Linear Algebra and the C Language/a0b4


Install and compile this file in your working directory.

/* ------------------------------------ */
/*  Save as :   c00b.c                  */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA   R4
#define CA   C3 
#define Cb   C1 
/* ------------------------------------ */       
/* ------------------------------------ */
int main(void)
{
double ta[RA*(CA+Cb)]={
//  x1   x3   x4       
    +1,  +0,  +0, // A
    +0,  +1,  +0, // B   
    +0,  -1,  +1, // C
    -1,  +0,  -1  // D  
};

double tb[RA*(CA+Cb)]={
        +20+10, // A
   +60 -10 -30, // B   
           +20, // C
       -100+30, // D  
};
                       
double **A      =   ca_A_mR(ta,i_mR(RA,CA));
double **b      =   ca_A_mR(tb,i_mR(RA,Cb));

double **Q      =              i_mR(RA,CA);
double **Q_T    =              i_mR(CA,RA);

double **R      =              i_mR(CA,CA);
double **invR   =              i_mR(CA,CA);

double **invR_Q_T = i_mR(CA,RA);
double **x        = i_mR(CA,C1); 

  clrscrn();
  printf(" Copy/Paste into the octave windows \n\n");
  p_Octave_mR(A,"a",P0);  
  printf(" [Q, R] = qr (a,0) \n\n");
  
  QR_mR(A,Q,R);    
  printf(" Q :");
  p_mR(Q, S10,P4, C10);  
  printf(" R :");
  p_mR(R, S10,P4, C10); 
  stop(); 
  
  clrscrn();
  transpose_mR(Q,Q_T);   
  printf(" Q_T :");
  pE_mR(Q_T,S9,P5, C3);
  invgj_mR(R,invR); 
  printf(" invR :");
  pE_mR(invR,S9,P5, C6);
  stop();

  clrscrn();
  printf(" Solving this system yields a unique\n"
         " least squares solution, namely   \n\n");
  mul_mR(invR,Q_T,invR_Q_T);
  mul_mR(invR_Q_T,b,x);
  printf(" x = invR * Q_T * b :");
  p_mR(x,S9,P5 ,C6);
  stop();
         
  f_mR(A);
  f_mR(b);
  f_mR(Q);
  f_mR(Q_T);
  f_mR(R);
  f_mR(invR);
  f_mR(x);
      
  return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */

Screen output example:

 Copy/Paste into the octave windows 

 a=[
+1,+0,+0;
+0,+1,+0;
+0,-1,+1;
-1,+0,-1]

 [Q, R] = qr (a,0) 

 Q :
   +0.7071    +0.0000    -0.5000 
   +0.0000    +0.7071    +0.5000 
   +0.0000    -0.7071    +0.5000 
   -0.7071    +0.0000    -0.5000 

 R :
   +1.4142    +0.0000    +0.7071 
   +0.0000    +1.4142    -0.7071 
   +0.0000    +0.0000    +1.0000 

 Press return to continue. 


 Q_T :
+7.07107e-01 +0.00000e+00 +0.00000e+00 
+0.00000e+00 +7.07107e-01 -7.07107e-01 
-5.00000e-01 +5.00000e-01 +5.00000e-01 

-7.07107e-01 
+0.00000e+00 
-5.00000e-01 

 invR :
+7.07107e-01 -0.00000e+00 -5.00000e-01 
-0.00000e+00 +7.07107e-01 +5.00000e-01 
-0.00000e+00 -0.00000e+00 +1.00000e+00 

 Press return to continue. 


 Solving this system yields a unique
 least squares solution, namely   

 x = invR * Q_T * b :
+30.00000 
+20.00000 
+40.00000 

 Press return to continue.