How Does Invertibility Affect the Determinant in Matrix Operations?

[RWalden21]

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.

 

[rock.freak667]

Right so how you would expand de(BAB^-1) ? And how does det(B) relate to det(B^-1)?

 

[RWalden21]

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

 

[rock.freak667]

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

Hint: B^-1B= I so if you find then the determinant of I should be the same as the determinant of B^-1B right?

 

[RWalden21]

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

 

[rock.freak667]

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

Yep, that will work. So now you know that det(B^-1) = 1/det(B), which may help you somewhere down the road.

 

[dimension10]

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.

Didn't you realize you've like already done the proof? You know that if $\mathbf A,\;\mathbf B\in\mathbb R$, 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?

Edit: Why is the $\LaTeX$ equation not rendering?
Edit2: Ok, now it renders. I just realized there is an even simpler method though. You could just break up the determinant into three parts and the join two of them and you are done.

 

[Mark44]

Didn't you realize you've like already done the proof? You know that if $\mathbf A,\;\mathbf B\in\mathbb R$

A and B aren't real numbers; they're matrices.

, 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?

 

[dimension10]

A and B aren't real numbers; they're matrices.

Sorry, I meant $\mathbf{A},\;\mathbf{B}\in\mathbb R^{n\times n}$