fun();
 return 0;
/* ------------------------------------ */
 Trouver les coefficients a, b, c, d, e du conique,
 ax**2 + by**2 + cx + dy + e = 0 
 qui passe par ces quatre points. 
 (x[1],y[1]) (x[2],y[2]) (x[3],y[3]) (x[4],y[4]) 
 En utilisant les quatre points nous obtenons la matrice.
 (a)x**2 (b)y**2 (c)x (d)y (e) = 0 
 x[1]**2 y[1]**2 x[1] y[1] 1 0
 x[2]**2 y[2]**2 x[2] y[2] 1 0
 x[3]**2 y[3]**2 x[3] y[3] 1 0
 x[4]**2 y[4]**2 x[4] y[4] 1 0
 Ce système à quatre lignes et cinq inconnus (a, b, c, d, e).
 C'est un système homogène, il a donc une infinité de solution.
 Pour trouver une solution j'ai choisi de poser que a = 1.
 Nous avons donc cinq lignes et cinq inconnus.
 1 0 0 0 0 1 
 x[1]**2 y[1]**2 x[1] y[1] 1 0 
 x[2]**2 y[2]**2 x[2] y[2] 1 0 
 x[3]**2 y[3]**2 x[3] y[3] 1 0 
 x[4]**2 y[4]**2 x[4] y[4] 1 0 
 Il suffit maintenant de résoudre le système. 
 Exemple de sortie écran :
 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 +0 
 +2 +3 
 +3 +4 
 +4 +0 
 Using the given points, we obtain this matrix.
 (a = 1. This is my choice)
 Ab :
 x**2 y**2 x y e = 0 
 +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 
 +1.00 +0.00 +1.00 +0.00 +1.00 +0.00 
 +4.00 +9.00 +2.00 +3.00 +1.00 +0.00 
 +9.00 +16.00 +3.00 +4.00 +1.00 +0.00 
 +16.00 +0.00 +4.00 +0.00 +1.00 +0.00 
 Press return to continue. 
 -----------------------------------
 A_T :
 +1.00 +1.00 +4.00 +9.00 +16.00 
 +0.00 +0.00 +9.00 +16.00 +0.00 
 +0.00 +1.00 +2.00 +3.00 +4.00 
 +0.00 +0.00 +3.00 +4.00 +0.00 
 +0.00 +1.00 +1.00 +1.00 +1.00 
 A_TA :
 +355.00 +180.00 +100.00 +48.00 +30.00 
 +180.00 +337.00 +66.00 +91.00 +25.00 
 +100.00 +66.00 +30.00 +18.00 +10.00 
 +48.00 +91.00 +18.00 +25.00 +7.00 
 +30.00 +25.00 +10.00 +7.00 +4.00 
 Press return to continue. 
 -----------------------------------
 inv(A_TA) :
 +1.0000 -0.1667 -5.0000 +1.1667 +4.0000 
 -0.1667 +0.2238 +0.7685 -0.9182 -0.4630 
 -5.0000 +0.7685 +25.2222 -5.6019 -20.5556 
 +1.1667 -0.9182 -5.6019 +4.1127 +3.7963 
 +4.0000 -0.4630 -20.5556 +3.7963 +17.8889 
 inv(A_TA)*A_T :
 +1.0000 +0.0000 -0.0000 -0.0000 -0.0000 
 -0.1667 +0.1389 -0.3333 +0.2500 -0.0556 
 -5.0000 -0.3333 +0.0000 +0.0000 +0.3333 
 +1.1667 -0.6389 +1.3333 -0.7500 +0.0556 
 +4.0000 +1.3333 -0.0000 -0.0000 -0.3333 
 x = inv(A_TA)*A_T*b :
 +1.0000 
 -0.1667 
 -5.0000 
 +1.1667 
 +4.0000 
 Press return to continue. 
 -----------------------------------
 x = inv(A_TA)*A_T*b :
 +1.00 
 -0.17 
 -5.00 
 +1.17 
 +4.00 
 The coefficients a, b, c, d, e, of the curve are : 
 +1.00x**2 -0.17y**2 -5.00x +1.17y +4.00 = 0
 Press return to continue.

Mathc matrices/c23n
Application ou QR décomposition 
Installer et compiler ces fichiers dans votre répertoire de travail.
