Gauss Elimination Method

# include<stdio.h>

# include<conio.h>

void main()

{

 int i,j,k,n;

 float a[10][10],x[10];

 float s,p; 

 printf("Enter the number of equations : ");

 scanf("%d",&n) ;

 printf("\nEnter the co-efficients of the equations :\n\n");

 for(i=0;i<n;i++)

 {

  for(j=0;j<n;j++)

  {

   printf("a[%d][%d]= ",i+1,j+1);

   scanf("%f",&a[i][j]);

  }

  printf("d[%d]= ",i+1);

  scanf("%f",&a[i][n]);

 }

 for(k=0;k<n-1;k++)

 {

  for(i=k+1;i<n;i++)

  {

   p = a[i][k] / a[k][k] ;

   for(j=k;j<n+1;j++)

    a[i][j]=a[i][j]-p*a[k][j];

  }

 }

 x[n-1]=a[n-1][n]/a[n-1][n-1];

 for(i=n-2;i>=0;i--)

 {

  s=0;

  for(j=i+1;j<n;j++)

  {

   s +=(a[i][j]*x[j]);

   x[i]=(a[i][n]-s)/a[i][i];

  }

 }

 printf("\nThe result is:\n");

 for(i=0;i<n;i++)

  printf("\nx[%d]=%f",i+1,x[i]);

 getch();

}

gauss seidel method

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define e 0.01

void main() {
  int i, j, k, n;
  float a[10][10], x[10];
  float sum, temp, error, big;
  printf("Enter the number of equations: ");
  scanf("%d", & n);
  printf("Enter the co-efficients of the equations: \n");
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n + 1; j++) {
      printf("a[%d][%d]= ", i, j);
      scanf("%f", & a[i][j]);
    }
  }
  for (i = 1; i <= n; i++) {
    x[i] = 0;
  }
  do {
    big = 0;
    for (i = 1; i <= n; i++) {
      sum = 0;
      for (j = 1; j <= n; j++) {
        if (j != i) {
          sum = sum + a[i][j] * x[j];
        }
      }
      temp = (a[i][n + 1] - sum) / a[i][i];
      error = fabs(x[i] - temp);
      if (error > big) {
        big = error;
      }
      x[i] = temp;
      printf("\nx[%d] =%f", i, x[i]);
    }
    printf("\n");
  }
  while (big >= e);
  printf("\n\nconverges to solution");
  for (i = 1; i <= n; i++) {
    printf("\nx[%d]=%f", i, x[i]);
  }
  getch();
}

Enter the number of equations: 3

Enter the co-efficients of the equations:

a[1][1]= 2

a[1][2]= 1

a[1][3]= 1

a[1][4]= 5

a[2][1]= 3

a[2][2]= 5

a[2][3]= 2

a[2][4]= 15

a[3][1]= 2

a[3][2]= 1

a[3][3]= 4

a[3][4]= 8

x[1] =2.500000

x[2] =1.500000

x[3] =0.375000

x[1] =1.562500

x[2] =1.912500

x[3] =0.740625

x[1] =1.173437

x[2] =1.999688

x[3] =0.913359

x[1] =1.043477

x[2] =2.008570

x[3] =0.976119

x[1] =1.007655

x[2] =2.004959

x[3] =0.994933

x[1] =1.000054

x[2] =2.001995

x[3] =0.999474

converges to solution

x[1]=1.000054

x[2]=2.001995

x[3]=0.999474