If $A=\begin{bmatrix}1 & 2 &-1\\-1 & 1 & 2\\2 & -1 & 1\end{bmatrix}$, then $det (adj (adj A))$, is
Answer & explanation
Correct answer: option 1
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 $|A|=14.$
$∴det(adj(adj\, A) )= 14^4$