Multiply the following determinants and obtain four different determinants by multiplying row to row, to column , column to row and column to column
`|{:(3,4),(1,2):}|xx|{:(6,2),(3,1):}|`

Answer

To solve the given problem, we need to multiply the two determinants and obtain four different determinants by performing the specified operations: row to row, row to column, column to row, and column to column.

Given determinants:
\[ D_1 = \begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix}, \quad D_2 = \begin{vmatrix} 6 & 2 \\ 3 & 1 \end{vmatrix} \]

### Step 1: Calculate the Determinants
First, we calculate the determinants \(D_1\) and \(D_2\).

1. **Calculate \(D_1\)**:
\[ D_1 = 3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2 \]

2. **Calculate \(D_2\)**:
\[ D_2 = 6 \cdot 1 - 2 \cdot 3 = 6 - 6 = 0 \]

### Step 2: Row to Row
To perform the row to row multiplication, we multiply each row of \(D_1\) with each row of \(D_2\).

\[ \text{Row to Row: } \begin{vmatrix} 3 \cdot 6 + 4 \cdot 3 & 3 \cdot 2 + 4 \cdot 1 \\ 1 \cdot 6 + 2 \cdot 3 & 1 \cdot 2 + 2 \cdot 1 \end{vmatrix} \]

Calculating the elements:
- First row, first column: \(3 \cdot 6 + 4 \cdot 3 = 18 + 12 = 30\)
- First row, second column: \(3 \cdot 2 + 4 \cdot 1 = 6 + 4 = 10\)
- Second row, first column: \(1 \cdot 6 + 2 \cdot 3 = 6 + 6 = 12\)
- Second row, second column: \(1 \cdot 2 + 2 \cdot 1 = 2 + 2 = 4\)

So, the resulting determinant is:
\[ \begin{vmatrix} 30 & 10 \\ 12 & 4 \end{vmatrix} \]

### Step 3: Row to Column
For row to column multiplication, we multiply each row of \(D_1\) with each column of \(D_2\).

\[ \text{Row to Column: } \begin{vmatrix} 3 \cdot 6 + 4 \cdot 2 & 3 \cdot 3 + 4 \cdot 1 \\ 1 \cdot 6 + 2 \cdot 2 & 1 \cdot 3 + 2 \cdot 1 \end{vmatrix} \]

Calculating the elements:
- First row, first column: \(3 \cdot 6 + 4 \cdot 2 = 18 + 8 = 26\)
- First row, second column: \(3 \cdot 3 + 4 \cdot 1 = 9 + 4 = 13\)
- Second row, first column: \(1 \cdot 6 + 2 \cdot 2 = 6 + 4 = 10\)
- Second row, second column: \(1 \cdot 3 + 2 \cdot 1 = 3 + 2 = 5\)

So, the resulting determinant is:
\[ \begin{vmatrix} 26 & 13 \\ 10 & 5 \end{vmatrix} \]

### Step 4: Column to Row
For column to row multiplication, we multiply each column of \(D_1\) with each row of \(D_2\).

\[ \text{Column to Row: } \begin{vmatrix} 3 \cdot 6 + 1 \cdot 2 & 3 \cdot 2 + 1 \cdot 1 \\ 4 \cdot 6 + 2 \cdot 2 & 4 \cdot 2 + 2 \cdot 1 \end{vmatrix} \]

Calculating the elements:
- First column, first row: \(3 \cdot 6 + 1 \cdot 2 = 18 + 2 = 20\)
- First column, second row: \(3 \cdot 2 + 1 \cdot 1 = 6 + 1 = 7\)
- Second column, first row: \(4 \cdot 6 + 2 \cdot 2 = 24 + 4 = 28\)
- Second column, second row: \(4 \cdot 2 + 2 \cdot 1 = 8 + 2 = 10\)

So, the resulting determinant is:
\[ \begin{vmatrix} 20 & 7 \\ 28 & 10 \end{vmatrix} \]

### Step 5: Column to Column
For column to column multiplication, we multiply each column of \(D_1\) with each column of \(D_2\).

\[ \text{Column to Column: } \begin{vmatrix} 3 \cdot 6 + 1 \cdot 3 & 3 \cdot 2 + 1 \cdot 2 \\ 4 \cdot 6 + 2 \cdot 3 & 4 \cdot 2 + 2 \cdot 1 \end{vmatrix} \]

Calculating the elements:
- First column, first row: \(3 \cdot 6 + 1 \cdot 3 = 18 + 3 = 21\)
- First column, second row: \(3 \cdot 2 + 1 \cdot 2 = 6 + 2 = 8\)
- Second column, first row: \(4 \cdot 6 + 2 \cdot 3 = 24 + 6 = 30\)
- Second column, second row: \(4 \cdot 2 + 2 \cdot 1 = 8 + 2 = 10\)

So, the resulting determinant is:
\[ \begin{vmatrix} 21 & 8 \\ 30 & 10 \end{vmatrix} \]

### Summary of Results:
1. Row to Row: \(\begin{vmatrix} 30 & 10 \\ 12 & 4 \end{vmatrix}\)
2. Row to Column: \(\begin{vmatrix} 26 & 13 \\ 10 & 5 \end{vmatrix}\)
3. Column to Row: \(\begin{vmatrix} 20 & 7 \\ 28 & 10 \end{vmatrix}\)
4. Column to Column: \(\begin{vmatrix} 21 & 8 \\ 30 & 10 \end{vmatrix}\)