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15-Dec-2024 16:08:51

rref2_test():
  MATLAB/Octave version 9.11.0.2358333 (R2021b) Update 7
  Test rref(), which analyzes matrices using the
  reduced row echelon form (RREF)

is_rref_test():
  is_rref() reports if a matrix is in reduced row echelon format.

  Matrix A0:
     1     0     0     9     4
     0     0     1     0     8
     0     0     0     0     0
     0     0     0     1     0

  is_rref(A0) = false

  Matrix A1:
     1     0     0     9     4
     0     0     0     1     0
     0     0     1     0     8
     0     0     0     0     0

  is_rref(A1) = false

  Matrix A2:
     1     0     0     9     4
     0     1     0     2     8
     0     0     3     0     0
     0     0     0     0     0

  is_rref(A2) = false

  Matrix A3:
     1     0     3     9     4
     0     1     0     2     8
     0     0     1     0     0
     0     0     0     0     0

  is_rref(A3) = false

  Matrix A4:
     1     0     3     0     4
     0     1     2     0     8
     0     0     0     1     0
     0     0     0     0     0

  is_rref(A4) = true

rref_columns_test():
  rref_columns() uses the reduced row echelon form (RREF)
  of a matrix to find the linearly independent columns.

  Matrix A:
     1     2     3     1
     2     4     9     3
     3     6     0     0
     4     8     0     2
     5    10     6     6
     6    12     6     3
     7    14     2     1


  Number of independent columns is 3

  Independent columns of A:
     1     3     1
     2     9     3
     3     0     0
     4     0     2
     5     6     6
     6     6     3
     7     2     1


rref_compute_test():
  rref_compute() is a user-written code to compute the
  reduced row echelon form (RREF) of a matrix.
  Compare it to MATLAB's built-in version.

  Matrix A:
     1     3     0     2     6     3     1
    -2    -6     0    -2    -8     3     1
     3     9     0     0     6     6     2
    -1    -3     0     1     0     9     3


  rref_compute(A):
    1.0000    3.0000         0         0    2.0000         0         0
         0         0         0    1.0000    2.0000         0         0
         0         0         0         0         0    1.0000    0.3333
         0         0         0         0         0         0         0


  Independent column indices:
     1     4     6


  rref(A) (MATLAB built-in code):
    1.0000    3.0000         0         0    2.0000         0         0
         0         0         0    1.0000    2.0000         0         0
         0         0         0         0         0    1.0000    0.3333
         0         0         0         0         0         0         0


rref_determinant_test():
  rref_determinant() uses the reduced row echelon form 
  of a square matrix to compute the determinant.

  Matrix A:
     5     7     6     5
     7    10     8     7
     6     8    10     9
     5     7     9    10


  Estimated determinant of A = 1:

  MATLAB det(A) = 1

rref_inverse_test():
  rref_inverse() uses the reduced row echelon form 
  of a square matrix to compute its inverse.

  Matrix A:
     5     7     6     5
     7    10     8     7
     6     8    10     9
     5     7     9    10


  Estimated inverse of A, A_inv:
   68.0000  -41.0000  -17.0000   10.0000
  -41.0000   25.0000   10.0000   -6.0000
  -17.0000   10.0000    5.0000   -3.0000
   10.0000   -6.0000   -3.0000    2.0000


  MATLAB inverse of A:
   68.0000  -41.0000  -17.0000   10.0000
  -41.0000   25.0000   10.0000   -6.0000
  -17.0000   10.0000    5.0000   -3.0000
   10.0000   -6.0000   -3.0000    2.0000


  Product A_inv * A:
    1.0000    0.0000   -0.0000    0.0000
         0    1.0000    0.0000         0
   -0.0000   -0.0000    1.0000   -0.0000
    0.0000    0.0000    0.0000    1.0000


rref_rank_test():
  rref_rank() uses the reduced row echelon form 
  of a matrix to estimate its rank.

  Matrix A:
     1    -2     3    -1
     3    -6     9    -3
     0     0     0     0
     2    -2     0     1
     6    -8     6     0
     3     3     6     9
     1     1     2     3


  A has rank 3
  MATLAB rank estimate is 3

rref_solve_test():
  rref_solve() uses the reduced row echelon form 
  of a square matrix to solve a linear system.

  Matrix A:
     5     7     6     5
     7    10     8     7
     6     8    10     9
     5     7     9    10

  Right hand side b:
    57
    79
    88
    86


  Estimated solution:
    1.0000
    2.0000
    3.0000
    4.0000


  MATLAB solution:
    1.0000
    2.0000
    3.0000
    4.0000


  Product A * x:
   57.0000
   79.0000
   88.0000
   86.0000


rref2_test():
  Normal end of execution.

15-Dec-2024 16:08:51