Linear Algebra and the C Language/a0jr

/* ------------------------------------ */
/*  Save as :   c00b.c                  */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define   RA R5
#define   CA C5      
/* ------------------------------------ */
int main(void)
{
double   xy[8] ={
   1,     -8,
   2,      2,
   3,      1,
   4,      2  };
   
double tA[RA*CA]={
/* x**2    y**2    x       y       e     */
  +1,     +0,     +0,     +0,     +0, 
  +1,    +64,     +1,     -8,     +1,     
  +4,     +4,     +2,     +2,     +1,      
  +9,     +1,     +3,     +1,     +1,    
 +16,     +4,     +4,     +2,     +1,         
};

double tb[RA*C1]={
/*    = 0   */
       +1,   
       +0,   
       +0,   
       +0,   
       +0 
};

double **XY      = ca_A_mR(xy,i_mR(R4,C2));
double **A       = ca_A_mR(tA,i_mR(RA,CA));
double **b       = ca_A_mR(tb,i_mR(RA,C1));
double **Inv    = i_mR(CA,RA);         
double **Invb   = i_mR(CA,C1);         

  clrscrn();
  printf("\n");
  printf(" Find the coefficients a, b, c, d, e, of the curve \n\n");
  printf("     ax**2 + by**2 + cx + dy + e  = 0 \n\n");
  printf(" that passes through these four points.\n\n");
  printf("         x          y");
  p_mR(XY,S10,P0,C6);
  stop();
  
  clrscrn();
  printf(" Using the given points, we obtain this matrix.\n");
  printf("  (a = 1. This is my choice)\n\n");  
  printf(" A :");
  p_mR(A,S10,P2,C7);
  printf(" b :");
  p_mR(b,S10,P2,C7);
   
  printf(" Inv ");
  invgj_mR(A,Inv);   
  pE_mR(Inv,S12,P4,C10); 
  stop();
  
  clrscrn(); 
  printf(" x = Inv * b ");   
  mul_mR(Inv,b,Invb); 
  p_mR(Invb,S10,P4,C10);
  printf(" The coefficients a, b, c, d, e, of the curve are : \n\n"
         "  %+.2fx**2 %+.2fy**2 %+.2fx %+.2fy %+.2f = 0\n\n"
            ,Invb[R1][C1],Invb[R2][C1],Invb[R3][C1],
             Invb[R4][C1],Invb[R5][C1]);      
  stop();  

  f_mR(XY);
  f_mR(b);
  f_mR(A);
  f_mR(Inv);
  f_mR(Invb); 

  return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */

 Find the coefficients a, b, c, d, e, of the curve 

     ax**2 + by**2 + cx + dy + e  = 0 

 that passes through these four points.

         x          y
        +1         -8 
        +2         +2 
        +3         +1 
        +4         +2 

 Press return to continue. 


 Using the given points, we obtain this matrix.
  (a = 1. This is my choice)

 A :
     +1.00      +0.00      +0.00      +0.00      +0.00 
     +1.00     +64.00      +1.00      -8.00      +1.00 
     +4.00      +4.00      +2.00      +2.00      +1.00 
     +9.00      +1.00      +3.00      +1.00      +1.00 
    +16.00      +4.00      +4.00      +2.00      +1.00 

 b :
     +1.00 
     +0.00 
     +0.00 
     +0.00 
     +0.00 

 Inv 
 +1.0000e+00  +8.6736e-19  +0.0000e+00  -1.3878e-17  -6.9389e-18 
 -1.4444e-01  +1.1111e-02  +3.8889e-02  -1.1111e-01  +6.1111e-02 
 -6.0000e+00  -1.7347e-18  -5.0000e-01  +0.0000e+00  +5.0000e-01 
 -5.6667e-01  -3.3333e-02  +3.8333e-01  -6.6667e-01  +3.1667e-01 
 +9.7111e+00  +2.2222e-02  +1.0778e+00  +1.7778e+00  -1.8778e+00 

 Press return to continue. 


 x = Inv * b 
   +1.0000 
   -0.1444 
   -6.0000 
   -0.5667 
   +9.7111 

 The coefficients a, b, c, d, e, of the curve are : 

  +1.00x**2 -0.14y**2 -6.00x -0.57y +9.71 = 0

 Press return to continue.