The sum of the cofactors of the elements of the second row of the determinant \(\begin{vmatrix}2 & -1 & 3\\ 4 & 0 & 1\\ -2 & 5 & 6\end{vmatrix}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 31
The cofactor of \(a_{ij}\) is \(C_{ij} = (-1)^{i + j}M_{ij}\); in the second row the signs are \(-, +, -\).
\(\displaystyle C_{21} = -\begin{vmatrix}-1 & 3\\ 5 & 6\end{vmatrix} = -(-6 - 15) = 21\)
\(\displaystyle C_{22} = +\begin{vmatrix}2 & 3\\ -2 & 6\end{vmatrix} = 12 + 6 = 18\)
\(\displaystyle C_{23} = -\begin{vmatrix}2 & -1\\ -2 & 5\end{vmatrix} = -(10 - 2) = -8\)
\(C_{21} + C_{22} + C_{23} = 21 + 18 - 8 = 31\)
Why the other options are incorrect
Option (1) 47 — Incorrect. This takes the cofactor of \(a_{23}\) as \(+8\); its sign \((-1)^{2 + 3}\) makes it \(-8\).
Option (2) \(-31\) — Incorrect. This uses the signs \(+, -, +\) of the first row instead of \(-, +, -\).
Option (3) 5 — Incorrect. This is the sum of the minors \(-21 + 18 + 8\); the signs \(-, +, -\) are not applied.