From 718b1abfd1d6367c3ab66fc135dd3712dc8a8435 Mon Sep 17 00:00:00 2001
From: Amit Subhash Chejara
 <125737375+amitsubhashchejara@users.noreply.github.com>
Date: Sat, 19 Oct 2024 21:38:14 +0530
Subject: [PATCH] Update documentation in _linalg.py for np.linalg.det() to
 clarify floating-point precision

This update modifies the documentation in _linalg.py for the np.linalg.det() function to clarify the effects of floating-point precision limitations when calculating the determinant of a matrix.

It includes an explanation that small numerical errors (i.e. floating point rounding error) may occur, leading to very small non-zero values instead of exactly zero for singular matrices. Additionally, an example is provided using np.isclose() to check if the determinant is effectively zero.
---
 numpy/linalg/_linalg.py | 13 +++++++++++++
 1 file changed, 13 insertions(+)

diff --git a/numpy/linalg/_linalg.py b/numpy/linalg/_linalg.py
index 70698c92fde7..932aef10f7d4 100644
--- a/numpy/linalg/_linalg.py
+++ b/numpy/linalg/_linalg.py
@@ -2357,6 +2357,12 @@ def det(a):
     The determinant is computed via LU factorization using the LAPACK
     routine ``z/dgetrf``.
 
+    Due to the limitations of floating-point arithmetic, small numerical
+    errors may occur when calculating the determinant of a matrix. As a result, 
+    the determinant may return very small non-zero values instead of exactly zero 
+    for singular matrices. To check if a determinant is effectively zero, use 
+    `np.isclose()`.
+
     Examples
     --------
     The determinant of a 2-D array [[a, b], [c, d]] is ad - bc:
@@ -2374,6 +2380,13 @@ def det(a):
     >>> np.linalg.det(a)
     array([-2., -3., -8.])
 
+    Checking if the determinant is effectively zero for singular matrices:
+
+    >>> a = np.array([[5, 5, 6], [7, 7, 5], [4, 4, 8]])
+    >>> det = np.linalg.det(a)
+    >>> np.isclose(det, 0)
+    True
+
     """
     a = asarray(a)
     _assert_stacked_2d(a)