# Linear algebra proof

## Homework Statement

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

## The Attempt at a Solution

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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)
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

##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})}##

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?

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.

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

is that the path you're leading me to?

jbunniii
Homework Helper
Gold Member
Only relationship I know that involves A and Ainverse is that (A-1)-1 =A

is that the path you're leading me to?
det(AB) = det(A)det(B). What does this tell you if you put B = A-1?

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

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

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)