naive gaussian elimination

Guri Bee

2 min read

·

16-10-2024

1

0

Share

Demystifying Naive Gaussian Elimination: A Step-by-Step Guide

Gaussian elimination is a fundamental technique in linear algebra used to solve systems of linear equations. While powerful, the naive version of Gaussian elimination is susceptible to numerical instability and can lead to inaccurate results, especially for ill-conditioned matrices. This article will delve into the workings of naive Gaussian elimination, highlighting its strengths and limitations.

What is Naive Gaussian Elimination?

Naive Gaussian elimination aims to transform a system of linear equations into an equivalent system in upper triangular form. This form makes it straightforward to solve for the variables using back-substitution. The method involves a series of elementary row operations on the augmented matrix of the system:

1. Forward Elimination: This phase systematically eliminates variables below the diagonal of the coefficient matrix. It involves performing row operations to create zeros in the lower triangular part of the matrix.

2. Back Substitution: With the matrix in upper triangular form, we can solve for the variables by starting from the last equation and working backwards, substituting the values of known variables.

Let's illustrate with an example:

Consider the following system of equations:

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

The augmented matrix for this system is:

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

Step 1: Forward Elimination

• Eliminate x from the second and third equations:
  • Divide the first row by 2 (R1/2).
  • Add 3 times the first row to the second row (R2 + 3*R1).
  • Add 2 times the first row to the third row (R3 + 2*R1).

This results in the following matrix:

[ 1  0.5  -0.5  |  4 ]
[ 0  0.5  0.5  |  1 ]
[ 0  2  1  |  5 ] 

• Eliminate y from the third equation:
  • Divide the second row by 0.5 (R2/0.5).
  • Subtract 2 times the second row from the third row (R3 - 2*R2).

This yields the upper triangular form:

[ 1  0.5  -0.5  |  4 ]
[ 0  1  1  |  2 ]
[ 0  0  -1  |  1 ]

Step 2: Back Substitution

• Solve for z from the third equation:

  • z = -1

• Substitute z = -1 into the second equation and solve for y:

  • y + (-1) = 2
  • y = 3

• Substitute z = -1 and y = 3 into the first equation and solve for x:

  • x + 0.5(3) - 0.5(-1) = 4
  • x = 2

Therefore, the solution to the system of equations is x = 2, y = 3, and z = -1.

Caveats of Naive Gaussian Elimination:

While effective for many systems, naive Gaussian elimination has some significant drawbacks:

• Division by Zero: The method can fail if a pivot element (the leading coefficient of a row) is zero. This can be mitigated by using pivoting techniques.
• Numerical Instability: Rounding errors can accumulate during the elimination process, particularly for ill-conditioned matrices. This can lead to inaccurate results.
• Computational Complexity: Naive Gaussian elimination has a time complexity of O(n^3), making it computationally expensive for large systems.

Key Points to Remember:

• Naive Gaussian elimination is a straightforward method for solving systems of linear equations.
• It relies on forward elimination to transform the system into upper triangular form and then back substitution to find the solutions.
• It's crucial to be aware of the potential for division by zero and numerical instability.

Additional Resources:

• Python implementation of Naive Gaussian Elimination on GitHub
• Detailed Explanation of Gaussian Elimination

By understanding the mechanics and limitations of naive Gaussian elimination, you can gain valuable insights into its strengths and weaknesses, enabling you to make informed decisions when choosing the appropriate method for solving your linear systems.