### 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.