Mathc complexes/a231
Apparence
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c00b.c |
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/* ------------------------------------ */
/* Save as : c00b.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R4
#define CA C5
#define Cb C1
/* ------------------------------------ */
void fun(void)
{
double ab[RA*((CA+Cb)*C2)] ={
-3,-1, +8,-3, +2,+8, +8,+6, +7,+8, 0,0,
-3,-1, +8,-3, +2,+8, +8,+6, +7,+8, 0,0,
-3,-1, +8,-3, -4,+2, -1,-3, +6,+5, 0,0,
-3,-1, +8,-3, -5,-3, -1,+0, +4,+3, 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, S6,P3, S5,P3, C5);
printf(" gj_PP_mZ(Ab) :");
p_mZ(gj_PP_mZ(Ab), S6,P3, S5,P3, C5);
stop();
clrscrn();
printf(" A :");
p_mZ(A, S6,P3, S5,P3, C5);
printf(" gj_PP_mZ(Ab) :");
p_mZ(Ab, S6,P3, S5,P3, C5);
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, C5);
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 :
-3 -1i +8 -3i +2 +8i +8 +6i +7 +8i
-3 -1i +8 -3i +2 +8i +8 +6i +7 +8i
-3 -1i +8 -3i -4 +2i -1 -3i +6 +5i
-3 -1i +8 -3i -5 -3i -1 +0i +4 +3i
b :
+0 +0i
+0 +0i
+0 +0i
+0 +0i
Ab :
-3 -1i +8 -3i +2 +8i +8 +6i +7 +8i +0 +0i
-3 -1i +8 -3i +2 +8i +8 +6i +7 +8i +0 +0i
-3 -1i +8 -3i -4 +2i -1 -3i +6 +5i +0 +0i
-3 -1i +8 -3i -5 -3i -1 +0i +4 +3i +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 :
-3.000-1.000i +8.000-3.000i +2.000+8.000i +8.000+6.000i +7.000+8.000i
-3.000-1.000i +8.000-3.000i +2.000+8.000i +8.000+6.000i +7.000+8.000i
-3.000-1.000i +8.000-3.000i -4.000+2.000i -1.000-3.000i +6.000+5.000i
-3.000-1.000i +8.000-3.000i -5.000-3.000i -1.000+0.000i +4.000+3.000i
gj_PP_mZ(Ab) :
+1.000+0.000i -2.100+1.700i -1.400-2.200i -3.000-1.000i -2.900-1.700i
-0.000+0.000i +0.000-0.000i +1.000+0.000i +0.759-0.335i +0.447+0.012i
-0.000+0.000i +0.000-0.000i +0.000-0.000i +1.000+0.000i -0.049+0.231i
+0.000+0.000i -0.000-0.000i +0.000+0.000i -0.000-0.000i -0.000+0.000i
-0.000+0.000i
-0.000+0.000i
-0.000+0.000i
+0.000+0.000i
Press return to continue.
------------------------------------
A :
-3.000-1.000i +8.000-3.000i +2.000+8.000i +8.000+6.000i +7.000+8.000i
-3.000-1.000i +8.000-3.000i +2.000+8.000i +8.000+6.000i +7.000+8.000i
-3.000-1.000i +8.000-3.000i -4.000+2.000i -1.000-3.000i +6.000+5.000i
-3.000-1.000i +8.000-3.000i -5.000-3.000i -1.000+0.000i +4.000+3.000i
gj_PP_mZ(Ab) :
+1.000+0.000i -2.100+1.700i -1.400-2.200i -3.000-1.000i -2.900-1.700i
+0.000+0.000i +0.000+0.000i +1.000-0.000i +0.759-0.335i +0.447+0.012i
+0.000+0.000i +0.000+0.000i +0.000-0.000i +1.000-0.000i -0.049+0.231i
+0.000+0.000i +0.000-0.000i +0.000+0.000i +0.000+0.000i +0.000+0.000i
+0.000+0.000i
+0.000+0.000i
+0.000+0.000i
+0.000+0.000i
B :
-3.0000 -1.0000i +2.0000 +8.0000i +8.0000 +6.0000i
-3.0000 -1.0000i +2.0000 +8.0000i +8.0000 +6.0000i
-3.0000 -1.0000i -4.0000 +2.0000i -1.0000 -3.0000i
-3.0000 -1.0000i -5.0000 -3.0000i -1.0000 +0.0000i
Press return to continue.
------------------------------------
Check if the columns of B are linearly independent
BT :
-3 -1i -3 -1i -3 -1i -3 -1i
+2 +8i +2 +8i -4 +2i -5 -3i
+8 +6i +8 +6i -1 -3i -1 +0i
BTb :
-3 -1i -3 -1i -3 -1i -3 -1i +0 +0i
+2 +8i +2 +8i -4 +2i -5 -3i +0 +0i
+8 +6i +8 +6i -1 -3i -1 +0i +0 +0i
gj_PP_mZ(BTb) :
+1.0000 +0.0000i +1.0000 +0.0000i -0.2600 -0.1800i -0.0800 +0.0600i
+0.0000 -0.0000i +0.0000 -0.0000i +1.0000 +0.0000i +0.2377 +0.7186i
+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
Press return to continue.