Question

Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

• A. \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
• B. \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
• C. \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
• D. \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)

Hint:

Be careful not to confuse the matrix \(A\) with its transpose \(A'\). Double-check the position swaps: the element at row 1, column 2 (\(a_{12}\)) shifts to row 2, column 1 (\(a_{21}\)) in the transposed matrix.

Answer 1

The Correct Option is A

Concept: A \(2 \times 2\) matrix \(A\) consists of elements arranged in two rows and two columns: \[ A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \] The transpose of a matrix, denoted as \(A'\) or \(A^T\), is obtained by swapping its rows with its columns: \[ A' = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix} \]

Step 1: Compute each element of matrix \(A\)
We are given the general rule for the entries: \(a_{ij} = \frac{(2i - j)^2}{3}\). Let us calculate each of the four components systematically.
• For element \(a_{11}\) (\(i=1, j=1\)): \[ a_{11} = \frac{(2(1) - 1)^2}{3} = \frac{(2 - 1)^2}{3} = \frac{(1)^2}{3} = \frac{1}{3} \]
• For element \(a_{12}\) (\(i=1, j=2\)): \[ a_{12} = \frac{(2(1) - 2)^2}{3} = \frac{(2 - 2)^2}{3} = \frac{(0)^2}{3} = 0 \]
• For element \(a_{21}\) (\(i=2, j=1\)): \[ a_{21} = \frac{(2(2) - 1)^2}{3} = \frac{(4 - 1)^2}{3} = \frac{(3)^2}{3} = \frac{9}{3} = 3 \]
• For element \(a_{22}\) (\(i=2, j=2\)): \[ a_{22} = \frac{(2(2) - 2)^2}{3} = \frac{(4 - 2)^2}{3} = \frac{(2)^2}{3} = \frac{4}{3} \]

Step 2: Construct matrix \(A\)
Combining these evaluated components together into the standard grid format gives: \[ A = \begin{bmatrix} \frac{1}{3} & 0 3 & \frac{4}{3} \end{bmatrix} \]

Step 3: Determine the transpose matrix \(A'\)
By definition, the row elements become column elements, and column elements become row elements: \[ A' = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix} \] Comparing this result with the given choices, it matches option (A).