Let \(A, B, C\) be three \(4 \times 4\) matrices such that \(det \ A = 5, det \ B = -3\), and \(det \ C = \frac{1}{2}\). Then the \(det\) \(2AB^{-1}C^3B^T\) is

The determinant is $10$. The inverse and transpose of $B$ contribute reciprocal factors, so the product is $2^4\det(A)(\det C)^3=16\cdot5\cdot(\frac12)^3$. Using the square of $\det C$ in that calculation gives $20$.