Mathc complexes/a230
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c00a.c |
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/* ------------------------------------ */
/* Save as : c00a.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R3
#define CA C4
#define Cb C1
/* ------------------------------------ */
void fun(void)
{
double ab[RA*((CA+Cb)*C2)] ={
1,2, 3,4, 5,6, 5,2, 0,0,
1,2, 3,4, 5,6, 1,3, 0,0,
1,2, 3,4, 1,1, 4,2, 0,0};
double **Ab = ca_A_mZ(ab,i_Abr_Ac_bc_mZ(RA,CA,Cb));
double **A = c_Ab_A_mZ(Ab,i_mZ(RA,CA));
double **b = c_Ab_b_mZ(Ab,i_mZ(RA,Cb));
double **B = i_mZ(RA,C3) ;
double **BT = i_mZ(C3,RA) ;
double **BTb = i_Abr_Ac_bc_mZ(C3,RA,Cb);
clrscrn();
printf("Basis for a Column Space by Row Reduction :\n\n");
printf(" A :");
p_mZ(A, S3,P0, S3,P0, C8);
printf(" b :");
p_mZ(b, S3,P0, S3,P0, C8);
printf(" Ab :");
p_mZ(Ab, S3,P0, S3,P0, C8);
stop();
clrscrn();
printf(" The leading 1’s of Ab give the position \n"
" of the columns of A which form a basis \n"
" for the column space of A \n\n"
" A :");
p_mZ(A, S8,P4, S8,P4, C4);
printf(" gj_PP_mZ(Ab) :");
p_mZ(gj_PP_mZ(Ab), S8,P4, S8,P4, C4);
c_c_mZ(A,C1,B,C1);
c_c_mZ(A,C3,B,C2);
c_c_mZ(A,C4,B,C3);
printf(" B :");
p_mZ(B, S8,P4, S8,P4, C4);
stop();
clrscrn();
printf(" Check if the columns of B are linearly independent\n\n"
" BT :");
p_mZ(transpose_mZ(B,BT), S4,P0, S3,P0, C4);
printf(" BTb :");
p_mZ(c_mZ(BT,BTb), S4,P0, S3,P0, C4);
printf(" gj_PP_mZ(BTb) :");
p_mZ(gj_PP_mZ(BTb), S8,P4, S8,P4, C4);
stop();
f_mZ(Ab);
f_mZ(A);
f_mZ(b);
f_mZ(B);
f_mZ(BT);
f_mZ(BTb);
}
/* ------------------------------------ */
int main(void)
{
fun();
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
La position des pivots de Ab donne la position des colonnes de A qui forment une base pour l'espace colonnes de A.
Exemple de sortie écran :
------------------------------------
Basis for a Column Space by Row Reduction :
A :
+1 +2i +3 +4i +5 +6i +5 +2i
+1 +2i +3 +4i +5 +6i +1 +3i
+1 +2i +3 +4i +1 +1i +4 +2i
b :
+0 +0i
+0 +0i
+0 +0i
Ab :
+1 +2i +3 +4i +5 +6i +5 +2i +0 +0i
+1 +2i +3 +4i +5 +6i +1 +3i +0 +0i
+1 +2i +3 +4i +1 +1i +4 +2i +0 +0i
Press return to continue.
------------------------------------
The leading 1’s of Ab give the position
of the columns of A which form a basis
for the column space of A
A :
+1.0000 +2.0000i +3.0000 +4.0000i +5.0000 +6.0000i +5.0000 +2.0000i
+1.0000 +2.0000i +3.0000 +4.0000i +5.0000 +6.0000i +1.0000 +3.0000i
+1.0000 +2.0000i +3.0000 +4.0000i +1.0000 +1.0000i +4.0000 +2.0000i
gj_PP_mZ(Ab) :
+1.0000 +0.0000i +2.2000 -0.4000i +3.4000 -0.8000i +1.8000 -1.6000i
-0.0000 +0.0000i +0.0000 -0.0000i +1.0000 +0.0000i +0.0976 -0.1220i
+0.0000 -0.0000i +0.0000 +0.0000i -0.0000 +0.0000i +1.0000 +0.0000i
+0.0000 +0.0000i
-0.0000 +0.0000i
+0.0000 -0.0000i
B :
+1.0000 +2.0000i +5.0000 +6.0000i +5.0000 +2.0000i
+1.0000 +2.0000i +5.0000 +6.0000i +1.0000 +3.0000i
+1.0000 +2.0000i +1.0000 +1.0000i +4.0000 +2.0000i
Press return to continue.
------------------------------------
Check if the columns of B are linearly independent
BT :
+1 +2i +1 +2i +1 +2i
+5 +6i +5 +6i +1 +1i
+5 +2i +1 +3i +4 +2i
BTb :
+1 +2i +1 +2i +1 +2i +0 +0i
+5 +6i +5 +6i +1 +1i +0 +0i
+5 +2i +1 +3i +4 +2i +0 +0i
gj_PP_mZ(BTb) :
+1.0000 +0.0000i +1.0000 +0.0000i +0.1803 -0.0164i +0.0000 +0.0000i
-0.0000 +0.0000i +1.0000 +0.0000i -0.6201 -0.5853i +0.0000 -0.0000i
+0.0000 -0.0000i +0.0000 -0.0000i +1.0000 +0.0000i +0.0000 +0.0000i
Press return to continue.