4

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?

k_rollo
  • 143

2 Answers2

2

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 $-+-+$.

Santiago Canez
  • 2,551
  • Ah! So I should use the sign patterns of my chosen row or column. Since I chose the 4th column, the pattern should be - + - + which will yield -[-2(0)] + [-5(-1)] - [0] + [0]. Is this correct? – k_rollo Feb 13 '14 at 17:26
  • Yes, and the same pattern works if you choose different rows instead. – Santiago Canez Feb 13 '14 at 17:28
  • Thank you very much! I will confirm with a different row and a different column with their corresponding sign patterns. Marked as accepted answer. – k_rollo Feb 13 '14 at 17:34
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.

user76568
  • 4,652