The matrix $A$ is a $2 \times 2$ matrix, so it has elements $a_{11}, a_{12}, a_{21}, a_{22}$. We use the formula $a_{ij} = \frac{|i-3j|}{2}$ to calculate each:
For $i=1, j=1$: $a_{11} = \frac{|1-3(1)|}{2} = \frac{|-2|}{2} = 1$
For $i=1, j=2$: $a_{12} = \frac{|1-3(2)|}{2} = \frac{|-5|}{2} = \frac{5}{2}$
For $i=2, j=1$: $a_{21} = \frac{|2-3(1)|}{2} = \frac{|-1|}{2} = \frac{1}{2}$
For $i=2, j=2$: $a_{22} = \frac{|2-3(2)|}{2} = \frac{|-4|}{2} = 2$
Substituting the calculated values into the matrix form:
$$A = \begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \end{bmatrix}$$The transpose $A'$ is obtained by swapping rows and columns:
$$A' = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{bmatrix}$$Final Answer: Option (B)
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