The cofactor of the element 6 in the determinant \(\begin{vmatrix}4 & -2 & 1\\ 3 & 0 & 5\\ -1 & 6 & 2\end{vmatrix}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-17\)
The element 6 is in row 3, column 2, so we need \(C_{32} = (-1)^{3+2}M_{32}\).
The minor \(M_{32}\) is the determinant left after deleting row 3 and column 2:
\( \displaystyle M_{32} = \begin{vmatrix}4 & 1\\ 3 & 5\end{vmatrix} \)
\(M_{32} = 20 - 3 = 17\)
Since \(3 + 2 = 5\) is odd, the cofactor has the opposite sign:
\(C_{32} = (-1)^{5} \times 17 = -17\)
Why the other options are incorrect
Option (1) 17 — Incorrect. This is the minor \(M_{32}\); the sign \((-1)^{5} = -1\) is left out.
Option (3) 23 — Incorrect. This adds the cross products, \(20 + 3\), instead of subtracting them.
Option (4) \(-23\) — Incorrect. The sign is right, but the minor is \(20 - 3 = 17\), not 23.