Construct a matrix of order `3xx2`, whose elements `a_(ij)` given by
aij = `{{:(2i-3j,igej),(3i+j,iltj):}`

Answer

To construct a matrix of order \(3 \times 2\) with elements defined by the relation:

\[ a_{ij} = \begin{cases} 2i - 3j & \text{if } i \geq j \\ 3i + j & \text{if } i < j \end{cases} \]

we will calculate each element \(a_{ij}\) for \(i = 1, 2, 3\) and \(j = 1, 2\).

### Step 1: Calculate \(a_{11}\)
For \(i = 1\) and \(j = 1\):
- Since \(i = j\), we use the first case:
\[ a_{11} = 2(1) - 3(1) = 2 - 3 = -1 \]

### Step 2: Calculate \(a_{12}\)
For \(i = 1\) and \(j = 2\):
- Since \(i < j\), we use the second case:
\[ a_{12} = 3(1) + 2 = 3 + 2 = 5 \]

### Step 3: Calculate \(a_{21}\)
For \(i = 2\) and \(j = 1\):
- Since \(i \geq j\), we use the first case:
\[ a_{21} = 2(2) - 3(1) = 4 - 3 = 1 \]

### Step 4: Calculate \(a_{22}\)
For \(i = 2\) and \(j = 2\):
- Since \(i = j\), we use the first case:
\[ a_{22} = 2(2) - 3(2) = 4 - 6 = -2 \]

### Step 5: Calculate \(a_{31}\)
For \(i = 3\) and \(j = 1\):
- Since \(i \geq j\), we use the first case:
\[ a_{31} = 2(3) - 3(1) = 6 - 3 = 3 \]

### Step 6: Calculate \(a_{32}\)
For \(i = 3\) and \(j = 2\):
- Since \(i \geq j\), we use the first case:
\[ a_{32} = 2(3) - 3(2) = 6 - 6 = 0 \]

### Final Matrix
Now we can construct the matrix using the calculated elements:
\[ A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 1 & -2 \\ 3 & 0 \end{bmatrix} \]