If \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\), then \(A \cdot \operatorname{adj}(A)\) is equal to

Given, \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\)
\[
\begin{aligned}
&C_{11}=8-6=2 \\
&C_{12}=-(0+9)=-9, C_{31}=+6-4=2 \\
&C_{13}=0-6=-6, C_{32}=-(-3-0)=3 \\
&C_{21}=-(-8+4)=4, C_{33}=2-0=2 \\
&C_{22}=4-6=-2, C_{23}=-(-2+6)=-4
\end{aligned}
\]
\(\operatorname{adj}(A)=\left[\begin{array}{ccc}2 & -9 & -6 \\ 4 & -2 & -4 \\ 2 & 3 & 2\end{array}\right]^{T}\)
\[
=\left[\begin{array}{ccc}
2 & 4 & 2 \\
-9 & -2 & 3 \\
-6 & -4 & 2
\end{array}\right]
\]
Now, \(A \cdot \operatorname{adj}(A)=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}2 & 4 & 2 \\ -9 & -2 & 3 \\ -6 & -4 & 2\end{array}\right]\)
\[
=\left[\begin{array}{lll}
8 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 8
\end{array}\right]
\]
Alternative : \(A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\)
\[
\begin{aligned}
\Rightarrow & A A^{-1} &=\frac{A \cdot[\operatorname{adj}(A)]}{|A|} \\
\Rightarrow & I|A| &=A \cdot[\operatorname{adj}(A)] \\
& A &=8 \\
& \therefore & A[\operatorname{adj}(A)] &=8 I=\left[\begin{array}{lll}
8 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 8
\end{array}\right]
\end{aligned}
\]