Verify A (adj A) = (adj A) A = I .

Solution:

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

$|A|=1(0-0)+1(9+2)+2(0-0)=11$

$\therefore|A| I=11\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$

Now,

$A_{11}=0, A_{12}=-(9+2)=-11, A_{13}=0$

$A_{21}=-(-3-0)=3, A_{22}=3-2=1, A_{23}=-(0+1)=-1$

$A_{31}=2-0=2, A_{32}=-(-2-6)=8, A_{33}=0+3=3$

$\therefore \operatorname{adj} A=\left[\begin{array}{lll}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]$

Now,

$A(\operatorname{adj} A)=\left[\begin{array}{ccr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\left[\begin{array}{lll}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]$

$=\left[\begin{array}{lll}0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right]$

$=\left[\begin{array}{lll}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$

Also,

$(\operatorname{adj} A) \cdot A=\left[\begin{array}{lll}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{lcr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$

$=\left[\begin{array}{lll}0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9\end{array}\right]$

$=\left[\begin{array}{lll}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$

Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I .$