find the problem
Question:

Find $A(\operatorname{adj} A)$ for the matrix $A=\left[\begin{array}{ccc}1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2\end{array}\right]$

Solution:

$A=\left[\begin{array}{ccc}1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}2 & -1 \\ 5 & 2\end{array}\right|=9, C_{12}=-\left|\begin{array}{cc}0 & -1 \\ -4 & 2\end{array}\right|=4$ and $C_{13}=\left|\begin{array}{cc}0 & 2 \\ -4 & 5\end{array}\right|=8$

$C_{21}=-\left|\begin{array}{cc}-2 & 3 \\ 5 & 2\end{array}\right|=19, C_{22}=\left|\begin{array}{cc}1 & 3 \\ -4 & 2\end{array}\right|=14$ and $C_{23}=-\left|\begin{array}{cc}1 & -2 \\ -4 & 5\end{array}\right|=3$

$C_{31}=\left|\begin{array}{cc}-2 & 3 \\ 2 & -1\end{array}\right|=-4, C_{32}=-\left|\begin{array}{cc}1 & 3 \\ 0 & -1\end{array}\right|=1$ and $C_{33}=\left|\begin{array}{cc}1 & -2 \\ 0 & 2\end{array}\right|=2$

$a d j A=\left[\begin{array}{ccc}9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2\end{array}\right]^{T}=\left[\begin{array}{ccc}9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2\end{array}\right]$

$\therefore A(\operatorname{adj} A)=\left[\begin{array}{ccc}25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25\end{array}\right]$

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