Determinant of 4x4 Matrix by Expansion Method

Question

Find det(B) = \begin{bmatrix} 2 & 5 & -3 & -2 \\ -2 & -3 & 2 & -5 \\ 1 & 3 & -2 & 0 \\ -1 & -6 & 4 & 0 \\ \end{bmatrix}

I chose the 4th column because it has the most 0s. Using basketweave, I solved for the determinants of the minor 3x3 matrices of entry B₁₄ and B₂₄.

det(B₁₄) = \begin{bmatrix} -2 & -3 & 2 \\ 1 & 3 & -2 \\ -1 & -6 & 4 \\ \end{bmatrix}

det(B₁₄)
= (-24 - 6 - 12) - (-12 - 24 - 6)
= -42 - (-42)
= 0

det(B₂₄) = \begin{bmatrix} 2 & 5 & -3 \\ 1 & 3 & -2 \\ -1 & -6 & 4 \\ \end{bmatrix}

det(B₂₄)
= (24 + 10 + 18) - (20 + 24 + 9)
= 52 - 53
= -1

I have checked with a matrix calculator and the the determinants of the 3x3 minor matrices are correct.

To find the det(B), I multiplied B₁₄ by det(B₁₄) and B₂₄ by det(B₂₄) and followed the + - + - pattern as showed by the formula here (scroll below for 4x4 formula). The rest will be 0s anyway.

det(B)
= [-2(0)] - [-5(-1)] + [0] - [0]
= -5

Checking again with the matrix calculator, the correct answer is +5.

I am confused as to how the signs apply. How did det(B) arrive to +5?

Answer 1

The pattern $+-+-$ works in the link you gave since they are expanding along the first row; it would be the same pattern when expanding along the first column. But then it changes as you change what column/row you're expanding along. For expanding along the second column it would be $-+-+$, along the third $+-+-$, and along the fourth $-+-+$.

Answer 2

The formula reads: $$|A|=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}M_{ij}$$ So the sign is positive/negative when the sum of the row index and the column index ($i+j$) is even/odd.