Let $A=\left\{a_{ij}\right\}$ be a $3\times 3$ matrix, where $a_{ij}=\left\{\begin{array}{c}{\left(-1\right)}^{j-i} \mathrm{if} i<j\\ 2 \mathrm{if} i=j\\ {\left(-1\right)}^{i+j} \mathrm{if} i>j\end{array}\right.$ then $\det \left(3Adj\left(2A^{-1}\right)\right)$ is equal to ________.

We have,

$a_{ij}=\left\{\begin{array}{c}{\left(-1\right)}^{j-i} \mathrm{if} i<j\\ 2 \mathrm{if} i=j\\ {\left(-1\right)}^{i+j} \mathrm{if} i>j\end{array}\right.$

The,

$A=\left[\begin{array}{rrr}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]$

$\Rightarrow \left|A\right|=4$

Now,

$\left|3adj\left(2{ A}^{-1}\right)\right|=3^3\left|adj\left(2A^{-1}\right)\right|$

$\because \left|kadj\left(A^{-1}\right)\right|=k^n\left|adj\left(A^{-1}\right)\right|$

$\Rightarrow \left|3adj\left(2{ A}^{-1}\right)\right|=3^3·2^6{\left|A^{-1}\right|}^2$

$\because \left|adj\left(k{A}^{-1}\right)\right|={\left(k^{\left(n-1\right)}\right)}^n\left|adj\left(A^{-1}\right)\right|={\left(k^{\left(n-1\right)}\right)}^n{\left|A^{-1}\right|}^{n-1}$

$\Rightarrow \left|3adj\left(2{ A}^{-1}\right)\right|={12}^3{\left|{\mathit{ }A}^{-1}\right|}^2=\frac{{12}^3}{| A{|}^2}=\frac{{12}^3}{16}=108$