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 3×3 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 3×3 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 4×4 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?
Best Answer
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 $-+-+$.