Fri Dec 27 09:31:17 2024

rref2_test():
  python version: 3.10.12
  numpy version:  1.26.4
  Test rref(), which analyzes matrices using the
  reduced row echelon form (RREF)

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

  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

  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

  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

  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

  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 MATLABs 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.         3.         0.         0.         2.         0.
  0.        ]
 [0.         0.         0.         1.         2.         0.
  0.        ]
 [0.         0.         0.         0.         0.         1.
  0.33333333]
 [0.         0.         0.         0.         0.         0.
  0.        ]]

  a_cols:
[0 3 5]

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 = : 1.0

  np.linalg.det(A) =  0.999999999999988

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. -41. -17.  10.]
 [-41.  25.  10.  -6.]
 [-17.  10.   5.  -3.]
 [ 10.  -6.  -3.   2.]]

  np.linalg.inv ( A ):
[[ 68. -41. -17.  10.]
 [-41.  25.  10.  -6.]
 [-17.  10.   5.  -3.]
 [ 10.  -6.  -3.   2.]]

  Product A_inv * A:
[[ 1.00000000e+00  7.46069873e-14 -3.19744231e-14  6.39488462e-14]
 [ 5.32907052e-15  1.00000000e+00  8.88178420e-15  3.55271368e-15]
 [-1.77635684e-15 -2.84217094e-14  1.00000000e+00 -3.55271368e-15]
 [ 4.66293670e-15  1.75415238e-14  5.55111512e-15  1.00000000e+00]]

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
  np.linalg.matrix_rank(A)  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.]
 [2.]
 [3.]
 [4.]]

  np.linalg.solve ( A, b ):
[[1.]
 [2.]
 [3.]
 [4.]]

  Product A * x:
[[57.]
 [79.]
 [88.]
 [86.]]

rref2_test():
  Normal end of execution.
Fri Dec 27 09:31:17 2024