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)

3×4 | x | y | z | = c |

R1: | 1 | -2 | 3 | 9 |

R2: | 0 | 1 | 3 | 5 |

R3: | 0 | 0 | 1 | 2 |

Row-Echelon Form Matrix

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

3×4 | x | y | z | = c |

R1: | 1 | 0 | 9 | 19 |

R2: | 0 | 1 | 3 | 5 |

R3: | 0 | 0 | 1 | 2 |

Matrix (Iteration 1)

2R2 + R1

3×4 | x | y | z | = c |

R1: | 1 | 0 | 0 | 1 |

R2: | 0 | 1 | 0 | -1 |

R3: | 0 | 0 | 1 | 2 |

Matrix (Iteration 2)

-9R3 + R1

-3R3 + R2

3×4 | x | y | z | = c |

R1: | 1 | 0 | 0 | 1 |

R2: | 0 | 1 | 0 | -1 |

R3: | 0 | 0 | 1 | 2 |

Reduced Row-Echelon Matrix

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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