A JEE Exam problem on determinants and matrices I am first stating the question:

Let $A=\{a_{ij}\}$ be a $3\times 3$ matrix, where
$$a_{ij}=\begin{cases} 
(-1)^{j-i}&\text{if $i<j$,}\\
2&\text{if $i=j$,}\\
(-1)^{i-j}&\text{if $i>j$,}
\end{cases}$$
then $\det(3\,\text{adj}(2A^{-1}))$ is equal to __________

I solved this in the following manner:
$$
A=\left[\begin{array}{lcc}
2 & (-1)^{2-1} & (-1)^{3-1} \\
(-1)^{2+1} & 2 & (-1)^{3-2} \\
(-1)^{3+1} & (-1)^{3 + 2} & 2
\end{array}\right]=\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]
$$
$$\begin{aligned}|A| &=2(4-1)+(-2+1)+(1-2) \\ &=6-1-1=4 \end{aligned}$$
$$
\begin{aligned}
& \operatorname{det}\left(3 \operatorname{adj}\left(2 A^{-1}\right)\right) \\
=& 3^{3}\left|\operatorname{adj}\left(2 A^{-1}\right)\right| \\
=& 3^{3}\left|2^{3} \operatorname{adj}\left(A^{-1}\right)\right| \\
=&(3 \times 2)^{3} \times\left(\left|A^{-1}\right|\right)^{2}\\=&6^3\times\Big(\frac14\Big)\\=&13.5
\end{aligned}
$$
Original image
Is my solution correct?
Note: The problem came in the JEE Main Exam of India, on the 20th of July. The answer given for this question in the Answer Key is 108.

No, your solution is not correct, but you are almost done.
Looking through the properties of adjugate matrix, we note that if $A$ is a $n\times n$ matrix then $\text{adj}(cA)=c^{n-1}\text{adj}(A)$ (not $c^n$ as you did) and $\text{adj}(A^{-1})=\det(A^{-1})A$. Therefore
$$\begin{align}\det(3\,\text{adj}(2A^{-1}))&=\det(3\cdot 2^{n-1}\det(A^{-1})A)\\
&=(3\cdot 2^{n-1}\det(A^{-1}))^{n}\det(A)
=
\frac{3^n2^{n(n-1)}}{\det(A)^{n-1}}.
\end{align}$$
Hence when $n=3$ and $\det(A)=4$, we find
$$\det(3\,\text{adj}(2A^{-1})=\frac{3^32^{6}}{4^{2}}=27\cdot 4=108$$
which is precisely the given answer.
