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
We know that for a square matrix of order n $adj (adj\, A) =|A|^{n-2} A$, if $|A|≠0$.
$⇒det (adj (adj\, A)) = ||A|^{n-2} A|$
$⇒det (adj (adj\, A)) =(|A|^{n-2})^n|A|$
$⇒det (adj (adj\, A)) =|A|^{n^2 - 2n +1}$
Here, $n = 3$ and $|A|= 14$
$∴ det (adj (adj\, A)) = 14^4$