Question:

Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=A \mid$.

$\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$

Solution:

$A=\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$

we have,

$|A|=-12-(-12)=-12+12=0$

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

Now,

$A_{11}=-6, A_{12}=4, A_{21}=-3, A_{22}=2$

$\therefore \operatorname{adj} A=\left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right]$

Now,

\begin{aligned} A(\operatorname{adj} A) &=\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]\left[\begin{array}{cr}-6 & -3 \\ 4 & 2\end{array}\right] \\ &=\left[\begin{array}{ll}-12+12 & -6+6 \\ 24-24 & 12-12\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \end{aligned}

Also, $($ adj $A) A=\left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$

$=\left[\begin{array}{cc}-12+12 & -18+18 \\ 8-8 & 12-12\end{array}\right]=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]$

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