To solve the problem $\det(2B^{-1}C^2A^TB)$, let’s break it down step by step.

Given:

• $\det(A) = 2$,
• $\det(B) = 6$,
• $\det(C) = 3$.

Key Properties of Determinants:

1. Scalar Multiplication: If $k$ is a scalar and $A$ is a matrix, then $\det(kA) = k^n \det(A)$, where $n$ is the dimension of $A$. Since the matrices are $3 \times 3$, $n = 3$.
2. Inverse Matrix: For an invertible matrix $B$, $\det(B^{-1}) = \frac{1}{\det(B)}$.
3. Product of Matrices: For matrices $X$ and $Y$, $\det(XY) = \det(X)\det(Y)$.
4. Transpose: For any matrix $A$, $\det(A^T) = \det(A)$.

Steps:

1. Start with the expression: $\det(2B^{-1}C^2A^TB).$

2. Break it into parts: Using the determinant property for matrix products: $\det(2B^{-1}C^2A^TB) = \det(2) \cdot \det(B^{-1}) \cdot \det(C^2) \cdot \det(A^T) \cdot \det(B).$

3. Compute each determinant:

   • $\det(2) = 2^3 = 8$ (since it’s a scalar multiplied to a $3 \times 3$ matrix).
   • $\det(B^{-1}) = \frac{1}{\det(B)} = \frac{1}{6}$.
   • $\det(C^2) = (\det(C))^2 = 3^2 = 9$.
   • $\det(A^T) = \det(A) = 2$ (property of the determinant under transpose).
   • $\det(B) = 6$ (given).

4. Substitute these values: $\det(2B^{-1}C^2A^TB) = 8 \cdot \frac{1}{6} \cdot 9 \cdot 2 \cdot 6.$

5. Simplify:

   • First, compute $8 \cdot \frac{1}{6} = \frac{8}{6} = \frac{4}{3}$,
   • Then, $\frac{4}{3} \cdot 9 = 12$,
   • Then, $12 \cdot 2 = 24$,
   • Finally, $24 \cdot 6 = 144$.

Final Answer:

The determinant $\det(2B^{-1}C^2A^TB) = 144$.

Correct Option:

b) 144

Would you like a further explanation, or do you have any questions?

Here are some related questions you might find helpful:

1. How is the determinant of a product of matrices computed?
2. Why does $\det(A^T) = \det(A)$ hold true?
3. What is the effect of a scalar on a determinant?
4. How does the determinant of an inverse matrix work?
5. Can the determinant be negative, and what does it imply?

Tip: When working with determinants, always break down the expression using determinant properties systematically for clarity.