/NumPy 1.17

# numpy.linalg.tensorinv

`numpy.linalg.tensorinv(a, ind=2)` [source]

Compute the ‘inverse’ of an N-dimensional array.

The result is an inverse for `a` relative to the tensordot operation `tensordot(a, b, ind)`, i. e., up to floating-point accuracy, `tensordot(tensorinv(a), a, ind)` is the “identity” tensor for the tensordot operation.

Parameters: `a : array_like` Tensor to ‘invert’. Its shape must be ‘square’, i. e., `prod(a.shape[:ind]) == prod(a.shape[ind:])`. `ind : int, optional` Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2. `b : ndarray` `a`’s tensordot inverse, shape `a.shape[ind:] + a.shape[:ind]`. LinAlgError If `a` is singular or not ‘square’ (in the above sense).

#### Examples

```>>> a = np.eye(4*6)
>>> a.shape = (4, 6, 8, 3)
>>> ainv = np.linalg.tensorinv(a, ind=2)
>>> ainv.shape
(8, 3, 4, 6)
>>> b = np.random.randn(4, 6)
>>> np.allclose(np.tensordot(ainv, b), np.linalg.tensorsolve(a, b))
True
```
```>>> a = np.eye(4*6)
>>> a.shape = (24, 8, 3)
>>> ainv = np.linalg.tensorinv(a, ind=1)
>>> ainv.shape
(8, 3, 24)
>>> b = np.random.randn(24)
>>> np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b))
True
```

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https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.linalg.tensorinv.html