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# numpy.linalg.tensorsolve

`numpy.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=len(b.shape))`.

  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

  `tensordot`, `tensorinv`

  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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