chrisplusian
  • chrisplusian
Another Linear algebra question.... see attachments please
Mathematics
katieb
  • katieb
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chrisplusian
  • chrisplusian
1 Attachment
chrisplusian
  • chrisplusian
1 Attachment
chrisplusian
  • chrisplusian
First one is the question, the second is what I did, but I don't feel like I have accomplished anything, or if I did I am not sure if I showed it properly.

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anonymous
  • anonymous
it is always true (matrices or whatever) that \[(AB)^{-1}=B^{-1}A^{-1}\]
chrisplusian
  • chrisplusian
ok
anonymous
  • anonymous
so \[B(AB)^{-1}=BB^{-1}A^{-1}=A^{-1}\]
chrisplusian
  • chrisplusian
I have that on my work
chrisplusian
  • chrisplusian
Did I do something wrong?
anonymous
  • anonymous
ok what are you trying to check?
chrisplusian
  • chrisplusian
I went to check to see if the A^-1 I got was correct and the work I did doesn't seem to match. Could you look at the document I uploaded?
anonymous
  • anonymous
you do not know what A is right? so there is no way to find \(A^{-1}\) other than to compute \[B(AB)^{-1}\]
chrisplusian
  • chrisplusian
That's what I did
anonymous
  • anonymous
i guess i am confused how are you trying to check your answer?
chrisplusian
  • chrisplusian
Are you able to see the picture I uploaded?
anonymous
  • anonymous
yeah i checked your first multiplication, it is correct
anonymous
  • anonymous
OOOH i see the mistake
chrisplusian
  • chrisplusian
I checked by multiplying B^(-1)A^(-1) to see if I got back (AB)^-1
anonymous
  • anonymous
your determinant is wrong
chrisplusian
  • chrisplusian
Which one?
anonymous
  • anonymous
should be \(16-28\) not \(16-14\)
chrisplusian
  • chrisplusian
wow
chrisplusian
  • chrisplusian
Your right....
anonymous
  • anonymous
actually no, i am wrong
anonymous
  • anonymous
it is the determinant of the original matrix hold on
anonymous
  • anonymous
lol no i am right
anonymous
  • anonymous
that is the determinant of the original matrix sorry, still should be \(16-28\)
chrisplusian
  • chrisplusian
Is what I did to check correct? (Not the numbers, I know they are wrong, but is the method correct?)
anonymous
  • anonymous
yeah that should work you are checking that \[(AB)^{-1}=A^{-1}B^{-1}\] it should work out
chrisplusian
  • chrisplusian
And The question kind of confused me.... they ask me to find the inverse of A "If it is Possible". My Professor is big on Proofs. How do I show this "IS" possible?
chrisplusian
  • chrisplusian
All I know is that if (AB)^-1 exists than B^-1 and A^-1 should exist right?
anonymous
  • anonymous
i don't think you need to bother to check the "if possible" business is because it is always possible that the inverse does not exist (if the determinant is zero) but in this case it does exist
anonymous
  • anonymous
yes you are right
chrisplusian
  • chrisplusian
Let me ask you this.... the only example I have been given to check if an inverse exists is in 2D..... would I check to see if the inverse of a 3x3 exists in the same way? Take the determinant of the 3x3, and be sure it doesn't equal 0? And if that is correct for a 3x3 is it correct for any mxn matrix as well?
anonymous
  • anonymous
you are told that \((AB)^{-1}\) exists, that means both \(A^{-1}\) and \(B^{-1}\) exist if either one had determinant zero, then so would the product
anonymous
  • anonymous
to answer your question, yes, but if you are told the product has an inverse, then so must the factors
chrisplusian
  • chrisplusian
I get that, but I just wanted to know if the concept of a non-zero determinant implying the inverse exists is true in general for any size matrix/

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