Question:
If $A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]$, then $\operatorname{ded}(\operatorname{adj}(\operatorname{adj} A))$ is
(a) $14^{4}$
(b) $14^{3}$
(c) $14^{2}$
(d) 14
Solution:
(a) $14^{4}$
Given :
$A=\left[\begin{array}{lll}1 & 2 & -1\end{array}\right.$
$\begin{array}{lll}-1 & 1 & 2\end{array}$
$\left.\begin{array}{lll}2 & -1 & 1\end{array}\right]$
$\therefore|A|=\mid 1 \quad 2 \quad-1$
$\begin{array}{lll}-1 & 1 & 2\end{array}$
$2-1 \quad 1 \mid=1(1+2)-2(-1-4)-1(1-2)=3+10+1=14$
We have
$|\operatorname{adj}(\operatorname{adj} A)|=|A|^{(n-1)^{2}}$
$\Rightarrow|\operatorname{adj}(\operatorname{adj} A)|=(14)^{2^{2}}=14^{4}$