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# Gauss Elimination Method      LIVE
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## What is the Gauss Elimination Method?

A system of linear equations with various unknown factors is known as a system of linear equations. Unknown factors can be found in various equations, as we all know. To check all of the equations that make up the system, you must find the value for the unknown factors. We can state that the given system is a consistent independent system if there is a single solution that indicates one value for each unknown factor. If there are several solutions, the system has an infinite number of solutions, then we call it a consistent dependent system.

## What is the Gauss Elimination Method?

The method we use to perform the three types of matrix row operations on an augmented matrix obtained from a linear system of equations to find the solutions for such a system is known as the Gaussian elimination method.

Row reduction is a technique that consists of two stages: forward elimination and back substitution. The difference between these two Gaussian elimination method phases is the output they create, not the activities you can perform through them. The row reduction required to simplify the matrix in question into its echelon form is referred to as the forward elimination step. A goal of this stage is to demonstrate if the system of equations represented in the matrix has a single feasible solution, an unlimited number of solutions, or no solution at all.

If it is discovered that the system has no solution, there is no need to proceed to the next stage of row reduction.

The Gaussian elimination with the back substitution step is carried out if solutions for the variables involved in the linear system can be found. This final step yields a reduced echelon form of the matrix, which yields the system of linear equations' general solution.

The Gaussian elimination rules are the same as the rules for the three basic row operations, in other words, you can algebraically act on a matrix's rows in the following three ways:

• Interchanging two rows, for example, R2 ↔ R3

• Multiplying a row by a constant, for example, R1 → kR1 where k is some nonzero number

• Adding a row to another row, for example, R2 → R2 + 3R1

## How to do Gaussian Elimination

It's all about the matrix you have in your hands and the necessary row operations to simplify it, not a series of Gaussian elimination processes to follow to solve a system of linear equations. Let's start with our first Gauss elimination method example with solution for a better understanding of the process and the intuition required to work through it:

1. Solve the given set of equations by using Gauss elimination method:

x + y + z = 4

x + 4y + 3z = 8

x + 6y + 2z = 6

Solution:

Given system of equations are as follows,

x + y + z = 4

x + 4y + 3z = 8

x + 6y + 2z = 6

Now convert the equations in matrix form.

$\begin{bmatrix}1 & 1 & 1 \\1 & 4 & 3\\ 1 &6 &2 \end{bmatrix} \begin{bmatrix}4 \\8\\6 \end{bmatrix}$

$R_{2}-R_{1}\longrightarrow R_{2}$

$R_{3}-R_{1}\longrightarrow R_{3}$

$\begin{bmatrix}1 & 1 & 1 \\0 & 3 & 2\\ 0 &5 &1 \end{bmatrix} \begin{bmatrix}4 \\4\\2 \end{bmatrix}$

$2R_{3}-R_{2}\longrightarrow R_{3}$

$\begin{bmatrix}1 & 1 & 1 \\0 & 3 & 2\\ 0 &7 &0 \end{bmatrix} \begin{bmatrix}4 \\4\\0 \end{bmatrix}$

The above matrix is the resulting matrix. Hence convert the above matrix in equation form to get the values of x, y, z.

The final equations are as follows,

x + y + z = 4

3y + 2z = 4

7y = 0

From this set, we can automatically observe that the value of the variable y is equal to 0. Now by using this knowledge to substitute it on the second equation to solve for z, and then substitute both y and z values on the first equation to solve for x.

1. y = 0

1. 30 + 2z = 4

2z = 4

z = 2

1. x + y + z = 4

x + 0 + 2 = 4

x = 4-2

x = 2

And the final solution for the set is:

x = 2 ;

y = 0 ;

z = 2 ;

Hence it is solved.

## Solved Gauss elimination method problems

1. Find the value of the given variables by using Gauss elimination method:

x + 3y + 6z = 12

x + 4y + 5z = 14

x + 6y + 7z = 18

Solution:

The given equations are,

x + 3y + 6z = 12

x + 4y + 5z = 14

x + 6y + 7z = 18

Now convert the equations in matrix form.

$\begin{bmatrix}1 & 3 & 6 \\1 & 4 & 5\\ 1 &6 &7 \end{bmatrix} \begin{bmatrix}12 \\14\\18 \end{bmatrix}$

$R_{2}-R_{1}\longrightarrow R_{2}$

$R_{3}-R_{1}\longrightarrow R_{3}$

$\begin{bmatrix}1 & 3 & 6 \\0 & 1 & -1\\ 6 &-1 &1 \end{bmatrix} \begin{bmatrix}12 \\2\\6 \end{bmatrix}$

$R_{3}-3R_{2}\longrightarrow R_{3}$

$\begin{bmatrix}1 & 3 & 6 \\0 & 1 & -1\\ 0 &0 &4 \end{bmatrix} \begin{bmatrix}12 \\2\\0 \end{bmatrix}$

The above matrix is the resulting matrix. Hence convert the above matrix in equation form to get the values of x, y, z.

