gaussian elimination and gauss-jordan elimination

3 min read 14-03-2025

Gaussian elimination and Gauss-Jordan elimination are fundamental methods in linear algebra used to solve systems of linear equations. They are efficient algorithms that transform a system of equations into a simpler form, making it easier to find the solution. While both methods achieve the same goal, they differ slightly in their approach and final result. This article will explore both techniques, highlighting their similarities and differences.

Understanding Systems of Linear Equations

Before diving into the elimination methods, let's clarify what we're dealing with. A system of linear equations is a set of equations where each equation is linear (meaning the highest power of the variables is 1). For example:

2x + y - z = 8
x - y + 2z = 3
3x + 2y - z = 11

Our goal is to find the values of x, y, and z that simultaneously satisfy all three equations.

Gaussian Elimination: Row Reduction to Echelon Form

Gaussian elimination, also known as row reduction, is a systematic process of transforming the augmented matrix of the system into row echelon form. The augmented matrix combines the coefficients of the variables and the constants. For the example above, the augmented matrix would be:

[ 2  1 -1 | 8 ]
[ 1 -1  2 | 3 ]
[ 3  2 -1 | 11]

The steps involved are:

Forward Elimination: Use elementary row operations (swapping rows, multiplying a row by a non-zero constant, adding a multiple of one row to another) to create zeros below the main diagonal. This process is done systematically, one column at a time.
Back Substitution: Once in row echelon form, the solution is obtained through back substitution. This involves solving for the variables starting from the last row and working upwards.

Example (Gaussian Elimination):

Let's reduce the augmented matrix to row echelon form:

Swap row 1 and row 2 to get a leading 1 in the first column:

[ 1 -1  2 | 3 ]
[ 2  1 -1 | 8 ]
[ 3  2 -1 | 11]

Subtract 2 times row 1 from row 2 and 3 times row 1 from row 3:

[ 1 -1  2 | 3 ]
[ 0  3 -5 | 2 ]
[ 0  5 -7 | 2 ]

Divide row 2 by 3:

[ 1 -1  2 | 3 ]
[ 0  1 -5/3 | 2/3 ]
[ 0  5 -7 | 2 ]

Subtract 5 times row 2 from row 3:

[ 1 -1  2 | 3 ]
[ 0  1 -5/3 | 2/3 ]
[ 0  0  4/3 | -4/3 ]

Now, we have row echelon form. Back substitution yields z = -1, y = 1, and x = 2.

Gauss-Jordan Elimination: Row Reduction to Reduced Row Echelon Form

Gauss-Jordan elimination extends Gaussian elimination by further reducing the matrix to reduced row echelon form. This means that not only are there zeros below the main diagonal, but also zeros above it, and the leading entries are all 1s.

Steps in Gauss-Jordan Elimination:

Follow the forward elimination steps of Gaussian elimination to achieve row echelon form.
Backward Elimination: Use row operations to create zeros above the main diagonal. This involves working from the last column to the first.
Normalization: Divide each row by its leading entry to make the leading entries 1.

Example (Gauss-Jordan Elimination):

Continuing from the row echelon form obtained in the Gaussian elimination example:

Subtract 2 times row 3 from row 1 and (5/3) times row 3 from row 2.
Add row 2 to row 1.
Divide row 3 by 4/3.

This will result in the reduced row echelon form, directly giving the solution: x=2, y=1, z=-1.

Comparison: Gaussian vs. Gauss-Jordan Elimination

Feature	Gaussian Elimination	Gauss-Jordan Elimination
Final Form	Row echelon form	Reduced row echelon form
Steps	Forward elimination, back substitution	Forward and backward elimination
Computational Cost	Less computationally expensive	More computationally expensive
Solution	Requires back substitution	Solution is directly read from the matrix

Applications

Both Gaussian and Gauss-Jordan elimination are widely used in various fields:

Solving systems of linear equations: This is the primary application.
Finding matrix inverses: The Gauss-Jordan method can efficiently compute the inverse of a square matrix.
Solving linear programming problems: These methods are used as part of the simplex method.
Computer graphics: Used in transformations and calculations.

Conclusion

Both Gaussian and Gauss-Jordan elimination are powerful tools for solving systems of linear equations. While Gauss-Jordan requires more steps, it provides the solution directly. Gaussian elimination is often preferred for its efficiency if back substitution isn't a major concern. The choice of method depends on the specific context and computational resources available. Understanding both methods provides a strong foundation in linear algebra.