$\det(ABC) = \det(BA^{-1}C^{T}A^{2})$

Question

Let $A$,$B$,$C$ three square matrices. Then I have to show that $$\det(ABC) = \det(BA^{-1}C^{T}A^{2})$$

I would start with the left side and proceed by rearranging and using some properties: \begin{align} \det(BA^{-1}C^{T}A^{2}) &= \det(B)\det(A^{-1})\det(C^{T})\det(A^{2}) \\ &= \det(B)\det(A)^{-1}\det(C)\det(A^{2}) \\ &= \det(A)\det(A)\det(A)^{-1}\det(B)\det(C) \\ &= \det(A)\det(B)\det(C) \end{align}

Is this proof correct?

Answer 1

Yes that is correct.

(I'm assuming that by "$*$" you mean multiplication. That ought to be notated as $\cdot$ or juxtaposition).