I am given the determinant of matrix $A$ and matrix $B$. Both are $3 \times 3$ matrices.
- $\det(A) = -5$
- $\det(B) = 5$
I need to find the determinant of $D = 6A^{-1}B^T$.
I calculated the determinant as follows:
$$ \begin{align} \det(D) & = \det(6A^{-1}B^T)\\ & = \det(6A^{-1}) \cdot \det(B^T)\\ & = 6 \cdot \det(A^{-1}) \cdot \det(B^T)\\ & = 6 \cdot \frac{1}{\det(A)} \cdot \det(B^T)\\ & = 6 \cdot \frac{1}{-5} \cdot (5)\\ & = -6 \end{align} $$
My guess is that the scalar multiplication within the determinant should be raised to the nth power, where n is the size of the n x n matrix.
So the correct answer should be -216?
Since $A^{-1}$ is a $3\times3$ matrix, $\det(6A^{-1})=6^3\det(A^{-1})$. Therefore, the answer is $-216$.