**Gaussian Elimination** Method in C++: A Quick and Efficient Solution for Linear Equations

Looking for a fast and efficient way to solve systems of linear equations?

Our blog post explores the Gaussian elimination method implemented in C++, providing you with a streamlined solution for your numerical analysis needs.

## Gauss Elimination Method C++

**Gauss Elimination Method** is a direct method to solve the system of linear equations.

It is quite general and well-adaptive to computer operations and Numerical Techniques.

Gauss Elimination Method gives us the exact value of variables.

Simultaneous linear equations occur quite often in computational processes in almost every field.

**In engineering and science****.****Economics and Statistics****Analysis of Electronic Circuits****Output and Cost in Chemical Plants****Optimizing Techniques****Graphical methods and illustrations****Numerical Techniques**

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Lu Decomposition Method C++

So the solutions to these problems can be found through **Direct and Iterative Methods**.

Also Read:

Partial Pivoting C++ Program

## Gauss Elimination Method C++ Program

```
//Gauss Elimination Method C++ code
//techindetail.com
#include<iostream>
#include<iomanip>
#include<cmath>
#define N 3
using namespace std;
int main()
{
float Matrix[N][N+1],x[N];
// Matrix = Augumented Matrix [Ad]
float temp,s;
//variables for loops
int i,j,k;
//Scan values of Matrix.
cout<<"Enter Elements of "<<N<<" Rows & "<<N+1<<" Columns\n";
cout<<fixed;
for(i=0; i<N; i++)
{
cout<<"\tEnter Row "<<i+1<<" & Press Enter\n";
for(j=0; j<N+1; j++)
cin>>Matrix[i][j];
}
//make above matrix upper triangular Matrix
for(j=0; j<N-1; j++)
{
for(i=j+1; i<N; i++)
{
temp=Matrix[i][j]/Matrix[j][j];
for(k=0; k<N+1; k++)
Matrix[i][k]-=Matrix[j][k]*temp;
}
}
//print the Upper Triangular matrix
cout<<"\n ---------------------------------\n";
cout<<"\n Upper Triangular Matrix is:\n";
for(i=0; i<N; i++)
{
for(j=0; j<N+1; j++)
cout<<setw(8)<<setprecision(4)<<Matrix[i][j];
cout<<endl;
}
//find values of x,y,z using back substitution
cout<<"\n ---------------------------------\n";
for(i=N-1; i>=0; i--)
{
s=0;
for(j=i+1; j<N; j++)
s += Matrix[i][j]*x[j];
x[i]=(Matrix[i][N]-s)/Matrix[i][i];
}
//print values of x,y,z
cout<<"\n The Solution is:\n";
for(i=0; i<N; i++)
cout<<"x["<<setw(3)<<i+1<<"]="<<setw(7)<<setprecision(4)<<x[i]<<endl;
return 0;
//techindetail.com
}
```

Code language: C++ (cpp)

### Gaussian Elimination Method C++ Program

Gaussian elimination is a powerful numerical method used to solve systems of linear equations.

By utilizing the Gaussian elimination algorithm, the program allows users to input a system of linear equations, and it quickly solves for the unknown variables, providing accurate solutions in a fraction of the time

With its speed and accuracy, this program is an indispensable tool for anyone dealing with linear algebra and numerical analysis.

Must Read:Gauss Jordan Method C++

**Example**

**Find the Solution of the following Linear Equations using the Gauss Elimination Method.**

### **x + y + z = 6**

x – y + z = 2

2x – y + 3z = 9

x – y + z = 2

2x – y + 3z = 9

*Sol: In this method, the variables are eliminated and the system is reduced to the upper triangular matrix from which the unknowns are found by back substitution.*

Related:

**Step 1:** Write the given System of Equations in the form of **AX = b**, i.e. Matrix Form.

Where as,**A** = Coefficient Matrix,**X** = variables (Column Matrix),**b** = constants (Column Matrix.

**Step 2: Find Augmented Matrix C = [ Ab ]**

**Step 3: Transform Augmented Matrix into Upper Triangular Matrix.**

In the upper Triangular matrix, all the elements below the Diagonal are zero

*Note: Only Row operations are allowed.*

Now as you can see that it is now reduced to an upper triangular matrix.

**Step 4: Find equations corresponding to upper triangular matrix.**

Now by using back Substitution reconstruct the equations and find the corresponding values of the variables x, y, and z.

**1.x + 1.y + 1.z = 60.x – 2y + 0.z = – 40.x – 0.y + 2z = 6**

Therefore,

**x + y + z = 6-2y = -42z = 6**

Now, Put the value of **y and z** into eq. 1.**X + 2 + 3 = 6 –> x = 1**

Therefore, the roots of the three equations are *x = 1 , y = 2, z= 3*

Suggestion:Gauss Elimination With Partial Pivoting C++