Is $\det(ABA ^{−1}) = \det B$?

Question

Suppose $A$ and $B$ are $n \times n$ matrices with $A$ invertible. Prove that $$\det (ABA^{−1}) = \det B$$

My Answer:

\begin{align}\det(ABA^{-1}) &= \det(B) \det (A)\det (A^{-1})\\ &= \det(B) \det(A\cdot A^{-1})\\ &=\det(B) \det(I_n)\\ &=\det(BI_n)\\ &=\det(B)\\ \end{align}

(The -1 is to the power -1 , i wasnt sure how to format it)

Answer 1

Alternatively, recognize that $\det(A^{-1})=\frac1{\det(A)}$.

Hence $$\det(ABA^{-1})=\det(A)\det(B)\det(A^{-1})=\det(A)\det(B)\frac1{\det(A)}=\det(B).$$

Edit:

Since $\det(A\cdot A^{-1})=\det(I_n)=1$ implies that $\det(A) \cdot \det(A^{-1})=1.$