# Homework Help: Linear algebra determinant proof

1. Feb 13, 2013

### Mdhiggenz

1. The problem statement, all variables and given/known data

Let A and B be nxn matrices. Prove that if AB=I, then BA=I

det(AB)=det(I)

1.det(A)*det(B)=1

det(BA)=det(I)

2.det(B)*det(A)=1

Equating 1 and two together I get det(B)*det(A)=det(A)*det(B)

Thus AB=I, then BA=I

Is this correct?

2. Relevant equations

3. The attempt at a solution

2. Feb 13, 2013

### Karnage1993

What you showed is that the $det(AB) = det(BA)$ but that doesn't really imply that $AB = BA$. If you want to show that $AB = BA$, you need to use the properties of inverses and facts about them.

What you did is a start, though. Since $det(AB) = det(A)det(B) = 1$, that means neither $det(A)$ or $det(B)$ are 0. Ie, A and B are non-singular and have inverses. Now all you need to do is show that $A$ and $B$ are inverses of each other.

Last edited: Feb 13, 2013
3. Feb 13, 2013

### Mdhiggenz

Could I say that if B=A-1 then

det(AB)=det(A*A-1)=1

?

4. Feb 13, 2013

### micromass

Sure, that is correct if $B=A^{-1}$. The problem is that you don't know that $B=A^{-1}$. You need to prove that.

5. Feb 13, 2013

### Mdhiggenz

Would it be enough if I maybe do something like this

assume A-1=B

then A*B=I == A*A-1=I
I=I

Which would make it true?

Because if B=A-1 then A*B=I would be true

6. Feb 13, 2013

### vela

Staff Emeritus
No, you can't start with an assumption, get a true result, and then claim that proves the assumption is true. It's possible to start with a false assumption and get a true result.

7. Feb 13, 2013

### Mdhiggenz

What you're saying makes sense but i'm a bit unsure on how to use a false assumption in this case. For instance if I assume that B~=A-1 what can I do with that?

8. Feb 13, 2013

### vela

Staff Emeritus
The point is you can't assume B=A-1. That's what you're trying to prove.

9. Feb 13, 2013

### Mdhiggenz

That's where i'm stuck, How do I prove the assumption I made without using another assumption? Can you help me get started in the direction of proving B=A-1 so I dont run in circles.

10. Feb 13, 2013

### Karnage1993

It's very simple. Start with your "If" statement and somehow isolate $B$. Remember that what you are given is assumed to be true.

11. Feb 13, 2013

### vela

Staff Emeritus
Let's back up for a second. You know what AB=I because that's a given. At this point, you don't know anything about A and B other than the fact that their product AB is the identity matrix. In particular, you don't know if A has an inverse or B has an inverse.

You then took the determinant of both sides and got to det(A)det(B)=1. Now det(A)det(B)=1 implies that det(A)≠0. See what Karnage1993 said up in post 2. det(A)≠0 tells you A does indeed have an inverse. So now you know that AB=I and A-1 exists. Can you use this knowledge to solve for B?

12. Feb 13, 2013

### Mdhiggenz

So the if statement is if AB=I

then A-1*AB=A-1I

Which is equal to B=A-1

13. Feb 13, 2013

### Karnage1993

Yes, that's correct, but you can only multiply both sides by $A^{-1}$ once you know for a fact that $A$ does indeed have an inverse, which we already went over above.

14. Feb 13, 2013

### Mdhiggenz

So I would have to state it like this

Since det(A)det(b)=det(I)=1
det(AB) is nonsingular meaning it has an inverse, and then I can show the proof for B=A^-1

15. Feb 13, 2013

### vela

Staff Emeritus
Not exactly. Try again. Your logic is a bit muddied still.

16. Feb 13, 2013

### Mdhiggenz

Looking at the book it says that a matrix A is termed invertable if there is a matrix B such that A*B=I.

So if i'm quoting the definition in the book technically i'm not assuming anything, just explaining the if statement. Right?