If \(A = \begin{bmatrix}1 & 2 & -1\\ 3 & 0 & 2\\ 4 & 1 & 5\end{bmatrix}\), then the element in the second row and third column of \(\mathrm{adj}\,A\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-5\)
The adjoint is the transpose of the cofactor matrix, so the element in row 2, column 3 of adj A is the cofactor \(A_{32}\), not \(A_{23}\).
Deleting row 3 and column 2 of A:
\(\displaystyle M_{32} = \begin{vmatrix}1 & -1\\ 3 & 2\end{vmatrix} = 2 + 3 = 5\)
\(A_{32} = (-1)^{3 + 2} \times 5 = -5\)
Why the other options are incorrect
Option (1) 7 — Incorrect. This is the cofactor \(A_{23} = -(1 - 8) = 7\); adj A takes the cofactors in transposed position.
Option (2) 5 — Incorrect. This is the minor \(M_{32}\) without the sign \((-1)^{5} = -1\).
Option (3) \(-7\) — Incorrect. This is the minor \(M_{23} = 1 - 8 = -7\), which belongs to row 2, column 3 of A, not of adj A.