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 ___.

We first write out the matrix $$A = \{a_{ij}\}$$ where $$a_{ij} = (-1)^{j-i}$$ for $$i < j$$, $$a_{ij} = 2$$ for $$i = j$$, and $$a_{ij} = (-1)^{i+j}$$ for $$i > j$$.

For $$i < j$$: $$a_{12} = (-1)^1 = -1$$, $$a_{13} = (-1)^2 = 1$$, $$a_{23} = (-1)^1 = -1$$.

For $$i > j$$: $$a_{21} = (-1)^3 = -1$$, $$a_{31} = (-1)^4 = 1$$, $$a_{32} = (-1)^5 = -1$$.

So $$A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix}$$

We compute $$\det(A)$$ by cofactor expansion along row 1: $$\det(A) = 2(4-1) - (-1)(-2+1) + 1(1-2) = 6 - 1 - 1 = 4$$

Now we need $$\det(3\,\text{Adj}(2A^{-1}))$$. For an $$n \times n$$ matrix $$M$$, $$\det(cM) = c^n \det(M)$$ and $$\det(\text{Adj}(M)) = (\det M)^{n-1}$$.

Since $$n = 3$$: $$\det(3\,\text{Adj}(2A^{-1})) = 3^3 \cdot \det(\text{Adj}(2A^{-1})) = 27 \cdot [\det(2A^{-1})]^2$$.

$$\det(2A^{-1}) = 2^3 \cdot \det(A^{-1}) = 8 \cdot \frac{1}{\det(A)} = 8 \cdot \frac{1}{4} = 2$$

Therefore $$\det(3\,\text{Adj}(2A^{-1})) = 27 \times 2^2 = 27 \times 4 = 108$$.