# numpy.linalg.tensorsolve¶

`numpy.linalg.``tensorsolve`(a, b, axes=None)[源代码]

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. x : ndarray, shape Q LinAlgError If a is singular or not ‘square’ (in the above sense).

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
```