NM_10

#include<stdio.h>
#include<conio.h>
#include<math.h>

#define f1(x,y,z)  (17-y+2*z)/20
#define f2(x,y,z)  (-18-3*x+z)/20
#define f3(x,y,z)  (25-2*x+3*y)/20
#define EPS 0.001

int main()
{
    float x0=0, y0=0, z0=0, x1, y1, z1, e1, e2, e3, e;
    int count=1;
    printf("\nCount\tx\ty\tz\n");
    do
        {
            /* Calculation */
            x1 = f1(x0,y0,z0);
            y1 = f2(x0,y0,z0);
            z1 = f3(x0,y0,z0);
            printf("%d\t%0.4f\t%0.4f\t%0.4f\n",count, x1,y1,z1);

            /* Error */
            e1 = fabs(x0-x1);
            e2 = fabs(y0-y1);
            e3 = fabs(z0-z1);

            count++;

            /* Set value for next iteration */
              x0 = x1;
              y0 = y1;
              z0 = z1;
        }while(e1>e && e2>e && e3>e);

        printf("\nSolution: x=%0.3f, y=%0.3f and z = %0.3f\n",x1,y1,z1);

    getch();
    return 0;
}



#include<stdio.h>
#include<conio.h>
#include<math.h>

/* Arrange systems of linear
   equations to be solved in
   diagonally dominant form
   and form equation for each
   unknown and define here
*/
/* In this example we are solving
   3x + 20y - z = -18
   2x - 3y + 20z = 25
   20x + y - 2z = 17
*/
/* Arranging given system of linear
   equations in diagonally dominant
   form:
   20x + y - 2z = 17
   3x + 20y -z = -18
   2x - 3y + 20z = 25
*/
/* Equations:
   x = (17-y+2z)/20
   y = (-18-3x+z)/20
   z = (25-2x+3y)/20
*/
/* Defining function */
#define f1(x,y,z)  (17-y+2*z)/20
#define f2(x,y,z)  (-18-3*x+z)/20
#define f3(x,y,z)  (25-2*x+3*y)/20
#define e 0.001
/* Main function */
int main()
{
 float x0=0, y0=0, z0=0, x1, y1, z1, e1, e2, e3;
 int count=1;

 printf("\nCount\tx\ty\tz\n");
 do
 {
  /* Calculation */
  x1 = f1(x0,y0,z0);
  y1 = f2(x1,y0,z0);
  z1 = f3(x1,y1,z0);
  printf("%d\t%0.4f\t%0.4f\t%0.4f\n",count, x1,y1,z1);

  /* Error */
  e1 = fabs(x0-x1);
  e2 = fabs(y0-y1);
  e3 = fabs(z0-z1);

  count++;

  /* Set value for next iteration */
  x0 = x1;
  y0 = y1;
  z0 = z1;

 }while(e1>e && e2>e && e3>e);

 printf("\nSolution: x=%0.3f, y=%0.3f and z = %0.3f\n",x1,y1,z1);

 getch();
 return 0;
}



#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAXIT 10
#define EPS 0.001
int main()
{
    int i,j,n,A[20][20],X[20][2],count=1;
    float Xo[20][2],Xn[20][2],Xmax=0;
    printf("Enter the size of the matrix A : ");scanf("%d",&n);
    printf("\nEnter the elements of the %d*%d matrix A\n",n,n);
    for(i=1;i<=n;i++)
    {for(j=1;j<=n;j++)
        {printf("\nA[%d][%d] : ",n,n);scanf("%d",&A[i][j]);}
    }

    printf("\nEnter the initial Guess Vector X : \n");
    for(i=1;i<=n;i++)
    {
        printf("Enter X[%d][1] : ",i);scanf("%d",&X[i][1]);
    }

    Xo[1][1]=A[1][1]*X[1][1]+A[1][2]*X[2][1]+A[1][3]*X[3][1];
    Xo[2][1]=A[2][1]*X[1][1]+A[2][2]*X[2][1]+A[2][3]*X[3][1];
    Xo[3][1]=A[3][1]*X[1][1]+A[3][2]*X[2][1]+A[3][3]*X[3][1];

    for(i=1;i<=n;i++)
    {
        if(Xo[i][1]>Xmax)
        {Xmax = Xo[i][1];}
    }

    Xn[1][1] = Xo[1][1]/Xmax;
    Xn[2][1] = Xo[2][1]/Xmax;
    Xn[3][1] = Xo[3][1]/Xmax;
    count++;
    if((fabs(Xn[1][1]-Xo[1][1])<EPS)&&(fabs(Xn[1][1]-Xo[1][1])<EPS)&&(fabs(Xn[1][1]-Xo[1][1])<EPS)||count<MAXIT)
    printf("\nThe Required Eigen Value is : %f",Xmax);
    printf("\nThe Required Eigen Vector is :\n");
    printf("\n|%.3f|",Xn[1][1]);
    printf("\n|%.3f|",Xn[2][1]);
    printf("\n|%.3f|",Xn[3][1]);
    getch();
    return 0;
}