Skip to main content
`Ctrl`+`K`

# scipy.linalg.solve_triangular

scipy.linalg.solve_triangular(a, b, trans=0, lower=False, unit_diagonal=False, overwrite_b=False, check_finite=True)

  Solve the equation a x = b for x, assuming a is a triangular matrix.

  Parameters:
    a(M, M) array_like

      A triangular matrix

    b(M,) or (M, N) array_like

      Right-hand side matrix in a x = b

    lower bool, optional

      Use only data contained in the lower triangle of a. Default is to use upper triangle.

    trans{0, 1, 2, ‘N’, ‘T’, ‘C’}, optional

      Type of system to solve:

      trans | system
      --- | ---
      0 or ‘N’ | a x = b
      1 or ‘T’ | a^T x = b
      2 or ‘C’ | a^H x = b

    unit_diagonal bool, optional

      If True, diagonal elements of a are assumed to be 1 and will not be referenced.

    overwrite_b bool, optional

      Allow overwriting data in b (may enhance performance)

    check_finite bool, optional

      Whether to check that the input matrices contain only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs.

  Returns:
    x(M,) or (M, N) ndarray

      Solution to the system a x = b. Shape of return matches b.

  Raises:
    LinAlgError

      If a is singular

  Notes

  New in version 0.9.0.

  Examples

  Solve the lower triangular system a x = b, where:

           [3  0  0  0]       [4]
      a =  [2  1  0  0]   b = [2]
           [1  0  1  0]       [4]
           [1  1  1  1]       [2]

      >>> import numpy as np
      >>> from scipy.linalg import solve_triangular
      >>> a = np.array([[3, 0, 0, 0], [2, 1, 0, 0], [1, 0, 1, 0], [1, 1, 1, 1]])
      >>> b = np.array([4, 2, 4, 2])
      >>> x = solve_triangular(a, b, lower=True)
      >>> x
      array([ 1.33333333, -0.66666667,  2.66666667, -1.33333333])
      >>> a.dot(x)  # Check the result
      array([ 4.,  2.,  4.,  2.])