Let $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$

We have,

$A_{11}=\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right|=3-0=3$

$A_{12}=\left|\begin{array}{cc}2 & 5 \\ -2 & 1\end{array}\right|=-(2+10)=-12$

$A_{13}=\left|\begin{array}{cc}2 & 3 \\ -2 & 0\end{array}\right|=0+6=6$

$A_{21}=\left|\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right|=-(-1-0)=1$

$A_{22}=\left|\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right|=1+4=5$

$A_{23}=-\left|\begin{array}{cc}1 & -1 \\ -2 & 0\end{array}\right|=-(0-2)=2$

$A_{31}=\left|\begin{array}{cc}-1 & 2 \\ 2 & 5\end{array}\right|=-5-6=-11$

$A_{32}=-\left|\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right|=-(5-4)=-1$

$A_{33}=\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|=3+2=5$

Hence, $a d j A=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right]$