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: Use the transpose rule directly:
The entry of $A^{\prime}$ in row $i$ and column $j$ is $a_{ji}$. So I can write the entries of $A^{\prime}$ without building $A$.
Let $c_{ij}$ stand for the entry of $A^{\prime}$. Then $c_{ij} = a_{ji} = \frac{(j + 2i)^2}{3}$.

Step 2: Compute the four entries:
$c_{11} = \frac{(1 + 2)^2}{3} = 3$
$c_{12} = \frac{(2 + 2)^2}{3} = \frac{16}{3}$
$c_{21} = \frac{(1 + 4)^2}{3} = \frac{25}{3}$
$c_{22} = \frac{(2 + 4)^2}{3} = 12$

Step 3: Match with the options:
The first row of $A^{\prime}$ is $3, \frac{16}{3}$. Options (B) and (D) have $\frac{16}{3}$ in that place, but (B) starts with 12, so (B) fails on $c_{11}$.
Options (A) and (C) have $\frac{25}{3}$ in the first row, which contradicts $c_{12} = \frac{16}{3}$.
Only (D) agrees with all four entries.

Final Answer:
The matrix with first row $3, \frac{16}{3}$ and second row $\frac{25}{3}, 12$ is option (D). \[ \boxed{\begin{bmatrix} 3 & \frac{16}{3} \\ \frac{25}{3} & 12 \end{bmatrix}} \]