Let A = [a_{ij}] be a 3 times 3 matrix,where | vedclass

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(D) First,we construct the matrix $A$ based on the given conditions:For $i=1, j=2: i < j \implies a_{12} = (-1)^{2-1} = -1$For $i=1, j=3: i < j \impli

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$108$

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(D) First,we construct the matrix $A$ based on the given conditions:For $i=1, j=2: i For $i=1, j=3: i For $i=2, j=1: i > j \implies a_{21} = (-1)^{2+1} = -1$For $i=2, j=3: i For $i=3, j=1: i > j \implies a_{31} = (-1)^{3+1} = 1$For $i=3, j=2: i > j \implies a_{32} = (-1)^{3+2} = -1$Diagonal elements $a_{ii} = 2$.Thus,$A = \begin{bmatrix} 2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2 \end{bmatrix}$.Calculating the determinant $|A| = 2(4-1) - (-1)(-2+1) + 1(1-2) = 2(3) - 1 - 1 = 6 - 2 = 4$.We need to find $\det(3 \operatorname{Adj}(2 A^{-1}))$.Using properties of determinants: $|kM| = k^n |M|$ for an $n 	imes n$ matrix,and $|\operatorname{Adj}(M)| = |M|^{n-1}$.Here $n=3$,so $|3 \operatorname{Adj}(2 A^{-1})| = 3^3 |\operatorname{Adj}(2 A^{-1})| = 27 |2 A^{-1}|^{3-1} = 27 |2 A^{-1}|^2$.Since $|2 A^{-1}| = 2^3 |A^{-1}| = 8 \cdot \frac{1}{|A|} = \frac{8}{4} = 2$.Therefore,$|3 \operatorname{Adj}(2 A^{-1})| = 27 \cdot (2)^2 = 27 \cdot 4 = 108$.

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 $\det(3 \operatorname{Adj}(2 A^{-1}))$ is equal to:

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