matrix is only N=5, might as well bruteforce recurse

A = [
  [3,0,0,-2,4],
  [0,2,0,0,0],
  [0,-1,0,5,-3],
  [-4,0,1,0,6],
  [0,-1,0,3,2]
]
def det(A):
  N = len(A)
  if N == 1: return A[0][0]
  return sum(
    (-1)**i * v * det([[v for j,v in enumerate(ls) if j != i] for ls in A[1:]])
    for i,v in enumerate(A[0])
    if v
  )

print(det(A))

It looks like my homework bruh print(-114)

-114

This is a relatively sparse matrix, so we might be able to solve it directly.

1. Find any row or column that has only one non-zero value. The remaining zeros nullify any sub-matrix they might multiply, so the determinant must go through that value. Accordingly, we can zero out the rest of the column or row (note the reversal of terms) where a value is the only non-zero in its column or row. (There are two such.)
2. Iterate #1 as far as possible. (This isolates a third row & column)
3. The determinant of the remaining matrix is (relatively) easy to compute.

The answer is -114.