SOURCE CODE FOR GAUSS ELIMINATION METHOD FOR FINDING THE SOLUTION OF LINEAR EQUATION

Gauss elimination method is one of the method of finding the solution of linear equationsin terms of matrix.
At first the equations are written in the form of matrix.Then the equations are solved.

HERE IS THE SOURCE CODE:



#include<stdlib.h>
#include<stdio.h>
#include<math.h>
int main()
{
        float **a,*temp,app,sum,mult;
        int i,j,k,n,p;
//take no of terms
        printf("Enter n : ");scanf("%d",&n);
//memory allocation
        a = (float**)malloc(n*sizeof(float*));
        for(i = 0; i < n; i++)
            a[i] = (float*)malloc(n*sizeof(float));
        temp = (float*)malloc(n*sizeof(float));
//take input from user
        printf("Enter the elements of augmended matrix rowwise\n");
        for(i=0;i<n;i++)
            for(j=0;j<=n;j++)
                scanf("%f",&a[i][j]);
//generation of scalar matrix
       for(i=0;i<(n);i++){
            app = a[i][i];
//initialization of p
            p = i;
//find largest no of the columns and row no. of largest no.
            for(k = i+1; k < n; k++)
            if(fabs(app) < fabs(a[k][i])){
               app = a[k][i] ;
               p = k;
            }
//swaping the elements of diagonal row and row containing largest no
            for(j = 0; j <= n; j++)
            {
                temp[j] = a[p][j];
                a[p][j] = a[i][j];
                a[i][j] = temp[j];
            }
//calculating triangular matrix
            for(j=i+1;j<n;j++){
                mult = a[j][i]/a[i][i];
                for(k=0;k<=n;k++)
                    a[j][k] -= mult*a[i][k];
            }
       }
//for calculating value of z,y,x via backward substitution method
        for(i=n-1;i>=0;i--)
        {
            sum = 0;
            for(j=i+1;j<n;j++)
                sum += a[i][j]*temp[j];
            temp[i] = (a[i][n]-sum)/a[i][i];
        }
        printf("****The matrix is : ***\n");
        for(i=0;i<n;i++){
            for(j=0;j<=n;j++)
                printf("%.2f\t",a[i][j]);
            printf("\n");
        }
//display solution
        printf("-------------The solution is ----------\n");
        for(i=0;i<n;i++)
        printf("X[%d] = %.2f\n",i+1,temp[i]);
//free allocated memory
        for(i = 0; i < n; i++)
            free(a[i]);
        free(a);
        free(temp);
    return 0;
}