Evaluate the following determinants by two method :
`|{:(1,2,4),(-1,3,0),(4,1,0):}|`

Answer

To evaluate the determinant \( D = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix} \) using two methods, we will proceed as follows:

### Method 1: General Method (Cofactor Expansion)

1. **Identify the determinant**:
\[ D = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix} \]

2. **Use cofactor expansion along the first row**:
\[ D = 1 \cdot D_{11} - 2 \cdot D_{12} + 4 \cdot D_{13} \]
where \( D_{ij} \) is the determinant of the matrix obtained by removing the \( i \)-th row and \( j \)-th column.

3. **Calculate \( D_{11} \)**:
\[ D_{11} = \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} = (3 \cdot 0) - (0 \cdot 1) = 0 \]

4. **Calculate \( D_{12} \)**:
\[ D_{12} = \begin{vmatrix} -1 & 0 \\ 4 & 0 \end{vmatrix} = (-1 \cdot 0) - (0 \cdot 4) = 0 \]

5. **Calculate \( D_{13} \)**:
\[ D_{13} = \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} = (-1 \cdot 1) - (3 \cdot 4) = -1 - 12 = -13 \]

6. **Substituting back into the determinant**:
\[ D = 1 \cdot 0 - 2 \cdot 0 + 4 \cdot (-13) = 0 - 0 - 52 = -52 \]

### Method 2: Shortcut Method (Column Repetition)

1. **Write the determinant and repeat the first two columns**:
\[ D = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 4 & 1 & 2 \\ -1 & 3 & 0 & -1 & 3 \\ 4 & 1 & 0 & 4 & 1 \end{vmatrix} \]

2. **Apply the formula**:
\[ D = 1 \cdot (3 \cdot 0 - 0 \cdot 1) - 2 \cdot (-1 \cdot 0 - 4 \cdot 1) + 4 \cdot (-1 \cdot 1 - 3 \cdot 4) \]

3. **Calculate each term**:
- First term: \( 1 \cdot 0 = 0 \)
- Second term: \( -2 \cdot (-1 \cdot 0 - 4 \cdot 1) = -2 \cdot (-4) = 8 \)
- Third term: \( 4 \cdot (-1 - 12) = 4 \cdot (-13) = -52 \)

4. **Combine the results**:
\[ D = 0 + 8 - 52 = -52 \]

### Final Result
Thus, the value of the determinant is:
\[ \boxed{-52} \]