Determinant by Elimination

Compute the determinant of a square matrix, without np.linalg.det.

The cofactor expansion you learned by hand costs O(n!), which is hopeless past about n = 10. Do it by elimination instead, which is O(n^3):

• Reduce to upper triangular form using row operations.
• Subtracting a multiple of one row from another does not change the determinant, so those steps are free.
• Swapping two rows flips the sign, so count the swaps.
• The determinant is the product of the diagonal, times -1 per swap.

Use partial pivoting for stability. If the best available pivot in a column is smaller than 1e-12, the matrix is singular and the determinant is 0.0.

The determinant of an empty (0, 0) matrix is 1.0 by convention.

Function signature

def determinant(A):

Example

Input

determinant(np.array([[1.0, 2.0], [3.0, 4.0]]))

Output

-2.0

Reasoning

For a 2 by 2 matrix this is 1 * 4 - 2 * 3 = -2. With elimination: pivoting swaps the rows (sign -1), the diagonal becomes 3 and 2/3, and -1 * 3 * 2/3 = -2.

Constraints and notes

• If the best pivot is below 1e-12, return 0.0.
• The determinant of a (0, 0) matrix is 1.0.
• Do not use np.linalg.det.