Construct a `3 xx 4`matrix, whose elements are given by: `a_(i j)=1/2|-3i+j|` `

Explanation

To construct a \(3 \times 4\) matrix whose elements are given by \(a_{ij} = \frac{1}{2} | -3i + j | \), we will calculate each element of the matrix step by step.

### Step-by-Step Solution:

1. **Define the Matrix Structure**:
The matrix \(A\) will have elements \(a_{ij}\) where \(i\) represents the row number and \(j\) represents the column number. The matrix will look like this:
\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix} \]

2. **Calculate Each Element**:
- **For \(a_{11}\)**:
\[ a_{11} = \frac{1}{2} | -3(1) + 1 | = \frac{1}{2} | -3 + 1 | = \frac{1}{2} | -2 | = \frac{1}{2} \times 2 = 1 \]

- **For \(a_{12}\)**:
\[ a_{12} = \frac{1}{2} | -3(1) + 2 | = \frac{1}{2} | -3 + 2 | = \frac{1}{2} | -1 | = \frac{1}{2} \times 1 = \frac{1}{2} \]

- **For \(a_{13}\)**:
\[ a_{13} = \frac{1}{2} | -3(1) + 3 | = \frac{1}{2} | -3 + 3 | = \frac{1}{2} | 0 | = 0 \]

- **For \(a_{14}\)**:
\[ a_{14} = \frac{1}{2} | -3(1) + 4 | = \frac{1}{2} | -3 + 4 | = \frac{1}{2} | 1 | = \frac{1}{2} \times 1 = \frac{1}{2} \]

- **For \(a_{21}\)**:
\[ a_{21} = \frac{1}{2} | -3(2) + 1 | = \frac{1}{2} | -6 + 1 | = \frac{1}{2} | -5 | = \frac{1}{2} \times 5 = \frac{5}{2} \]

- **For \(a_{22}\)**:
\[ a_{22} = \frac{1}{2} | -3(2) + 2 | = \frac{1}{2} | -6 + 2 | = \frac{1}{2} | -4 | = \frac{1}{2} \times 4 = 2 \]

- **For \(a_{23}\)**:
\[ a_{23} = \frac{1}{2} | -3(2) + 3 | = \frac{1}{2} | -6 + 3 | = \frac{1}{2} | -3 | = \frac{1}{2} \times 3 = \frac{3}{2} \]

- **For \(a_{24}\)**:
\[ a_{24} = \frac{1}{2} | -3(2) + 4 | = \frac{1}{2} | -6 + 4 | = \frac{1}{2} | -2 | = \frac{1}{2} \times 2 = 1 \]

- **For \(a_{31}\)**:
\[ a_{31} = \frac{1}{2} | -3(3) + 1 | = \frac{1}{2} | -9 + 1 | = \frac{1}{2} | -8 | = \frac{1}{2} \times 8 = 4 \]

- **For \(a_{32}\)**:
\[ a_{32} = \frac{1}{2} | -3(3) + 2 | = \frac{1}{2} | -9 + 2 | = \frac{1}{2} | -7 | = \frac{1}{2} \times 7 = \frac{7}{2} \]

- **For \(a_{33}\)**:
\[ a_{33} = \frac{1}{2} | -3(3) + 3 | = \frac{1}{2} | -9 + 3 | = \frac{1}{2} | -6 | = \frac{1}{2} \times 6 = 3 \]

- **For \(a_{34}\)**:
\[ a_{34} = \frac{1}{2} | -3(3) + 4 | = \frac{1}{2} | -9 + 4 | = \frac{1}{2} | -5 | = \frac{1}{2} \times 5 = \frac{5}{2} \]

3. **Construct the Matrix**:
Now that we have all the elements, we can construct the matrix \(A\):
\[ A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix} \]

### Final Matrix:
\[ A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix} \]