Gauss Elimination Method 

The Gauss Elimination Method is one of the most commonly used techniques for solving systems of linear equations. It’s a systematic way to eliminate variables, transforming the original system of equations into a form that can be easily solved. In this blog, we will delve into the Gauss Elimination method, its applications, and how to use it to solve a system of equations. Let’s break down everything you need to know!

1.0What is the Gauss Elimination Method?

The Gauss Elimination method is an algorithm used to solve systems of linear equations. It involves performing a sequence of operations on the augmented matrix of a system of equations, gradually reducing it to a simpler form. This form, often called row echelon form, can then be solved easily through back-substitution. The method systematically eliminates variables to reduce the complexity of solving for unknowns.

Steps of the Gauss Elimination Method:

1. Form the augmented matrix of the system of equations.
2. Use row operations to transform the augmented matrix into an upper triangular form (row echelon form).
3. Once the matrix is in row echelon form, use back-substitution to find the values of the unknowns.

2.0Gauss Elimination Method Formula

The core of the Gauss Elimination method lies in performing three basic operations on rows of the matrix:

1. Swapping two rows.
2. Multiplying a row by a non-zero scalar.
3. Adding a multiple of one row to another row.

These operations are done to make the matrix into an upper triangular form (where all the elements below the main diagonal are zero). Afterward, the system can be solved through back-substitution.

For example, consider the system of equations:

$\begin{array}{r}x+2y+3z=9\\ 2x+3y+3z=8\\ 3x+3y+2z=7\end{array}$

This system can be written as the augmented matrix:

$\left[\begin{array}{c}3\; 4\; 1\; 1\\ 2\; 3\; 3\; 5\\ 1\; 2\; 3\; 3\end{array}\right]$

Through the Gauss Elimination method, you would use the steps outlined to manipulate this matrix, leading to an upper triangular matrix, which can then be solved for x, y, and z.

3.0Application of Gauss Elimination Method

The Gauss Elimination method is widely used in various fields, including:

1. Engineering: For solving systems that arise in structural analysis, electrical circuits, and fluid dynamics.
2. Physics: For solving problems related to motion, forces, or electrical systems.
3. Computer Science: In algorithms related to linear algebra, data fitting, and machine learning models.
4. Economics: Solving systems related to economic models and optimization problems.

By reducing the complexity of solving simultaneous equations, the Gauss Elimination method has made a significant impact in mathematical and applied fields.

4.0Example of Gauss Elimination Method

Let’s consider an example of solving a system of linear equations:

$\begin{array}{r}x+2y+3z=9\\ 2x+3y+3z=8\\ 3x+3y+2z=7\end{array}$

Step 1: Write the augmented matrix:

$\left[\begin{array}{c}1\; 2\; 3\; 9\\ 2\; 3\; 3\; 8\\ 3\; 1\; 2\; 7\end{array}\right]$

Step 2: Apply row operations to eliminate variables below the first pivot (the 1 in the first row, first column):

Subtract 22 times row 1 from row 2, and subtract 33 times row 1 from row 3:

$\left[\begin{array}{c}123|9\\ 0-1-3\; |-10\\ 0-5-7\; |-20\end{array}\right]$

Step 3: Now, make the second pivot a 1 by dividing row 2 by −1-1:

$\left[\begin{array}{c}123|9\\ 013|10\\ 0-5-7\; |-20\end{array}\right]$

Step 4: Eliminate the y-term from the third row by adding 5 times row 2 to row 3:

$\left[\begin{array}{c}1\; 2\; 3\; 9\\ 0\; 1\; 3\; 10\\ 0\; 0\; 8\; 30\end{array}\right]$

Step 5: Solve for z from the third row:

$z=\frac{30}{8}=3.75$

Step 6: Substitute z = 3.75 into the second row to solve for y:

y + 3(3.75) = 10 

⇒ y = −1.25 

Step 7: Finally, substitute y = –1.25 and z = 3.75 into the first row to solve for x:

x + 2(−1.25) + 3(3.75) = 9

⇒ x = 1.5 

Thus, the solution to the system of equations is:

x = 1.5, y = −1.25, z = 3.75