/* Save as : c00o.c */
/* ------------------------------------ */
/* ------------------------------------ */
void fun(void)
double xy[8] ={
 1, 1,
 2, 4,
 3, 9,
 4, 16 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y e = 0 */
 +1.00, +0.00, +0.00, +0.00, +0.00, +1.00, 
 +1.00, +1.00, +1.00, +1.00, +1.00, +0.00, 
 +4.00, +16.00, +2.00, +4.00, +1.00, +0.00, 
 +9.00, +81.00, +3.00, +9.00, +1.00, +0.00, 
 +16.00, +256.00, +4.00, +16.00, +1.00, +0.00 
double **XY = ca_A_mR(xy,i_mR(R4,C2));
double **Ab = ca_A_mR(ab,i_Abr_Ac_bc_mR(RA,CA,Cb));
double **A = c_Ab_A_mR(Ab,i_mR(RA,CA));
double **b = c_Ab_b_mR(Ab,i_mR(RA,Cb));
double **Q = i_mR(RA,CA);
double **R = i_mR(CA,CA);
double **invR = i_mR(CA,CA);
double **Q_T = i_mR(CA,RA);
double **invR_Q_T = i_mR(CA,RA);
double **x = i_mR(CA,Cb); // x invR * Q_T * b
 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,S5,P0,C6);
 printf(" Using the given points, we obtain this matrix.\n");
 printf(" (a = 1. This is my choice)\n\n"); 
 printf(" Ab :\n");
 printf(" x**2 y**2 x y e = 0 ");
 p_mR(Ab,S7,P2,C6);
 stop();
 clrscrn();
 QR_uv_mR(A,Q,R); 
 printf(" Q :");
 p_mR(Q,S10,P4,C6); 
 printf(" R :");
 p_mR(R,S10,P4,C6);
 stop();
 clrscrn();
 transpose_mR(Q,Q_T); 
 printf(" Q_T :");
 pE_mR(Q_T,S12,P4,C6); 
 inv_mR(R,invR); 
 printf(" invR :");
 pE_mR(invR,S12,P4,C6);
 stop();
 clrscrn();
 printf(" Solving this system yields a unique\n"
 " least squares solution, namely \n\n");
 mul_mR(invR,Q_T,invR_Q_T);
 mul_mR(invR_Q_T,b,x);
 printf(" x = invR * Q_T * b :");
 p_mR(x,S10,P2,C6);
 printf(" The coefficients a, b, c, d, e, of the curve are : \n\n"
 " %+.2fx**2 %+.2fy = 0\n\n"
 ,x[R1][C1],x[R4][C1]); 
 stop();
 f_mR(XY); 
 f_mR(A);
 f_mR(b);
 f_mR(Ab);
 f_mR(Q);
 f_mR(Q_T);
 f_mR(R);
 f_mR(invR); 
 f_mR(invR_Q_T); 
 f_mR(x); 
/* ------------------------------------ */
int main(void)

Exemple de sortie écran :
 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 +1 
 +2 +4 
 +3 +9 
 +4 +16 
 Using the given points, we obtain this matrix.
 (a = 1. This is my choice)
 Ab :
 x**2 y**2 x y e = 0 
 +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 
 +1.00 +1.00 +1.00 +1.00 +1.00 +0.00 
 +4.00 +16.00 +2.00 +4.00 +1.00 +0.00 
 +9.00 +81.00 +3.00 +9.00 +1.00 +0.00 
 +16.00 +256.00 +4.00 +16.00 +1.00 +0.00 
 Press return to continue. 
 Q :
 +0.0531 -0.1949 -0.6241 -0.7548 +0.0000 
 +0.0531 -0.1807 +0.6555 -0.4915 +0.5415 
 +0.2123 -0.5532 +0.3364 -0.1204 -0.7220 
 +0.4777 -0.6080 -0.2539 +0.4005 +0.4212 
 +0.8492 +0.5037 +0.0568 -0.1173 -0.0902 
 R :
 +18.8414 +259.5343 +5.3074 +18.7884 +1.5922 
 +0.0000 +70.6822 -1.0960 +0.1949 -0.8381 
 -0.0000 -0.0000 +0.7936 +0.6241 +0.7948 
 +0.0000 +0.0000 +0.0000 +0.7548 -0.3287 
 +0.0000 +0.0000 +0.0000 +0.0000 +0.1504 
 Press return to continue. 
 -----------------------------------
 Q_T :
 +5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01 
 -1.9488e-01 -1.8073e-01 -5.5316e-01 -6.0796e-01 +5.0374e-01 
 -6.2405e-01 +6.5551e-01 +3.3642e-01 -2.5391e-01 +5.6750e-02 
 -7.5483e-01 -4.9155e-01 -1.2040e-01 +4.0047e-01 -1.1727e-01 
 +3.1923e-14 +5.4149e-01 -7.2199e-01 +4.2116e-01 -9.0249e-02 
 invR :
 +5.3074e-02 -1.9488e-01 -6.2405e-01 -7.5483e-01 -0.0000e+00 
 +0.0000e+00 +1.4148e-02 +1.9537e-02 -1.9805e-02 -6.7686e-02 
 +0.0000e+00 -0.0000e+00 +1.2600e+00 -1.0417e+00 -8.9346e+00 
 -0.0000e+00 +0.0000e+00 -0.0000e+00 +1.3248e+00 +2.8955e+00 
 -0.0000e+00 +0.0000e+00 -0.0000e+00 -0.0000e+00 +6.6483e+00 
 Press return to continue. 
 -----------------------------------
 Solving this system yields a unique
 least squares solution, namely 
 x = invR * Q_T * b :
 +1.00 
 -0.00 
 -0.00 
 -1.00 
 +0.00 
 The coefficients a, b, c, d, e, of the curve are : 
 +1.00x**2 -1.00y = 0
 Press return to continue.