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int main(void)
double ab[RA*(CA+Cb)]={
-0.34800000, -0.32100000, -0.23300000, -0.58616667,
-0.03200000, +0.89700000, +0.82100000, +0.69016667,
+0.04500000, -0.55900000, +0.28100000, -0.14083333
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));
clrscrn();
printf(" You have this nonlinear system of equations :\n\n");
printf(" (x, y, z not 0)\n");
printf("\n");
printf(" a 1/x + b 1/y + c 1/z = d \n");
printf(" e 1/x + f 1/y + g 1/z = h \n");
printf(" i 1/x + j 1/y + k 1/z = l \n\n");
printf(" We write 1/x = X, 1/y = Y, 1/z = Z\n\n");
printf(" a X + b Y + c Z = d\n");
printf(" e X + f Y + g Z = h\n");
printf(" i X + j Y + k Z = l\n\n");
printf(" Ab :\n");
printf(" a b c d\n");
printf(" e f g h\n");
printf(" i j k l\n\n\n");
stop();
clrscrn();
printf(" With A :");
p_mR(A,S8,P4,C7);
printf(" And b :");
p_mR(b,S8,P4,C7);
printf(" Ab :");
p_mR(Ab,S8,P4,C7);
getchar();
clrscrn();
printf(" Copy/Past into the octave window.\n\n");
p_Octave_mR(Ab,"Ab",P4);
printf("\n rref(Ab.00000000001)\n\n");
printf(" gj_TP_mR(Ab) :");
gj_TP_mR(Ab);
p_mR(Ab,S10,P4,C7);
printf(" Then 1/x = X = 1. , 1/y = Y = .5, 1/z = Z = .333 \n\n");
stop();
f_mR(Ab);
f_mR(b);
f_mR(A);
return 0;
/* ------------------------------------ */
Exemple de sortie écran :
You have this nonlinear system of equations :
(x, y, z not 0)
a 1/x + b 1/y + c 1/z = d
e 1/x + f 1/y + g 1/z = h
i 1/x + j 1/y + k 1/z = l
We write 1/x = X, 1/y = Y, 1/z = Z
a X + b Y + c Z = d
e X + f Y + g Z = h
i X + j Y + k Z = l
Ab :
a b c d
e f g h
i j k l
Press return to continue.
With A :
-0.3480 -0.3210 -0.2330
-0.0320 +0.8970 +0.8210
+0.0450 -0.5590 +0.2810
And b :
-0.5862
+0.6902
-0.1408
Ab :
-0.3480 -0.3210 -0.2330 -0.5862
-0.0320 +0.8970 +0.8210 +0.6902
+0.0450 -0.5590 +0.2810 -0.1408
Copy/Past into the octave window.
Ab=[
-0.3480,-0.3210,-0.2330,-0.5862;
-0.0320,+0.8970,+0.8210,+0.6902;
+0.0450,-0.5590,+0.2810,-0.1408]
rref(Ab.00000000001)
gj_TP_mR(Ab) :
+1.0000 -0.0000 -0.0000 +1.0000
+0.0000 +1.0000 +0.0000 +0.5000
+0.0000 +0.0000 +1.0000 +0.3333
Then 1/x = X = 1. , 1/y = Y = .5, 1/z = Z = .333
Press return to continue.
Mathc matrices/a122
Application
Installer et compiler ces fichiers dans votre répertoire de travail.
/* Save as : c00a.c */
int main(void)
double ab[RA*(CA+Cb)]={
-2.0, -9.0, +4.0, -202.0,
-1.0, +1.0, -1.0, +3.0,
+8.0, +6.0, -2.0, +236.0
double **Ab = ca_A_mR(ab,i_Abr_Ac_bc_mR(RA,CA,Cb));