is an indication that we have an infinite number of solutions.

The first row of the last augmented matrix tells us that the solution is given by or (after solving for y) Since there are an infinite number of points on this line, we say that there are an infinite number of solutions.

Summary : When do you have no solution? When do you have an infinite number of solutions?

No Solution ( also called inconsistent)

The augmented matrix has a row with zeros followed by a nonzero number.

Infinite Number of Solutions ( also called dependent)

The augmented matrix has a row consisting of all zeros.