#include <math.h>
#define n 5
/* This uses twice the necessary storage and does perhaps twice the
   necessary computation. Many of the results are known to be zero
   or one. These are stored as well. */
double ab(double); double ab(double x){return x<0?-x:x;}
int invrt(double A[n][n]);
int invrt(double A[n][n]){int i, j, k, im; double a[n][2*n];
for(i=0; i<n; i++) for(j=0; j<n; j++)
{a[j][i] = A[j][i]; a[j][n+i]= i==j;}
for(i=0; i<n; i++) {double big=0; im=n;
  for(j=i; j<n; j++) if(big<ab(a[j][i])) {im=j; big=ab(a[j][i]);}
  if(im==n) return 1;
  if(im!=i) for (j=0; j<2*n; j++)
     {double q = a[i][j]; a[i][j]=a[im][j]; a[im][j]=q;}
for(j=i+1; j<n; j++) {double g = a[j][i]/a[i][i];
    for(k=0; k<2*n; k++) a[j][k] -= g*a[i][k];}}
for(i=0; i<n; i++) {double g = 1.0/a[i][i];
for(j=0; j<2*n; j++) a[i][j]*=g; 
for(j=0; j<i; j++) {double g=a[j][i];
for(k=0; k<2*n; k++) a[j][k] -=g*a[i][k];}}
for(i=0; i<n; i++) for(j=0; j<n; j++) A[i][j]=a[i][j+n]; return 0;}

#include <stdio.h>
int main(){double a[5][5] = {
 {1, 4, 3, 6, 5},
 {3, 4, 1, 2.32, 7},
 {7, 4, 5, 2, 1},
 {4, 6, 1, 8, 5},
 {2, 7, 4, 1, 2}}, x[5][5], p[5][5];
 {int i, j; for(i=0; i<5; ++i) for (j=0; j<5; ++j)
    a[i][j] = i==0?-1:(i==j);}
 {int i, j; for(i=0; i<5; ++i) for (j=0; j<5; ++j) x[i][j] = a[i][j];}
 invrt(a);
 printf("One\n");
 {int i, j; for(i=0; i<5; ++i) for (j=0; j<5; ++j)
   {int k; p[i][j] = 0; for(k=0; k<5; ++k) p[i][j] += a[i][k]*x[k][j];}}
 {int i, j;
   for(i=0; i<5; ++i) {for (j=0; j<5; ++j) printf("%.9f ", x[i][j]);
      printf("\n");}
   for(i=0; i<5; ++i) {for (j=0; j<5; ++j) printf("%.9f ", a[i][j]);
      printf("\n");}
   for(i=0; i<5; ++i) {for (j=0; j<5; ++j) printf("%.9f ", p[i][j]);
      printf("\n");}}}