Step-by-Step Solution

1. First, we need to find the elements of the matrix \(A=[a_{ij}]\) using the given rule: \(a_{ij} = \text{maximum}(i, j) - \text{minimum}(i, j)\).

2. For \(a_{11}\), \(i = 1\) and \(j = 1\). So, \(a_{11} = \text{maximum}(1, 1) - \text{minimum}(1, 1) = 1 - 1 = 0\).

3. For \(a_{12}\), \(i = 1\) and \(j = 2\). So, \(a_{12} = \text{maximum}(1, 2) - \text{minimum}(1, 2) = 2 - 1 = 1\).

4. For \(a_{21}\), \(i = 2\) and \(j = 1\). So, \(a_{21} = \text{maximum}(2, 1) - \text{minimum}(2, 1) = 2 - 1 = 1\).

5. For \(a_{22}\), \(i = 2\) and \(j = 2\). So, \(a_{22} = \text{maximum}(2, 2) - \text{minimum}(2, 2) = 2 - 2 = 0\).

6. Therefore, the matrix \(A\) is given by: \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\).

7. Now, we need to find \(A^2\), which is \(A \times A\):

   \(A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\)

8. Performing the matrix multiplication:

   \(A^2 = \begin{bmatrix} (0 \times 0) + (1 \times 1) & (0 \times 1) + (1 \times 0) \\ (1 \times 0) + (0 \times 1) & (1 \times 1) + (0 \times 0) \end{bmatrix}\)

   \(A^2 = \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix}\)

   \(A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)