1. The problem asks to find $3A$ where $A$ is the matrix $\begin{bmatrix} 2 & -4 \\ 1 & 0 \end{bmatrix}$. 2. The formula for scalar multiplication of a matrix is $kA = k \times A$, where each element of matrix $A$ is multiplied by scalar $k$. 3. Multiply each element of $A$ by 3: $$3A = 3 \times \begin{bmatrix} 2 & -4 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 \times 2 & 3 \times (-4) \\ 3 \times 1 & 3 \times 0 \end{bmatrix}$$ 4. Calculate each element: $$3A = \begin{bmatrix} 6 & -12 \\ 3 & 0 \end{bmatrix}$$ 5. So, the final answer is: $$\boxed{\begin{bmatrix} 6 & -12 \\ 3 & 0 \end{bmatrix}}$$