Question

If \(A = [a_{ij}]\) is a \(2 \times 2\) matrix whose elements are given by \(a_{ij} = \frac{|i - 3j|}{2}\), then \(A'\) is :

• A. \(\begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \end{bmatrix}\)
• B. \(\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{bmatrix}\)
• C. \(\begin{bmatrix} 2 & \frac{5}{2} \\ \frac{1}{2} & 1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 1 \end{bmatrix}\)

Hint:

Always double-check if the question asks for matrix \(A\) or its transpose \(A'\).
Transpose operation: Row 1 becomes Column 1, and Row 2 becomes Column 2.

Answer 1

The Correct Option is B

Concept:
A \(2\times2\) matrix \(A\) is represented as: \[ A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \] Here, \(i\) denotes the row number and \(j\) denotes the column number.
The transpose of \(A\), denoted by \(A'\), is obtained by interchanging its rows and columns. 
Step 1: Calculate \(a_{11}\)
For \(i=1\) and \(j=1\): \[ a_{11}=\frac{|1-3(1)|}{2} \] \[ =\frac{|-2|}{2}=1 \] 
Step 2: Calculate \(a_{12}\)
For \(i=1\) and \(j=2\): \[ a_{12}=\frac{|1-3(2)|}{2} \] \[ =\frac{|-5|}{2}=\frac{5}{2} \] 
Step 3: Calculate \(a_{21}\)
For \(i=2\) and \(j=1\): \[ a_{21}=\frac{|2-3(1)|}{2} \] \[ =\frac{|-1|}{2}=\frac{1}{2} \] 
Step 4: Calculate \(a_{22}\)
For \(i=2\) and \(j=2\): \[ a_{22}=\frac{|2-3(2)|}{2} \] \[ =\frac{|-4|}{2}=2 \] 
Step 5: Construct matrix \(A\) and find \(A'\)
Therefore: \[ A= \begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \end{bmatrix} \] Interchanging the rows and columns: \[ A'= \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{bmatrix} \] 
Final Answer:
Therefore, \[ \boxed{ A'= \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{bmatrix} } \] Hence, the correct answer is Option (B).