# numpy.linalg.slogdet#

linalg.slogdet(a)[source]#

Compute the sign and (natural) logarithm of the determinant of an array.

If an array has a very small or very large determinant, then a call to `det` may overflow or underflow. This routine is more robust against such issues, because it computes the logarithm of the determinant rather than the determinant itself.

Parameters:
a(…, M, M) array_like

Input array, has to be a square 2-D array.

Returns:
A namedtuple with the following attributes:
sign(…) array_like

A number representing the sign of the determinant. For a real matrix, this is 1, 0, or -1. For a complex matrix, this is a complex number with absolute value 1 (i.e., it is on the unit circle), or else 0.

logabsdet(…) array_like

The natural log of the absolute value of the determinant.

If the determinant is zero, then `sign` will be 0 and logabsdet
will be -inf. In all cases, the determinant is equal to
`sign * np.exp(logabsdet)`.

Notes

New in version 1.8.0.

Broadcasting rules apply, see the `numpy.linalg` documentation for details.

New in version 1.6.0.

The determinant is computed via LU factorization using the LAPACK routine `z/dgetrf`.

Examples

The determinant of a 2-D array `[[a, b], [c, d]]` is `ad - bc`:

```>>> a = np.array([[1, 2], [3, 4]])
>>> (sign, logabsdet) = np.linalg.slogdet(a)
>>> (sign, logabsdet)
(-1, 0.69314718055994529) # may vary
>>> sign * np.exp(logabsdet)
-2.0
```

Computing log-determinants for a stack of matrices:

```>>> a = np.array([ [[1, 2], [3, 4]], [[1, 2], [2, 1]], [[1, 3], [3, 1]] ])
>>> a.shape
(3, 2, 2)
>>> sign, logabsdet = np.linalg.slogdet(a)
>>> (sign, logabsdet)
(array([-1., -1., -1.]), array([ 0.69314718,  1.09861229,  2.07944154]))
>>> sign * np.exp(logabsdet)
array([-2., -3., -8.])
```

This routine succeeds where ordinary `det` does not:

```>>> np.linalg.det(np.eye(500) * 0.1)
0.0
>>> np.linalg.slogdet(np.eye(500) * 0.1)
(1, -1151.2925464970228)
```