Linear Algebra and the C Language/a06t

/* ------------------------------------ */
/*  Save as :  c00e.c                   */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
void fun(int r)
{
double s = r_I(9);
int    c = C2;

double **A =    r_mR(        i_mR(r,r),9.);
double **B = mulC_mR(c_mR(A, i_mR(r,r)),s,c);

  clrscrn();
  printf(" A:");
  p_mR(A, S3,P0,C6);
  printf(" The column[%d] of A is muliply by %+.3f:\n",c,s);
  printf(" B:");
  p_mR(B, S3,P0,C6);

  printf("  If B is the matrix that  results when \n");
  printf("  a single column of A is muliplied by  \n");
  printf("  a scalar s, then det(B) = s det(A)  \n\n");
  printf("    det_R(B) = %+.0f\n",       det_R(B));
  printf(" %+.0f det_R(A) = %+.0f\n",s,s*det_R(A));
  printf("    det_R(A) = %+.0f\n",       det_R(A));

  f_mR(A);
  f_mR(B);
}
/* ------------------------------------ */
int main(void)
{
time_t t;

  srand(time(&t));
  
do
{
  fun(rp_I(RC3)+RC1);
  
} while(stop_w());

  return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */

 A:
 +3  +3  -6 
 +6  -4  -4 
 -1  +7  +9 

 The column[3] of A is muliply by +6.000:
 B:
 +3  +3 -36 
 +6  -4 -24 
 -1  +7 +54 

  If B is the matrix that  results when 
  a single column of A is muliplied by  
  a scalar s, then det(B) = s det(A)  

    det_R(B) = -2412
 +6 det_R(A) = -2412
    det_R(A) = -402

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