c
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
We have,
$A_{11}=4, A_{12}=-3, A_{13}=-2, A_{22}=1$
$\therefore a d j A=\left[\begin{array}{ll}A_{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right]=\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$