Construct a matrix of order `2xx3`, whose elements are given by (a) `aij=((i-2j)^(2))/(2)`

Answer

To construct a matrix of order \(2 \times 3\) with elements defined by the formula \(a_{ij} = \frac{(i - 2j)^2}{2}\), we will follow these steps:

### Step 1: Identify the matrix dimensions
The matrix is of order \(2 \times 3\), which means it has 2 rows and 3 columns.

### Step 2: Define the indices
We will denote the elements of the matrix as follows:
- Row 1: \(a_{11}, a_{12}, a_{13}\)
- Row 2: \(a_{21}, a_{22}, a_{23}\)

### Step 3: Calculate each element using the formula
We will substitute the values of \(i\) (row index) and \(j\) (column index) into the formula \(a_{ij} = \frac{(i - 2j)^2}{2}\).

#### For Row 1 (\(i = 1\)):
1. **Calculate \(a_{11}\)**:
\[ a_{11} = \frac{(1 - 2 \cdot 1)^2}{2} = \frac{(1 - 2)^2}{2} = \frac{(-1)^2}{2} = \frac{1}{2} \]

2. **Calculate \(a_{12}\)**:
\[ a_{12} = \frac{(1 - 2 \cdot 2)^2}{2} = \frac{(1 - 4)^2}{2} = \frac{(-3)^2}{2} = \frac{9}{2} \]

3. **Calculate \(a_{13}\)**:
\[ a_{13} = \frac{(1 - 2 \cdot 3)^2}{2} = \frac{(1 - 6)^2}{2} = \frac{(-5)^2}{2} = \frac{25}{2} \]

#### For Row 2 (\(i = 2\)):
1. **Calculate \(a_{21}\)**:
\[ a_{21} = \frac{(2 - 2 \cdot 1)^2}{2} = \frac{(2 - 2)^2}{2} = \frac{0^2}{2} = 0 \]

2. **Calculate \(a_{22}\)**:
\[ a_{22} = \frac{(2 - 2 \cdot 2)^2}{2} = \frac{(2 - 4)^2}{2} = \frac{(-2)^2}{2} = \frac{4}{2} = 2 \]

3. **Calculate \(a_{23}\)**:
\[ a_{23} = \frac{(2 - 2 \cdot 3)^2}{2} = \frac{(2 - 6)^2}{2} = \frac{(-4)^2}{2} = \frac{16}{2} = 8 \]

### Step 4: Construct the matrix
Now we can construct the matrix using the calculated elements:
\[ A = \begin{pmatrix} \frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\ 0 & 2 & 8 \end{pmatrix} \]

### Final Answer:
The required matrix is:
\[ A = \begin{pmatrix} \frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\ 0 & 2 & 8 \end{pmatrix} \]