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14.20 Examples

The following program solves the linear system A x = b. The system to be solved is, 

[ 0.18 0.60 0.57 0.96 ] [x0]   [1.0]
[ 0.41 0.24 0.99 0.58 ] [x1] = [2.0]
[ 0.14 0.30 0.97 0.66 ] [x2]   [3.0]
[ 0.51 0.13 0.19 0.85 ] [x3]   [4.0]

and the solution is found using LU decomposition of the matrix A. 

#include <stdio.h>
#include <gsl/gsl_linalg.h>

int
main (void)
{
  double a_data[] = { 0.18, 0.60, 0.57, 0.96,
                      0.41, 0.24, 0.99, 0.58,
                      0.14, 0.30, 0.97, 0.66,
                      0.51, 0.13, 0.19, 0.85 };

  double b_data[] = { 1.0, 2.0, 3.0, 4.0 };

  gsl_matrix_view m 
    = gsl_matrix_view_array (a_data, 4, 4);

  gsl_vector_view b
    = gsl_vector_view_array (b_data, 4);

  gsl_vector *x = gsl_vector_alloc (4);
  
  int s;

  gsl_permutation * p = gsl_permutation_alloc (4);

  gsl_linalg_LU_decomp (&m.matrix, p, &s);

  gsl_linalg_LU_solve (&m.matrix, p, &b.vector, x);

  printf ("x = \n");
  gsl_vector_fprintf (stdout, x, "%g");

  gsl_permutation_free (p);
  gsl_vector_free (x);
  return 0;
}

Here is the output from the program, 

x = 
-4.05205
-12.6056
1.66091
8.69377

This can be verified by multiplying the solution x by the original matrix A using GNU OCTAVE, 

octave> A = [ 0.18, 0.60, 0.57, 0.96;
              0.41, 0.24, 0.99, 0.58; 
              0.14, 0.30, 0.97, 0.66; 
              0.51, 0.13, 0.19, 0.85 ];

octave> x = [ -4.05205; -12.6056; 1.66091; 8.69377];

octave> A * x
ans =
  1.0000
  2.0000
  3.0000
  4.0000

This reproduces the original right-hand side vector, b, in accordance with the equation A x = b.