Cofactor of a12 in \(A=\begin{bmatrix}1 & 2 & 3\\ 4 & -5 & 6\\ 0 & 1 & -1\end{bmatrix}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4
The minor \(M_{12}\) is the determinant left after deleting row 1 and column 2:
\( M_{12} = \begin{vmatrix}4 & 6\\ 0 & -1\end{vmatrix} \)
\(M_{12} = 4(-1) - 6(0) = -4\)
The cofactor is \(A_{ij} = (-1)^{i+j}M_{ij}\):
\(A_{12} = (-1)^{1+2}(-4)\)
\(A_{12} = 4\)
Why the other options are incorrect
Option (1) 5 — Incorrect. The cofactor works out to 4.
Option (3) −4 — Incorrect. This is the minor \(M_{12}\); the cofactor multiplies it by \((-1)^{1+2} = -1\).
Option (4) −5 — Incorrect. The cofactor works out to 4.