The final equations are as follows,

x + 3y + 6z = 12

y - z = 2

4z = 0

From this set, we can automatically observe that the value of the variable z is equal to 0. Now by using this knowledge to substitute it on the second equation to solve for y, and then substitute both y and z values on the first equation to solve for x.

1. z = 0

2. y -0 = 2

y = 2

1. x + 3y + 6z = 12

x + 32 + 60 = 12

x = 12- 6

x = 6

And the final solution for the set is:

x = 6 ;

y = 2 ;

z = 0 ;

Hence it is solved.

2. Solve the given set of equations by using Gauss elimination method:

x - y + z = 4

x - 4y + 2z = 8

x + 2y + 8z = 12

Solution:

The given equations are as follows,

x - y + z = 4

x - 4y + 2z = 8

x + 2y + 8z = 12

Now convert the equations in matrix form.

$\begin{bmatrix}1 & -1 & 1 \\1 & -4 & 2\\ 1 &2 &8 \end{bmatrix} \begin{bmatrix}4 \\8\\12 \end{bmatrix}$

$R_{2}-R_{1}\longrightarrow R_{2}$

$R_{3}-R_{1}\longrightarrow R_{3}$

$\begin{bmatrix}1 & -1 & 1 \\0 & -3 & 1\\ 0 &3 &7 \end{bmatrix} \begin{bmatrix}4 \\4\\8 \end{bmatrix}$

$R_{3}+R_{2}\longrightarrow R_{3}$

$\begin{bmatrix}1 & -1 & 1 \\0 & -3 & 1\\ 0 &0 &8 \end{bmatrix} \begin{bmatrix}4 \\4\\12 \end{bmatrix}$

The above matrix is the resulting matrix. Hence convert the above matrix in equation form to get the values of x, y, z.

The final equations are as follows,

x - y + z = 4

-3y + z = 4

8z = 12

From this set, we can observe that the value of the variable 8z is equal to 12. Firstly calculate the value of z in fraction form then use it for the rest of the variables.

8z = 12

z = 128 = 32

Hence the value of z is 32.

Now by using this knowledge to substitute it on the second equation to solve for y, and then substitute both y and z values on the first equation to solve for x.

1. z = 32

2. -3y + z = 4

-3y +32= 4

-6y + 3= 8

-6y = 5

y = -56

1. x - y + z = 4

x - 1(-56) + 1 32 = 4

6x + 5 + 9 = 24

6x = 24 - 14

x = 106

x = 53

And the final solution for the set is:

x = 53 ;

y = -56 ;

z = 32;

Hence it is solved.

## Practice Questions MCQs

1. In the Gauss Elimination Method, which of the following steps is not included?

• Elimination of unknowns

• Elimination of unknowns

• Reduction to an upper triangular system

• Back substitution to find unknowns

• Cofactor evaluation

Answer: Option d; The basic phases in Gauss Elimination are the elimination of unknowns, reduction to an upper triangular system, and back substitution to find unknowns.

2. For which of the following applications is the Gauss Elimination Method widely used?

• Computer operations

• Network circuit difficulties

• MATLAB operations

• Telecommunication operations

Answer: Option a); For computer processes, the Gauss Elimination Method is widely used. It's a quick and efficient method.

3. What is the name given to the coefficients of the equation produced during the elimination?

• Joints

• Pivots

• Calculated coefficients

• Operative coefficients

Answer: Option b); The pivots are the coefficients of the equation produced during the elimination.

## Conclusion

The Gaussian elimination technique is the most ancient and extensively used approach in linear algebra. We will examine the applications of the Gaussian Elimination technique in this paper, as it may be used in any sector. The Gaussian elimination method, which is the oldest and most extensively used, is not just limited to figures and numbers, but can also be utilised in a variety of sectors.

The key worry is whether it is becoming more complicated and whether more fresh and inventive approaches have been created. According to current analysts, there are many less viable ways that are still enticing to people who choose intricacy over solution, but in the long term, the Gaussian method still reigns supreme until further research is completed.

Last updated date: 21st Sep 2023
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## FAQs on Gauss Elimination Method

1. What is the Gauss elimination method in numerical methods?

Gauss elimination is a well-known numerical method that is used to solve a variety of scientific problems.

[A]{x}={C}

Where [A] is the coefficient matrix, x is the unknown vector, and C is the constant vector. The Gauss elimination method works by removing each unknown's coefficient from the equation one by one.

2. Why do we use the Gauss elimination method?

Gauss elimination method is used to find the value of variables. It's divided into a series of operations on the corresponding coefficient matrix. The rank of a matrix, the determinant of a square matrix, and the inverse of an invertible matrix can all be computed using this method.

3. What exactly is the distinction between Gauss Elimination and Gauss-Jordan?

Gaussian Elimination assists in converting a matrix to row-echelon form, whereas Gauss-Jordan Elimination converts a matrix to reduced row-echelon form. It is frequently more convenient for small systems to apply Gauss-Jordan elimination and directly solve for each variable represented in the matrix system.