code: def _remove_redundancy_id(A, rhs, rank=None, randomized=True):
      """Eliminates redundant equations from a system of equations.
  
      Eliminates redundant equations from system of equations defined by Ax = b
      and identifies infeasibilities.
  
      Parameters
      ----------
      A : 2-D array
          An array representing the left-hand side of a system of equations
      rhs : 1-D array
          An array representing the right-hand side of a system of equations
      rank : int, optional
          The rank of A
      randomized: bool, optional
          True for randomized interpolative decomposition
  
      Returns
      -------
      A : 2-D array
          An array representing the left-hand side of a system of equations
      rhs : 1-D array
          An array representing the right-hand side of a system of equations
      status: int
          An integer indicating the status of the system
          0: No infeasibility identified
          2: Trivially infeasible
      message : str
          A string descriptor of the exit status of the optimization.
  
      """
  
      status = 0
      message = ""
      inconsistent = (
          "There is a linear combination of rows of A_eq that "
          "results in zero, suggesting a redundant constraint. "
          "However the same linear combination of b_eq is "
          "nonzero, suggesting that the constraints conflict "
          "and the problem is infeasible."
      )
  
      A, rhs, status, message = _remove_zero_rows(A, rhs)
  
      if status != 0:
          return A, rhs, status, message
  
      m, n = A.shape
  
      k = rank
      if rank is None:
          k = np.linalg.matrix_rank(A)
  
      idx, proj = interp_decomp(A.T, k, rand=randomized)
  
      # first k entries in idx are indices of the independent rows
      # remaining entries are the indices of the m-k dependent rows
      # proj provides a linear combinations of rows of A2 that form the
      # remaining m-k (dependent) rows. The same linear combination of entries
      # in rhs2 must give the remaining m-k entries. If not, the system is
      # inconsistent, and the problem is infeasible.
      if not np.allclose(rhs[idx[:k]] @ proj, rhs[idx[k:]]):
          status = 2
          message = inconsistent
  
      # sort indices because the other redundancy removal routines leave rows
      # in original order and tests were written with that in mind
      idx = sorted(idx[:k])
      A2 = A[idx, :]
      rhs2 = rhs[idx]
      return A2, rhs2, status, message
  
file_path: scipy/scipy/optimize/_remove_redundancy.py
imports: import numpy
  from scipy.linalg import svd
  from scipy.linalg.interpolative import interp_decomp
  import scipy
  from scipy.linalg.blas import dtrsm
label_desc: Use image node_id 8 for calling a global function with example usage: _remove_redundancy_id(A, rhs, rank, randomized) and returns: A2,  rhs2,  status,  message, A,  rhs,  status,  message