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 $-216$.


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