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jax.scipy.linalg.inv

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jax.scipy.linalg.inv#

jax.scipy.linalg.inv(a, overwrite_a=False, check_finite=True)[source]#

Return the inverse of a square matrix

JAX implementation of scipy.linalg.inv().

Parameters:

  • a (jax.typing.ArrayLike) – array of shape (..., N, N) specifying square array(s) to be inverted.

  • overwrite_a (bool) – unused in JAX

  • check_finite (bool) – unused in JAX

Returns:

Array of shape (..., N, N) containing the inverse of the input.

Return type:

Array

Notes

In most cases, explicitly computing the inverse of a matrix is ill-advised. For example, to compute x = inv(A) @ b, it is more performant and numerically precise to use a direct solve, such as jax.scipy.linalg.solve().

See also

Example

Compute the inverse of a 3x3 matrix

>>> a = jnp.array([[1., 2., 3.],
...                [2., 4., 2.],
...                [3., 2., 1.]])
>>> a_inv = jax.scipy.linalg.inv(a)
>>> a_inv  
Array([[ 0.        , -0.25      ,  0.5       ],
       [-0.25      ,  0.5       , -0.25000003],
       [ 0.5       , -0.25      ,  0.        ]], dtype=float32)

Check that multiplying with the inverse gives the identity:

>>> jnp.allclose(a @ a_inv, jnp.eye(3), atol=1E-5)
Array(True, dtype=bool)

Multiply the inverse by a vector b, to find a solution to a @ x = b

>>> b = jnp.array([1., 4., 2.])
>>> a_inv @ b
Array([ 0.  ,  1.25, -0.5 ], dtype=float32)

Note, however, that explicitly computing the inverse in such a case can lead to poor performance and loss of precision as the size of the problem grows. Instead, you should use a direct solver like jax.scipy.linalg.solve():

>>> jax.scipy.linalg.solve(a, b)
 Array([ 0.  ,  1.25, -0.5 ], dtype=float32)
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