Question

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

• A. ${\left(14\right)}^4$
• B. ${\left(14\right)}^3$
• C. ${\left(14\right)}^2$
• D. ${\left(14\right)}^1$

Answer 1

The correct option is A ${\left(14\right)}^4$

We know that $adj\left(adjA\right)={\left|A\right|}^{n-2}A$ if $\left|A\right|\ne 0$, provided order of $A$ is $n$.

$\therefore adj\left(adjA\right)=\left|A\right|A$ $($ as $n=3)$

$\therefore det\left(adj\left(adjA\right)\right)={\left|A\right|}^{3}detA={\left|A\right|}^4$

But $\left|A\right|=\left|\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right|=14$

$\therefore det\left(adj\left(adjA\right)\right)={\left(14\right)}^4$