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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, similarly as in tensordot(a, x, axes=len(b.shape)).

  Parameters: | a : array_like, shape b.shape+Q Coefficient tensor. Shape Q of the rightmost indices of a must be such that a is ‘square’, ie., prod(Q) == prod(b.shape). b : array_like, any shape Right-hand tensor. axes : tuple of integers Axes in a to reorder to the right, before inversion. If None (default), no reordering is done.
  Returns: | x : array, shape Q

  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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