Gauss Jordan Method

Consider a system of two equations in the two variables x and y

The Gauss Jordan method uses the elementary row operations to try to transform the initial augmented matrix

into one of the form

This gives the solution and

The next few examples will illustrate the method.

Example 1

Solve the system using Gauss Jordan .

( i ) 4x + 7y = 13

( ii ) -3x + 5y = 9

Solution .

The initital augmented matrix is

The first goal is to eliminate x from the second equation i.e. get a 0 where the -3 is. To do this we will first get a 1 where the 4 is.

Multiply row 1 by 1/4

The third type of row operation can now be used to get a 0 where the -3 is. We need to add 3 times row one to row two.

Add to row 2 3 times row 1

Note that column one now has the form we are after. Now start in column two. Get a one where the 41/4 is first.

Multiply row 2 by 4/41

Note that row two now tells us what y is.

Now eliminate y from equation one. We do this by adding to row one -7/4 times row two.

Add to row 1 -7/4 times row 2

This tells us that and

Example 2

Use Gauss Jordan to solve

( i ) 2x + 4y + 6z = 14

( ii ) -3x + y + 2z = -3

( iii ) 4x - 2y + z = 7

Solution .

Our goal is to use row operations to reach a matrix of the form

When we do, we will have the solution for x, y, and z gvien by the entries in the last column.

First enter the augmented matrix.

To get the first column in the desired form, first get a one where the 2 is. Then use the one to help get zeros where the -3 and 4 are. This will be done in the next three steps.

Multiply row 1 by 1/2

Add to row 2 3 times row 1

Add to row 3 -4 times row 1

Now that column one looks like the one in our goal work to get column two to have the desired form. To do this we need to get a one where the 7 is in the second row. Then use this one to eliminate the 2 and -10 in the first and third rows.

Multiply row 2 by 1/7

Add to row 1 -2 times row 2

Add to row 3 10 times row 2

The first two columns now look like the first two columns of our goal. Now move on to the third column. As before establish the one where it is needed first and then use it to get zeros in the other two rows.

Multiply row 3 by 7/33

Add to row 1 1/7 times row 3

Add to row 2 -11/7 times row 3

This last augmented matrix gives us our solution of x = 2, y =1, and z = 1

Example 3

Use Gauss Jordan to solve

( i ) x + 2y + 3z = 6

( ii ) 3x + 7y + 10z = 22

( iii) 2x + 5y + 7z = 14

Solution .

The augmented matrix is

Since the first column has the one in the correct location, use it to get zeros in the second and third rows of the first column.

Add to row 2 -3 times row 1

Add to row 3 -2 times row 1

Since column two has the one in the second row, we now use it to get zeros in the other rows of the second column.

Add to row 3 -1 times row 2

The last row tells us that this system has no solution. (The equation corresponding to the last row says 0 = -2 which we know is false.)