# numpy.linalg.tensorsolve

linalg.tensorsolve(a, b, axes=None)

  Solve the tensor equation `a x = b` for x.

  It is assumed that all indices of x are summed over in the product, together with the rightmost indices of a, as is done in, for example, `tensordot(a, x, axes=b.ndim)`.

  Parameters
    a array_like

      Coefficient tensor, of shape `b.shape + Q`. Q, a tuple, equals the shape of that sub-tensor of a consisting of the appropriate number of its rightmost indices, and must be such that `prod(Q) == prod(b.shape)` (in which sense a is said to be ‘square’).

    b array_like

      Right-hand tensor, which can be of any shape.

    axes tuple of ints, optional

      Axes in a to reorder to the right, before inversion. If None (default), no reordering is done.

  Returns
    x ndarray, shape Q
  Raises
    LinAlgError

      If a is singular or not ‘square’ (in the above sense).

  See also

  `numpy.tensordot`, `tensorinv`, `numpy.einsum`

  Examples

      >>> a = np.eye(2*3*4)
      >>> a.shape = (2*3, 4, 2, 3, 4)
      >>> b = np.random.randn(2*3, 4)
      >>> x = np.linalg.tensorsolve(a, b)
      >>> x.shape
      (2, 3, 4)
      >>> np.allclose(np.tensordot(a, x, axes=3), b)
      True

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