Answer to Question #115308 in Linear Algebra for matome

Question

Suppose A and B are n x n matrices with A invertible. Prove that det ABA^-1 = det B

Answer 1

matrices are not commutative. that is AB is not equal to BA

$det(BA)=det(B)det(A)\\det(BA)=det(A)det(B)\\det(BA)=det(AB)$

$BB^{-1}=I \implies det(I)=1$

hence to show $det (ABA^{-1}) = det( B)$

$\implies det (ABA^{-1}) =det(A)det(B)det(A^{-1})\\det (ABA^{-1}) =det(A)det(A^{-1})det(B)\\ det (ABA^{-1})=(det(A)det(A^{-1}))det(B)\\det (ABA^{-1})=(det(AA^{-1}))det(B)$

but $AA^{-1}=I$

$det (ABA^{-1})=(det(I))det(B)\\$ $det(I)=1$

hence:

$det (ABA^{-1})=det(B)$