Mathc complexes/a232
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c00a.c |
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/* ------------------------------------ */
/* Save as : c00a.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R4
#define CA C5
#define Cb C1
/* ------------------------------------ */
#define CB C2 /* B : a basis for the column space of A */
/* ------------------------------------ */
void fun(void)
{
double ab[RA*((CA+Cb)*C2)] ={
-3*1,-1*1, +8*1,-3*1, +2*1,+8*1, +8*1,+6*1, +7*1,+8*1, 0,0,
-3*2,-1*2, +8*2,-3*2, +2*2,+8*2, +8*2,+6*2, +7*2,+8*2, 0,0,
-3*7,-1*7, +8*7,-3*7, +2*7,+8*7, +8*7,+6*7, +7*7,+8*7, 0,0,
+2*1,-1*1, -6*1,+2*1, -7*1,-6*1, -1*1,+0*1, +4*1,+3*1, 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,CB) ;
double **BT = i_mZ(CB,RA) ;
double **BTb = i_Abr_Ac_bc_mZ(CB,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, S7,P3, S7,P3, C5);
printf(" gj_PP_mZ(Ab) :");
p_mZ(gj_PP_mZ(Ab), S7,P3, S7,P3, C5);
c_c_mZ(A,C1,B,C1);
c_c_mZ(A,C2,B,C2);
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
-6 -2i +16 -6i +4+16i +16+12i +14+16i
-21 -7i +56-21i +14+56i +56+42i +49+56i
+2 -1i -6 +2i -7 -6i -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
-6 -2i +16 -6i +4+16i +16+12i +14+16i +0 +0i
-21 -7i +56-21i +14+56i +56+42i +49+56i +0 +0i
+2 -1i -6 +2i -7 -6i -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
-6.000 -2.000i +16.000 -6.000i +4.000+16.000i +16.000+12.000i +14.000+16.000i
-21.000 -7.000i +56.000-21.000i +14.000+56.000i +56.000+42.000i +49.000+56.000i
+2.000 -1.000i -6.000 +2.000i -7.000 -6.000i -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 +1.000 +0.000i +0.714 +0.143i -0.714 +1.000i -2.143 +1.143i
+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 +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 +8.0000 +6.0000i
-6.0000 -2.0000i +16.0000+12.0000i
-21.0000 -7.0000i +56.0000+42.0000i
+2.0000 -1.0000i -1.0000 +0.0000i
Press return to continue.
Check if the columns of B are linearly independent
BT :
-3 -1i -6 -2i -21 -7i +2 -1i
+8 +6i +16+12i +56+42i -1 +0i
BTb :
-3 -1i -6 -2i -21 -7i +2 -1i +0 +0i
+8 +6i +16+12i +56+42i -1 +0i +0 +0i
gj_PP_mZ(BTb) :
+1.0000 +0.0000i +2.0000 +0.0000i +7.0000 +0.0000i -0.0800 +0.0600i
+0.0000 +0.0000i +0.0000 +0.0000i +0.0000 +0.0000i +1.0000 -0.0000i
+0.0000 +0.0000i
+0.0000 +0.0000i
Press return to continue.