Find the adjoint of the matrix
$\mathrm{A}=\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right]$  and hence show that A(adj A) = |A|$I_3$

Given, $A=\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right]$
$\mathrm{Let} A_{\mathrm{ij}} \mathrm{be} \mathrm{the} \mathrm{cofactor} \mathrm{of} \mathrm{an} \mathrm{element} a_{ij} \mathrm{of} A. \mathrm{Then}, \mathrm{cofactors} \mathrm{of} \mathrm{elements} \mathrm{of} \left|\mathrm{A}\right| \mathrm{are}$$\begin{array}{l}{\mathrm{A}}_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}1& -2\\ -2& 1\end{array}\right|=(1-4)=-3\\ {\mathrm{A}}_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}2& -2\\ 2& 1\end{array}\right|=-(2+4)=-6\\ {\mathrm{A}}_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}2& 1\\ 2& -2\end{array}\right|=(-4-2)=-6\\ {\mathrm{A}}_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}-2& -2\\ -2& 1\end{array}\right|=-(-2-4)=6\\ {\mathrm{A}}_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}-1& -2\\ 2& 1\end{array}\right|=(-1+4)=3\\ {\mathrm{A}}_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}-1& -2\\ 2& -2\end{array}\right|=-(2+4)=-6\\ {\mathrm{A}}_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}-2& -2\\ 1& -2\end{array}\right|=(4+2)=6\\ {\mathrm{A}}_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}-1& -2\\ 2& -2\end{array}\right|=-(2+4)=-6\\ {\mathrm{A}}_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}-1& -2\\ 2& 1\end{array}\right|=(-1+4)=3\end{array}$
Clearly, the adjoint of the matrix A is given by
$\mathrm{adj} \mathrm{A}=\left[\begin{array}{l}{\mathrm{A}}_{11}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{A}}_{21}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{A}}_{31}\\ {\mathrm{A}}_{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{A}}_{22}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{A}}_{32}\\ {\mathrm{A}}_{13}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{A}}_{23}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{A}}_{33}\end{array}\right]=\left[\begin{array}{ccc}-3& 6& 6\\ -6& 3& -6\\ -6& -6& 3\end{array}\right]$
Now, $\left|\mathrm{A}\right|=\left|\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right|$
$\begin{array}{l}=-1(1-4)+2(2+4)-2(-4-2)\\ =-1(-3)+2\left(6\right)-2(-6)\\ =3+12+12=27\end{array}$
and $\mathrm{A}\cdot (\mathrm{adj}\mathrm{A})=\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right]\left[\begin{array}{ccc}-3& 6& 6\\ -6& 3& -6\\ -6& -6& 3\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{lcc}3+12+12& -6-6+12& -6+12-6\\ -6-6+12& 12+3+12& 12-6-6\\ -6+12-6& 12-6-6& 12+12+3\end{array}\right]\\ =\left[\begin{array}{ccc}27& 0& 0\\ 0& 27& 0\\ 0& 0& 27\end{array}\right]=27\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$
$=27{\mathrm{I}}_3=\left|\mathrm{A}\right|{\mathrm{I}}_{3 } \mathrm{Hence} \mathrm{proved}.$