Simultaneous equation using gauss elimination method
Program:
#include<stdio.h>
#include<conio.h>
void main()
{
int i, j, k, n;
float a[20][20], x[20];
double s, p ;
clrscr();
printf("Enter the number of equations : ");
scanf("%d", &n);
printf("\nEnter the coefficients 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("b[%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[n1] = a[n1][n] / a[n1][n1];
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] = %.2f", i + 1, x[i]);
getch();
}
Output:
Enter the number of equations : 3
Enter the coefficients of the equations :
a[1][1] = 10
a[1][2] = 1
a[1][3] = 1
b[1] = 12
a[2][1] = 2
a[2][2] = 10
a[2][3] = 1
b[2] = 13
a[3][1] = 1
a[3][2] = 1
a[3][3] = 5
b[3] = 7
The result is :
x[1] = 1.00
x[2] = 1.00
x[3] = 1.00

CreatedOct 15, 2014

UpdatedDec 30, 2019

Views4,502
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