To solve this problem, let's analyze the information given:

1. We are provided with the determinant of a matrix: $\begin{pmatrix} 1 & 3 & 0 \\ 2 & 6 & 4 \\ -1 & 0 & 2 \end{pmatrix}$ This determinant is given as $-12$.

2. We need to determine the determinant of the matrix: $\begin{pmatrix} 2 & 6 & 0 \\ 4 & 12 & 8 \\ -2 & 0 & 4 \end{pmatrix}$

Step-by-Step Solution:

Notice that the second matrix is a scalar multiple of the first matrix. Specifically, each element in the second matrix is twice the corresponding element in the first matrix. Thus, the second matrix can be represented as: $2 \cdot \begin{pmatrix} 1 & 3 & 0 \\ 2 & 6 & 4 \\ -1 & 0 & 2 \end{pmatrix}$

Determinant Scaling Property

When a matrix is scaled by a factor, $k$, the determinant of the matrix is scaled by $k^n$, where $n$ is the order of the matrix. Since this is a $3 \times 3$ matrix, $n = 3$, so the determinant of the second matrix is: $2^3 \times \text{determinant of the first matrix} = 8 \times (-12) = -96$

Conclusion:

The determinant of the matrix $\begin{pmatrix} 2 & 6 & 0 \\ 4 & 12 & 8 \\ -2 & 0 & 4 \end{pmatrix}$ is $-96$.

Answer:

The correct choice is (a) $-96$.

Would you like more details on matrix properties or additional help? Here are some related questions to deepen your understanding:

1. How does the determinant change if each row of a matrix is multiplied by a different scalar?
2. What is the effect on the determinant if only one row of a matrix is scaled?
3. How can determinants help in finding the inverse of a matrix?
4. What is the significance of a zero determinant in a matrix?
5. How do you calculate the determinant of larger matrices, such as $4 \times 4$?

Tip: Remember, multiplying each entry of a $3 \times 3$ matrix by a scalar multiplies the determinant by that scalar raised to the power of three.