Gauss-Jordan EliminationIn this example we solve a system of linear equations by writing the system as an “augmented” matrix and reducing thatmatrix to Reduced Row Echelon Form.

2 x1 + 4 x2 + 6 x3 = 184 x1 + 5 x2 + 6 x3 = 243 x1 + x2 - 2 x3 = 4

2 4 6 184 5 6 243 1 -2 4

Our first goal is to get aleading one in the first row;divide the first row by 2.

x1 + 2 x2 + 3 x3 = 94 x1 + 5 x2 + 6 x3 = 243 x1 + x2 - 2 x3 = 4

1 2 3 94 5 6 243 1 -2 4

R1 ®1

2R1

Now we use this leadingone to eliminate the entriesbelow it.

x1 + 2 x2 + 3 x3 = 9- 3 x2 - 6 x3 = -12

3 x1 + x2 - 2 x3 = 4

1 2 3 90 -3 -6 -123 1 -2 4

R2 ® R2 - 4 R1 Multiply the first row by - 4and add it to the second row.

x1 + 2 x2 + 3 x3 = 9- 3 x2 - 6 x3 = -12- 5 x2 - 11 x3 = -23

1 2 3 90 -3 -6 -120 -5 -11 -23

R3 ® R3 - 3 R1 Multiply the first row by - 3and add it to the third row.

x1 + 2 x2 + 3 x3 = 9x2 + 2 x3 = 4

- 5 x2 - 11 x3 = -23

1 2 3 90 1 2 40 -5 -11 -23

R2 ® -1

3R2

Now we get a leading onein the second row; dividethe second row by - 3.

x1 + 2 x2 + 3 x3 = 9x2 + 2 x3 = 4

- x3 = -3

1 2 3 90 1 2 40 0 -1 -3

R3 ® R3 + 5 R2 Multiply the second row by5 and add it to the third row.

x1 + 2 x2 + 3 x3 = 9x2 + 2 x3 = 4

x3 = 3

1 2 3 90 1 2 40 0 1 3

R3 ® -R3 Multiply the third rowby - 1 to get a leading one;this is Row Echelon Form.

x1 - x3 = 1x2 + 2 x3 = 4

x3 = 3

1 0 -1 10 1 2 40 0 1 3

R1 ® R1 - 2 R2 Now eliminate entries abovethe leading ones. Multiplythe second row by - 2 andadd it to the first row.

x1 - x3 = 1x2 = -2

x3 = 3

1 0 -1 10 1 0 -20 0 1 3

R2 ® R2 - 2 R3 Multiply the third rowby -2 and add it to thesecond row.

x1 = 4x2 = -2

x3 = 3

1 0 0 40 1 0 -20 0 1 3

R1 ® R1 + R3 Add the third row to the firstrow; this is Reduced Row

Echelon Form.

The solution to the system is the single point H4, -2, 3L. Some of the steps could be carried out in a different order. Forexample, you could eliminate the entry above the leading one in the second row before creating a leading one in the thirdrow.