Mathc matrices/a46
Installer et compiler ces fichiers dans votre répertoire de travail.
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c00.c |
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/* ------------------------------------ */
/* Save as : c00.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R3
#define CA C3
#define Cb C3
/* ------------------------------------ */
/* ------------------------------------ */
#define B (1.)
#define C (1.)
/* ------------------------------------ */
//#define B (-4./5.)
//#define C (-3./5.)
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
/*
* Find a value for "B" so that the system is compatible.
*/
double ab[RA*(CA+Cb)]={
// A = x*B + y*C + z
+9, -2, +1, +3*B, +1*C, +2,
-27, +6, -3, +0*B, +0*C, +3,
+18, -4, +2, +1*B, +2*C, +0,
};
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(" A :");
p_mR(A,S3,P0,C7);
printf(" b :");
p_mR(b,S3,P0,C7);
printf(" Ab :");
p_mR(Ab,S3,P0,C7);
getchar();
clrscrn();
printf(" Copy/Past into the octave window.\n\n");
p_Octave_mR(Ab,"Ab",P0);
printf("\n rref(Ab,.00000000001)\n\n");
printf(" gj_PP_mR(Ab,YES) :");
gj_PP_mR(Ab,YES);
p_mR(Ab,S8,P4,C7);
stop();
f_mR(Ab);
f_mR(b);
f_mR(A);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Calculer la valeur de B et de C pour que le système soit compatible. Exemple de sortie écran :
------------------------------------
A :
+9 -2 +1
-27 +6 -3
+18 -4 +2
b :
+3 +1 +2
+0 +0 +3
+1 +2 +0
Ab :
+9 -2 +1 +3 +1 +2
-27 +6 -3 +0 +0 +3
+18 -4 +2 +1 +2 +0
------------------------------------
Copy/Past into the octave window.
Ab=[
+9,-2,+1,+3,+1,+2;
-27,+6,-3,+0,+0,+3;
+18,-4,+2,+1,+2,+0]
rref(Ab,.00000000001)
gj_PP_mR(Ab,YES) :
+1.0000 -0.2222 +0.1111 -0.0000 -0.0000 -0.1111
+0.0000 +0.0000 +0.0000 +3.0000 +1.0000 +3.0000
+0.0000 +0.0000 +0.0000 +1.0000 +2.0000 +2.0000
Press return to continue.
Les deux dernières lignes donnent :
+0.0000 +0.0000 +0.0000 +3.0000 +1.0000 +3.0000 +0.0000 +0.0000 +0.0000 +1.0000 +2.0000 +2.0000 ou bien +0.0000 = +3.0000 +1.0000 +3.0000 +0.0000 = +1.0000 +2.0000 +2.0000 en introduisant B et C +0.0000 = +3.0000 B +1.0000 C +3.0000 +0.0000 = +1.0000 B +2.0000 C +2.0000 soit le système 3.0000 B +1.0000 C = -3.0000 +1.0000 B +2.0000 C = -2.0000 cela donne B = -4./5. C = -3./5.