After applying elementary transformation `R_(1) - 3R_(2)` on matrix `[{:(3,-2),(1,4):}]` we get `[{:(0,-12),(1,4):}]`. True or False.

Answer

To determine whether the statement is true or false, we will apply the elementary transformation \( R_1 - 3R_2 \) on the given matrix \( \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Identify the Rows**:
- Let \( R_1 = \begin{pmatrix} 3 & -2 \end{pmatrix} \)
- Let \( R_2 = \begin{pmatrix} 1 & 4 \end{pmatrix} \)

2. **Calculate \( 3R_2 \)**:
- Multiply \( R_2 \) by 3:
\[ 3R_2 = 3 \times \begin{pmatrix} 1 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 12 \end{pmatrix} \]

3. **Perform the Transformation**:
- Now apply the transformation \( R_1 - 3R_2 \):
\[ R_1 - 3R_2 = \begin{pmatrix} 3 & -2 \end{pmatrix} - \begin{pmatrix} 3 & 12 \end{pmatrix} \]
- This results in:
\[ R_1 - 3R_2 = \begin{pmatrix} 3 - 3 & -2 - 12 \end{pmatrix} = \begin{pmatrix} 0 & -14 \end{pmatrix} \]

4. **Construct the New Matrix**:
- After the transformation, the new matrix becomes:
\[ \begin{pmatrix} 0 & -14 \\ 1 & 4 \end{pmatrix} \]

5. **Compare with the Given Matrix**:
- The given matrix after transformation is \( \begin{pmatrix} 0 & -12 \\ 1 & 4 \end{pmatrix} \).
- Our calculated matrix is \( \begin{pmatrix} 0 & -14 \\ 1 & 4 \end{pmatrix} \).

6. **Conclusion**:
- Since \( -14 \neq -12 \), the statement is **False**.