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.
 A: 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.
A: Alternatively, use the properties:
$$\begin{align}\text{adj}(cA)&=c^{n-1}\text{adj}(A) \quad (1)\\
\det(cA)&=c^n\det(A) \quad (2)\\
\det(\text{adj}(A))&=(\det(A))^{n-1} \quad (3)\\
\det(A^{-1})&=(\det(A))^{-1} \quad (4)
\end{align}$$
to get:
$$\det(3\,\text{adj}(2A^{-1}))\stackrel{(1)}{=}\\
\det(3\cdot 2^2\,\text{adj}(A^{-1}))\stackrel{(2)}{=}\\
12^3\det(\text{adj}(A^{-1}))\stackrel{(3)}{=}\\
12^3(\det(A^{-1}))^2\stackrel{(4)}{=}\\
12^3((\det(A))^{-1})^2=\\
12^3\cdot 4^{-2}=\\
108.$$
