If $A=\begin{bmatrix}1 & 2 &-1\\-1 & 1 & 2\\2 & -1 & 1\end{bmatrix}$, then $det (adj (adj A))$, is

$14^4$

The correct answer is option (1) : $14^4$

We know that for a square matrix of order n

$adj (adj\, A)= |A|^{n-2}A,$ if $|A|≠0$.

$⇒det(adj(adj\, A))= \left|A|^{n-2}A\right|$

$⇒det(adj(adj\, A))= \left(|A|^{n-2}\right)^n |A|$

$⇒det(adj(adj\, A))= |A|^{n^2-2n+1}$

Here, $n=3$ and $|A|=14.$

$∴det(adj(adj\, A) )= 14^4$