If \(A = \begin{bmatrix}1 & 0 & 2\\ 0 & 2 & 1\\ 1 & 0 & 3\end{bmatrix}\), then \(\mathrm{adj}\,A\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}6 & 0 & -4\\ 1 & 1 & -1\\ -2 & 0 & 2\end{bmatrix}\)
The adjoint is the transpose of the matrix of cofactors, where \(C_{ij} = (-1)^{i + j}M_{ij}\).
\(C_{11} = 6 - 0 = 6,\ \ C_{12} = -(0 - 1) = 1,\ \ C_{13} = 0 - 2 = -2\)
\(C_{21} = -(0 - 0) = 0,\ \ C_{22} = 3 - 2 = 1,\ \ C_{23} = -(0 - 0) = 0\)
\(C_{31} = 0 - 4 = -4,\ \ C_{32} = -(1 - 0) = -1,\ \ C_{33} = 2 - 0 = 2\)
The rows of cofactors become the columns of adj A:
\(\displaystyle \mathrm{adj}\,A = \begin{bmatrix}6 & 0 & -4\\ 1 & 1 & -1\\ -2 & 0 & 2\end{bmatrix}\)
Check: \(|A| = 1(6) + 0 + 2(-2) = 2\), and \(A\,(\mathrm{adj}\,A)\) has first row \((6 - 4,\ 0,\ -4 + 4) = (2, 0, 0)\), as \(A\,(\mathrm{adj}\,A) = 2I\) requires.
Why the other options are incorrect
Option (1) \(\begin{bmatrix}6 & 1 & -2\\ 0 & 1 & 0\\ -4 & -1 & 2\end{bmatrix}\) — Incorrect. This is the cofactor matrix itself; it has not been transposed.
Option (3) \(\displaystyle\frac{1}{2}\begin{bmatrix}6 & 0 & -4\\ 1 & 1 & -1\\ -2 & 0 & 2\end{bmatrix}\) — Incorrect. This is \(A^{-1} = \frac{1}{2}\,\mathrm{adj}\,A\), not the adjoint.
Option (4) \(\begin{bmatrix}6 & 0 & -4\\ -1 & 1 & 1\\ -2 & 0 & 2\end{bmatrix}\) — Incorrect. This uses the minors without the signs \((-1)^{i + j}\), so \(C_{12}\) and \(C_{32}\) have the wrong sign.