Gauss-Jordan Elimination applies elementary row operations to a matrix until a row-echelon form of the matrix is obtained; the same matrix process used for Gaussian Elimination. The reduction process of Gauss-Jordan continues the reduction process beyond Gaussian Elimination until a reduced row-echelon form is attained.

The advantage of using matrices to solve systems of linear equations is that it is a procedural and rule-based process. When the rules are followed exactly, the solving of a complex system of linear equations, equations that each has multiple terms with coefficients, is simplified.

To solve the following system of linear equations using Gauss-Jordan Elimination:

First write the linear equations system to be solved.

Linear Equations System to be Solved

x − 2y + 3z = 9

−x + 3y = −4

2x − 5y + 5z = 17

Next use Gaussian Elimination to obtain the row-echelon form of the linear system.

(Review Gaussian Elimination if you need to see how to transform a system of linear equations to row-echelon form)

Row-Echelon Form Matrix

3×4 |
x |
y |
z |
= c |

R1: |
1 |
−2 |
3 |
9 |

R2: |
0 |
1 |
3 |
5 |

R3: |
0 |
0 |
1 |
2 |

Now apply elementary row operations to obtain zeros above each of the leading 1’s.

Matrix (Iteration 1)

3×4 |
x |
y |
z |
= c |

R1: |
1 |
0 |
9 |
19 |

R2: |
0 |
1 |
3 |
5 |

R3: |
0 |
0 |
1 |
2 |

2R2 + R1

Matrix (Iteration 2)

3×4 |
x |
y |
z |
= c |

R1: |
1 |
0 |
0 |
1 |

R2: |
0 |
1 |
0 |
−1 |

R3: |
0 |
0 |
1 |
2 |

−9R3 + R1

−3R3 + R2

Reduced Row-Echelon Matrix

3×4 |
x |
y |
z |
= c |

R1: |
1 |
0 |
0 |
1 |

R2: |
0 |
1 |
0 |
−1 |

R3: |
0 |
0 |
1 |
2 |

This provides us the values to each variable x, y, and z of the system of linear equations.

1x = 1

1y = −1

1z = 2

x − 2y + 3z = 9

−x + 3y = −4

2x − 5y + 5z = 17

1 − 2(−1) + 3(2) = 9

−1 + 3(−1) = −4

2(1) − 5(−1) + 5(2) = 17

The system of equations is now solved by Gauss-Jordan Elimination.

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