Question

Let \( A = [a_{ij}] \) be a \( 2 \times 2 \) matrix whose elements are given by \( a_{ij} = \frac{(i + 2j)^2}{3} \). Then \( A^{\prime} \) is given by :

• A. \( \begin{bmatrix} 12 & \frac{25}{3} \\ \frac{16}{3} & 3 \end{bmatrix} \)
• B. \( \begin{bmatrix} 12 & \frac{16}{3} \\ \frac{25}{3} & 3 \end{bmatrix} \)
• C. \( \begin{bmatrix} 3 & \frac{25}{3} \\ \frac{16}{3} & 12 \end{bmatrix} \)
• D. \( \begin{bmatrix} 3 & \frac{16}{3} \\ \frac{25}{3} & 12 \end{bmatrix} \)

Hint:

Find a_11 = 3, a_12 = 25/3, a_21 = 16/3, a_22 = 12, then swap rows and columns.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The matrix \(A\) has 2 rows and 2 columns. The entry in row \(i\) and column \(j\) is \(a_{ij} = \frac{(i + 2j)^2}{3}\). We build \(A\) first and then take its transpose.

Step 2: Find each entry of A:
\(a_{11} = \frac{(1 + 2)^2}{3} = \frac{9}{3} = 3\)
\(a_{12} = \frac{(1 + 4)^2}{3} = \frac{25}{3}\)
\(a_{21} = \frac{(2 + 2)^2}{3} = \frac{16}{3}\)
\(a_{22} = \frac{(2 + 4)^2}{3} = \frac{36}{3} = 12\)
\[ A = \begin{bmatrix} 3 & \frac{25}{3} \\ \frac{16}{3} & 12 \end{bmatrix} \]

Step 3: Take the transpose:
The transpose \(A^{\prime}\) swaps rows and columns. So the first row of \(A\) becomes the first column of \(A^{\prime}\).
\[ A^{\prime} = \begin{bmatrix} 3 & \frac{16}{3} \\ \frac{25}{3} & 12 \end{bmatrix} \]

Step 4: Check the options:
Option (A) has 12 and 3 on the diagonal, which is the wrong way round, so it is wrong. Option (B) also has the diagonal reversed.
Option (C) is the matrix \(A\) itself, not its transpose.
Option (D) matches \(A^{\prime}\) exactly.

Final Answer:
The transpose is the matrix in option (D). \[ \boxed{A^{\prime} = \begin{bmatrix} 3 & \frac{16}{3} \\ \frac{25}{3} & 12 \end{bmatrix}} \]