Question

If $A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right]$, then $\text{det}\left(\text{adj}\right(\text{adj}A\left)\right)$ is equal to.

• A. ${14}^4$
• B. ${14}^3$
• C. ${14}^2$
• D. $14$

Answer 1

The correct option is A ${14}^4$

We have, $A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right]$
$\Rightarrow |A|=1(1+2)-2(-1-4)-1(1-2)$
$=3+10+1=14$
We know that, for a square matrix of order $n$,
$\text{adj}\left(\text{adj}A\right)=|A{|}^{n-2}A,$ if $|A|\ne 0$
$\Rightarrow \text{det}\left(\text{adj}\right(\text{adj}A\left)\right)=||A{|}^{n-2}A|$
$\Rightarrow \text{det}\left(\text{adj}\right(\text{adj}A\left)\right)=(|A{|}^{n-2}{)}^n|A|$
$\Rightarrow \text{det}\left(\text{adj}\right(\text{adj}A\left)\right)=|A{|}^{n^2-2n+1}$
Here, $n=3$ and $|A|=14$
Therefore, $\text{det}\left(\text{adj}\right(\text{adj}A\left)\right)=(14{)}^{3^2-2\times 3+1}={14}^4$.