$\det(P^{-1}AP) = \det(A)$ for any invertible $P$ and square matrix $A$

$\det(P^{-1}AP) = \det(P^{-1}) \cdot \det(A) \cdot \det(P)$

If $\det(A) = 3$, then $\det(P^{-1}AP) = 3$

The determinant is invariant under similarity transformations