$AA^{-1} =I $ and $A^{-1}=E_{1}\dots E_{n}A$ does this prove that $\det(E_{1}\dots E_{n}) = \det(1/A^{2})$

Question

I was answering a question on a test and I happened by this.

$AA^{-1} = I$

and $A^{-1} = E_{1}\dots E_{n}A$

from the first I know that $\det(A^{-1})= \det(1/A)$

from the second I know that $\det(A^{-1}) = \det(E_{1}\dots E_{n})\cdot\det(A)$

Have i made a mistake or does this mean that the elementary row operation required to change $A$ to $A^{-1}$ have the determinat equal to $\det(1/A^{2})$