C Algorithm for solving Linear Equations using Gauss Elimination Method

In this tutorial we will be learning how to use Matrix Elimination to solve a linear system. Matrix Elimination is one of many techniques that can be used to solve systems of linear equation.

Here is an sample equation with solution

3x-5y+z=-23

-2x+3y-3z=19

4x+5z=-18

Step 1: Write the equation in Augmented Matrix.

Step 2: Use rows one and two to create the first in row two and we need to multiply row one by 2 and row two by 3  → 2R1 + 3R2 = R2

Step 3: Use rows one and three to create the second zero in row three and we need to multiply row one by -4 and row three by 3 → –4R1 + 3R3 = R3

Step 4: Use rows two and three to create the final zero in row three and we need to multiply row two by 20 and row three by 1 → 20R2 + R3 = R3

Step 5: To find the values of x, y, and z we want to use back substitution.

Using row three:     –129z = 258

                                      z = –2

Using row two:   –y – 7z = 11

                            –y – 7(–2) = 11       put z=-2 and 

                            –y + 14 = 11            solve for y

                            –y = –3

                              y = 3

Using row one: 3x – 5y + z = –23

                          3x – 5(3) + (–2) = –23

                          3x – 17 = –23              put z=-2, y=3 

                          3x = –6                        and solve x

                            x = –2

Program:

#include<stdio.h>

int main()

{

    int i,j,r,n;

   

    double a[15][15][15],x[15],sum=0;

   

    printf("\n Enter the order of matrix :");

   

    scanf("%d",&n);

   

    printf("\n Enter the elements of the augmented matrix row-wise :\n\n");

   

    for(i=0;i<n;i++)

    {

        for(j=0;j<n+1;j++)

        {

            printf("a[%d][%d]:",i+1,j+1);

            scanf("%lf",&a[i][j][0]);

        }

    }

    // Here is loop for the generation of upper triangular matrix

    for(r=0;r<=n-1;r++)

    {

        for(i=r+1;i<=n-1;i++)

        {

            for(j=r+1;j<=n;j++)

            {

                a[i][j][r+1] = a[i][j][r] - a[r][j][r]*a[i][r][r]/a[r][r][r];

             }

        }

    }

    x[n]=a[n-1][n][n-1]/a[n-1][n-1][n-1];  

   

// Here is loop for backward substitution

   

    for(i=1;i<=n-1;i++)

    {

        sum=a[n-i-1][n][n-i-1];

        for(j=1;j<=i;j++)

        {

            sum = sum - a[n-i-1][n-j][n-i-1] * x[n-j+1];

        }

        x[n-i]=sum/a[n-i-1][n-i-1][n-i-1];

    }

    printf("\n The solution is :\n");

    for(i=0;i<=n-1;i++)

    {

        printf("\n x%d= %13.8lf\t",i+1,x[i+1]);

    }

   

    return 0;

}

OUTPUT: