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

void gauss(double lin[][12],int n)
{ 
	double x[11],       // solution matrix
	       temp,        // temporary storage to exchange rows in matrix
	       epsil=1.e-6, // check criteria
	       mult,        // multiplier used to elimnate values
	       abspiv;      // absolute value of pivot element

	int    i,j,k,       // indices
	       pivrow,      // row containing pivot element
	       singul=0;    // singular matrix indicator

	for (i=1 ; i<=n ; ++i)
	{

// Locate the optimal pivot element

		abspiv = fabs(lin[i][i]);
		pivrow = i;
		for (k=(i+1) ; k<=n ; ++k)
		{
			if (fabs(lin[k][i]) > abspiv)
			{
				abspiv = fabs(lin[k][i]);
				pivrow = k;
			}
		}

// Check if matrix is nearly singular

		if (abspiv < epsil)
		{
			singul = 1;
			printf("Problems with singular matrix.\n");
			break;
		}

// Interchange rows if necessary

		if (pivrow != i)
		{
			for (j=1 ; j<=(n+1) ; ++j)
			{
				temp = lin[i][j];
				lin[i][j] = lin[pivrow][j];
				lin[pivrow][j] = temp;
			}
		}

// Eliminate coefficients below

		for (j=(i+1) ; j<=n ; ++j)
		{
			mult = -lin[j][i]/lin[i][i];
			for (k=i ; k<=(n+1) ; ++k)
				lin[j][k] = lin[j][k] + mult * lin[i][k];
		}
	}

// Find the solution by back substitution 

	x[n] = lin[n][n+1]/lin[n][n];

	for (j=(n-1) ; j>=1 ; --j)
	{
		x[j] = lin[j][n+1];

		for (k=(j+1) ; k<=n ; ++k)
			x[j] = x[j] - lin[j][k] * x[k];
	
		x[j] = x[j] / lin[j][j];
	}

// Print out solution or if singular

	if (singul == 1)
		printf("Singular matrix problem.  No solution.\n");
	else 
		for (i=1 ; i<=n ; ++i)
			printf("Unknown #%d is...> %lf\n",i,x[i]);
    return;
}