Let $A$ be an $n × n$ matrix and let $P$ be an $n × n$ invertible matrix. Prove that $\det( P^{-1} AP) = \det(A)$
Pretty lost on this one, partially because I don't understand the relationship between the determinant of a matrix and the determinant of its inverse. Thanks.
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$\begingroup$Hint: $\det(AB)=\det(A)\det(B)$. Apply this to $PP^{-1}=I$ and to $P^{-1}AP$
$\endgroup$ $\begingroup$I don't know how much you know here, but if you know the determinants are multiplicative, that is $\det(AB) = \det(A)\det(B)$, then you can do the following:
$$\det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P)=\det(P^{-1})\det(P)\det(A)=\det(P^{-1}P)\det(A).$$
Now you are left with showing that $\det(P^{-1}P)=1$ and you are done.
If you don't know that $\det(AB)=\det(A)\det(B)$, then here's a proof.
$\endgroup$ $\begingroup$\begin{align}\det(P^{-1}AP)&=\det(P^{-1})\det(A)\det(P) \\&= \det(P^{-1})\det(P)\det(A) \\&= \det(P^{-1}P)\det(A)\\&= \det(I)\det(A)\\& = 1\det(A)\\&=\det(A)\end{align}
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