I suppose you can expand by det(B)det(A)det(B^-1) but I don't know how det(B) relates to det(B^-1). What I was thinking was by expanding and somehow getting B and B^-1 to = I somehow det(A) would remain
Homework Statement
Let A and B be square matrices, with B invertible. Show that det(BAB^-1) = det(A)
Homework Equations
I think its based off the theorem: If A and B are nxn matrices, then det(AB)= det(A)det(B)
The Attempt at a Solution
I started by simplifying BAB^-1det(A)...