Mathc matrices/05j
Apparence
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c00a.c |
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/* ------------------------------------ */
/* Save as : c00a.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
#define RCA RC5
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double a[RCA*RCA] ={
+0.329527604306, -0.365886364507, -0.067047239569, +0.001915635416, -0.287345312440,
-0.365886364507, +0.800330583940, -0.036588636451, +0.001045389613, -0.156808441932,
-0.067047239569, -0.036588636451, +0.993295276043, +0.000191563542, -0.028734531244,
+0.001915635416, +0.001045389613, +0.000191563542, +0.999994526756, +0.000820986607,
-0.287345312440, -0.156808441932, -0.028734531244, +0.000820986607, +0.876852008954
};
double v[RCA*RCA] ={
-0.479027840775, -0.099503719020, +0.002857131195, -0.393919298579, +0.818823787938,
+0.877799708227, +0.000000000000, +0.000000000000, +0.000000000000, +0.446843838561,
+0.000000000000, +0.995037190210, +0.000000000000, +0.000000000000, +0.081882378794,
+0.000000000000, +0.000000000000, +0.999995918392, +0.000000000000, -0.002339496537,
+0.000000000000, +0.000000000000, +0.000000000000, +0.919145030018, +0.350924480545
};
double **A = ca_A_mR(a, i_mR(RCA,RCA));
double **V = ca_A_mR(v, i_mR(RCA,RCA));
double **invV = invgj_mR(V, i_mR(RCA,RCA));
double **EValue = i_mR(RCA,RCA);
double **T = i_mR(RCA,RCA);
clrscrn();
printf(" A :");
p_mR(A, S8,P6, C5);
printf(" V :");
p_mR(V, S9,P6, C5);
printf(" EValue = invV * A * V");
mul_mR(invV,A,T);
mul_mR(T,V,EValue);
p_mR(EValue, S9,P6, C5);
printf(" A = V * EValue * invV");
mul_mR(V,EValue,T);
mul_mR(T,invV,A);
p_mR(A, S8,P6, C5);
stop();
clrscrn();
printf(" The matrix A projects the space in the direction\n"
" of the eigenvector V5 on a hyperplan determined\n"
" by the eigenvector V1,V2,V3 and V4 if :\n\n"
" The eigenvector V1 has its eigenvalue equal to one and \n"
" The eigenvector V2 has its eigenvalue equal to one and \n"
" The eigenvector V3 has its eigenvalue equal to one and \n"
" The eigenvector V4 has its eigenvalue equal to one and \n"
" The eigenvector V5 has its eigenvalue equal to zero and \n\n"
" If The vectors V1,V2,V3,V4 and V5 are linearly independent\n\n"
" det(V) = %.5e\n\n",det_R(V));
stop();
f_mR(A);
f_mR(V);
f_mR(invV);
f_mR(T);
f_mR(EValue);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Vérifier les calculs.
Exemple de sortie écran :
A :
+0.329528 -0.365886 -0.067047 +0.001916 -0.287345
-0.365886 +0.800331 -0.036589 +0.001045 -0.156808
-0.067047 -0.036589 +0.993295 +0.000192 -0.028735
+0.001916 +0.001045 +0.000192 +0.999995 +0.000821
-0.287345 -0.156808 -0.028735 +0.000821 +0.876852
V :
-0.479028 -0.099504 +0.002857 -0.393919 +0.818824
+0.877800 +0.000000 +0.000000 +0.000000 +0.446844
+0.000000 +0.995037 +0.000000 +0.000000 +0.081882
+0.000000 +0.000000 +0.999996 +0.000000 -0.002339
+0.000000 +0.000000 +0.000000 +0.919145 +0.350924
EValue = invV * A * V
+1.000000 -0.000000 +0.000000 -0.000000 +0.000000
-0.000000 +1.000000 +0.000000 -0.000000 +0.000000
+0.000000 +0.000000 +1.000000 +0.000000 -0.000000
-0.000000 -0.000000 +0.000000 +1.000000 -0.000000
-0.000000 -0.000000 +0.000000 -0.000000 -0.000000
A = V * EValue * invV
+0.329528 -0.365886 -0.067047 +0.001916 -0.287345
-0.365886 +0.800331 -0.036589 +0.001045 -0.156808
-0.067047 -0.036589 +0.993295 +0.000192 -0.028735
+0.001916 +0.001045 +0.000192 +0.999995 +0.000821
-0.287345 -0.156808 -0.028735 +0.000821 +0.876852
Press return to continue.
The matrix A projects the space in the direction
of the eigenvector V5 on a hyperplan determined
by the eigenvector V1,V2,V3 and V4 if :
The eigenvector V1 has its eigenvalue equal to one and
The eigenvector V2 has its eigenvalue equal to one and
The eigenvector V3 has its eigenvalue equal to one and
The eigenvector V4 has its eigenvalue equal to one and
The eigenvector V5 has its eigenvalue equal to zero and
If The vectors V1,V2,V3,V4 and V5 are linearly independent
det(V) = 9.80453e-01
Press return to continue.