RWalden21 said: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
RWalden21 said:is this correct?
det(BAB^-1)= det(B)det(A)det(B^-1)= det(BB^-1)det(A)= det(I)det(A)= det(A)
det(I) = 1
RWalden21 said: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) to just try to get det(A) but I'm just not sure how to do the proof.
A and B aren't real numbers; they're matrices.dimension10 said:Didn't you realize you've like already done the proof? You know that if ##\mathbf A,\;\mathbf B\in\mathbb R##
dimension10 said:, then ##\det\mathbf{AB}=\det\mathbf A\det\mathbf B## and ##\det\left(\mathbf{A}^{-1}\right)=\det\left(\mathbf{A}\right)^{-1}##. So if you just simplify ##\det\left(\mathbf{BAB^{-1}}\right)##, what do you get?
Mark44 said:A and B aren't real numbers; they're matrices.
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