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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

$|\operatorname{adj}(\operatorname{adj} A)|=|A|^{(n-1)^{2}}$

$\Rightarrow|\operatorname{adj}(\operatorname{adj} A)|=(14)^{2^{2}}=14^{4}$

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