Program to find simultaneous linear equations using Jordan Elimination method

#include<stdio.h>

#include<stdlib.h>

main()

{

int a[10][10],n,i,j,k;

printf("\n\tEnter the no.of variables: ");

scanf("%d", &n);

for(i=0;i<n;i++)

{

printf("\n\tEQUATION %d\n\t", i+1);

for(j=0;j<n;j++)

{

printf("\n\tEnter the coefficients of x%d : ", j+1);

scanf("%d", &a[i][j]);

}

printf("\nEnter the constant: ");

scanf("%d", &a[i][n]);

}

for(k=0;k<n;k++)

{

for(i=0;i<n;i++)

{

l=a[i][k];

for(j=0;j<=n;j++)

{

if(i!=k)

a[i][j]=a[k][k]*a[i][j]-(l*a[k][j]);

}

}

}

printf("\n\tSolutions are: \n");

for(i=0;i<n;i++)

{

if(a[i][i]==0)

{

printf("\n\tNOn trivial solution");

exit(0);

}

printf("\n\tThe value of %d

is %.3f",i+1,(float)(a[i][n]/(float)a[i][i]));

}

}

/*

-----------------OUTPUT------------------------

Enter the no.of variables: 3

EQUATION 1

Enter the coefficients of x1 : 3

Enter the coefficients of x2 : 2

Enter the coefficients of x3 : 1

Enter the constant: 6

EQUATION 2

Enter the coefficients of x1 : 1

Enter the coefficients of x2 : 2

Enter the coefficients of x3 : 3

Enter the constant: 6

EQUATION 3

Enter the coefficients of x1 : 2

Enter the coefficients of x2 : 3

Enter the coefficients of x3 : 1

Enter the constant: 6

Solutions are:

The value of 1 is 1.000

The value of 2 is 1.000

The value of 3 is 1.000

-----------------------------------------------

*/