Question

Verify $A\left(adj\right)\left(A\right)-\left(adjA\right)A=|A|I_n|)in$
$\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right]$
Now, $|A|=\left|\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right|=1(0-0)-(-1)(9+2)+2(0-0)=0+11+0=11$
$\therefore |A|I=11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right]$

Cofactors of A are
$A_{11}=0,A_{12}=-(9+2)=-11,A_{13}=0,A_{21}=-(-3-0)=3,A_{22}=3-2=1,A_{23}=-(0+1)=-1,A_{31}=2-0=2,A_{32}=-(-2-6)=8A_{33}=0+3=3$
$\therefore adj\left(A\right)={\left[\begin{array}{ccc}0& -11& 0\\ 3& 1& -1\\ 2& 8& 3\end{array}\right]}^T=\left[\begin{array}{ccc}0& 3& 2\\ -11& 1& 8\\ 0& -1& 3\end{array}\right]$
Now, $A\left(adjA\right)=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right]\left[\begin{array}{ccc}0& 3& 2\\ -11& 1& 8\\ 0& -1& 3\end{array}\right]$
$=\left[\begin{array}{ccc}0+11+0& 3-1-2& 2-8+6\\ 0+0+0& 9+0+2& 6+0-6\\ 0+0+0& 3+0-3& 2+0+9\end{array}\right]=\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right]$
Also $\left(adjA\right)A=\left[\begin{array}{ccc}0& 3& 2\\ -11& 1& 8\\ 0& -1& 3\end{array}\right]\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right]$
$=\left[\begin{array}{ccc}0+9+2& 0+0+0& 0-6+6\\ -11+3+8& 11+0+0& -22-2+24\\ 0-3+2& 0+0+0& 0+2+9\end{array}\right]=\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right]$
Hence, $A\left(adjA\right)=\left(adjA\right)A=|A|I_3$