# Linear algebra proof

1. Feb 13, 2013

### Mdhiggenz

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

Show that if the det(A)=1

Given Goal

if det(A)=1 then det(A-1)=1

Not sure if this proof is correct, TA explained it rather quickly so im running on pure memory.

I'm not sure about a few things

1.Why can we assume that det(A

2. why can we use the opposite relation to the equation A=adj(A-1)/det(A-1)

That's where I get a bit lost on the logic.

thanks

2. Relevant equations

3. The attempt at a solution

2. Feb 13, 2013

### Karnage1993

1) ??

2) You start off with $\displaystyle A^{-1} = \frac{adj(A)}{det(A)}$

Let the matrix $\displaystyle B = A^{-1}$. Plug B into the above formula and you have

$\displaystyle B^{-1} = \frac{adj(B)}{det(B)}$

$B$ is just $A^{-1}$ so

$\displaystyle (A^{-1})^{-1} = \frac{adj(A^{-1})}{det(A^{-1})}$, and $(A^{-1})^{-1}$ is simply $A$. Therefore, $\displaystyle A = \frac{adj(A^{-1})}{det(A^{-1})}$

3. Feb 13, 2013

### Mdhiggenz

sorry i did not know that 1 got cut off I meant why can we assume that the det(A)=det(ainverse)

Also would you agree that my proof is correct?

4. Feb 13, 2013

### Karnage1993

In this case, the two determinants are the same. But in general, they are not. There's a relation between the determinant of $A$ and $A^{-1}$, and I'll let you figure out what that is. You have probably seen it already.

You'll understand your proof more once you figure out that.

5. Feb 13, 2013

### Mdhiggenz

Only relationship I know that involves A and Ainverse is that (A-1)-1 =A

is that the path you're leading me to?

6. Feb 13, 2013

### jbunniii

det(AB) = det(A)det(B). What does this tell you if you put B = A-1?

7. Feb 13, 2013

### Mdhiggenz

We would get det(A*A^-1) which equals 1

8. Feb 13, 2013

### Karnage1993

Yes, it equals 1 but more importantly, you can split the det(A*A^-1) as jubunniii suggested.

9. Feb 13, 2013

### Mdhiggenz

Where would that relationship help with the proof?

Right now I know det(A)=1

From 1 and 2 I get A-1=adj(A)

Here is where I get completely lost.

10. Feb 13, 2013

### Karnage1993

Try manipulating the A^-1 and A formulas involving the adjoint in note #2 of the 2nd post in this thread. Then, take the adjoint of both sides of one formula and use the other formula to conclude that adj(adj(A)) = A. Remember that you're given det(A) = 1 which makes both formulas simple.

11. Feb 13, 2013

### Mdhiggenz

I figured it out, it was pretty obvious haha what had to be done. Thanks