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x + 3y = 7 size 12{x+3y=7} {}
3x + 4y = 11 size 12{3x+4y="11"} {}

If we multiply the first row by – 3, and add it to the second row, we get,

x + 3y = 7 size 12{x+3y=7} {}
5y = 10 size 12{ - 5y= - "10"} {}

And once again, the same solution is maintained.

Now that we understand how the three row operations work, it is time to introduce the Gauss-Jordan method to solve systems of linear equations.

As mentioned earlier, the Gauss-Jordan method starts out with an augmented matrix, and by a series of row operations ends up with a matrix that is in the reduced row echelon form . A matrix is in the reduced row echelon form if the first nonzero entry in each row is a 1, and the columns containing these 1's have all other entries as zeros. The reduced row echelon form also requires that the leading entry in each row be to the right of the leading entry in the row above it, and the rows containing all zeros be moved down to the bottom.

We state the Gauss-Jordan method as follows.

Gauss-jordan method

  1. Write the augmented matrix.
  2. Interchange rows if necessary to obtain a non-zero number in the first row, first column.
  3. Use a row operation to make the entry in the first row, first column, a 1.
  4. Use row operations to make all other entries as zeros in column one.
  5. Interchange rows if necessary to obtain a nonzero number in the second row, second column. Use a row operation to make this entry 1. Use row operations to make all other entries as zeros in column two.
  6. Repeat step 5 for row 3, column 3. Continue moving along the main diagonal until you reach the last row, or until the number is zero. The final matrix is called the reduced row-echelon form.
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Solve the following system by the Gauss-Jordan method.

2x + y + 2z = 10 size 12{2x+y+2z="10"} {}
x + 2y + z = 8 size 12{x+2y+z=8} {}
3x + y z = 2 size 12{3x+y - z=2} {}

We write the augmented matrix.

2 1 2 10 1 2 1 8 3 1 1 2 size 12{ left [ matrix { 2 {} # 1 {} # 2 {} # \lline {} # "10" {} ##1 {} # 2 {} # 1 {} # \lline {} # 8 {} ## 3 {} # 1 {} # - 1 {} # \lline {} # 2{}} right ]} {}

We want a 1 in row one, column one. This can be obtained by dividing the first row by 2, or interchanging the second row with the first. Interchanging the rows is a better choice because that way we avoid fractions.

1 2 1 8 2 1 2 10 3 1 1 2 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ##2 {} # 1 {} # 2 {} # \lline {} # "10" {} ## 3 {} # 1 {} # - 1 {} # \lline {} # 2{}} right ]} {}

we interchanged row 1(R1) and row 2(R2)

We need to make all other entries zeros in column 1. To make the entry (2) a zero in row 2, column 1, we multiply row 1 by - 2 and add it to the second row. We get,

1 2 1 8 0 3 0 6 3 1 1 2 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ##0 {} # - 3 {} # 0 {} # \lline {} # - 6 {} ## 3 {} # 1 {} # - 1 {} # \lline {} # 2{}} right ]} {}       − 2R1 + R2 size 12{ -2R1+R2} {}

To make the entry (3) a zero in row 3, column 1, we multiply row 1 by –3 and add it to the third row. We get,

1 2 1 8 0 3 0 6 0 5 4 22 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ##0 {} # - 3 {} # 0 {} # \lline {} # - 6 {} ## 0 {} # - 5 {} # - 4 {} # \lline {} # - "22"{}} right ]} {}       − 3R1 + R3 size 12{ -3R1+R3} {}

So far we have made a 1 in the left corner and all other entries zeros in that column. Now we move to the next diagonal entry, row 2, column 2. We need to make this entry(–3) a 1 and make all other entries in this column zeros. To make row 2, column 2 entry a 1, we divide the entire second row by –3.

1 2 1 8 0 1 0 2 0 5 4 22 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ##0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # - 5 {} # - 4 {} # \lline {} # - "22"{}} right ]} {}       R2 ÷ 3 size 12{R2 div left ( -3 right )} {}

Next, we make all other entries zeros in the second column.

1 0 1 4 0 1 0 2 0 0 4 12 size 12{ left [ matrix { 1 {} # 0 {} # 1 {} # \lline {} # 4 {} ##0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # 0 {} # - 4 {} # \lline {} # - "12"{}} right ]} {}       − 2R2 + R1 size 12{ - 2R2+R1} {} and 5 R 2 + R 3

We make the last diagonal entry a 1, by dividing row 3 by – 4.

1 0 1 4 0 1 0 2 0 0 1 3 size 12{ left [ matrix { 1 {} # 0 {} # 1 {} # \lline {} # 4 {} ##0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # 3{}} right ]} {}       R3 ÷ 4 size 12{R3 div left ( - 4 right )} {}

Finally, we make all other entries zeros in column 3.

1 0 0 1 0 1 0 2 0 0 1 3 size 12{ left [ matrix { 1 {} # 0 {} # 0 {} # \lline {} # 1 {} ##0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # 3{}} right ]} {}       − R3 + R1 size 12{ -R3+R1} {}

Clearly, the solution reads x = 1 size 12{x=1} {} , y = 2 size 12{y=2} {} , and z = 3 size 12{z=3} {} .

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Before we leave this section, we mention some terms we may need in the fourth chapter. The process of obtaining a 1 in a location, and then making all other entries zeros in that column, is called pivoting . The number that is made a 1 is called the pivot element , and the row that contains the pivot element is called the pivot row . We often multiply the pivot row by a number and add it to another row to obtain a zero in the latter. The row to which a multiple of pivot row is added is called the target row .

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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