Linear algebra proof

  • Thread starter Mdhiggenz
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  • #1
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Homework Statement



Show that if the det(A)=1
then adj(adj(A))=A

Given Goal
det(A)=1 adj(adj(A))=A

Using the following formula A-1=adj(A)/det(A)

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

likewise A=adj(A-1)/det(A-1)

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

Thus A=adj(A-1)
A=adj(A-1)=adj(A)adj=adj(adj(A))

What i'm confused about is

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

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
133
1
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

##\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
327
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?
 
  • #4
133
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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
327
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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
jbunniii
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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?
 
  • #7
327
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We would get det(A*A^-1) which equals 1
 
  • #8
133
1
Yes, it equals 1 but more importantly, you can split the det(A*A^-1) as jubunniii suggested.
 
  • #9
327
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Still pretty confused on how that would show that the adj(adj(A))=A

Where would that relationship help with the proof?

Right now I know det(A)=1

1.A-1=adj(A)/det(A)

and 2. A=adj(A-1)/det(A-1)

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

and A=adj(A-1)

Here is where I get completely lost.
 
  • #10
133
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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
327
1
I figured it out, it was pretty obvious haha what had to be done. Thanks
 

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