Question

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

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

Answer 1

Answer: (A)

${\left(14\right)}^4$

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

Here $n=3$,

$\Rightarrow adj\left(adjA\right)=\left|A\right|A$

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

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

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