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   >>> 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) >>