Another Linear algebra question.... see attachments please

- chrisplusian

Another Linear algebra question.... see attachments please

- katieb

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- chrisplusian

##### 1 Attachment

- chrisplusian

##### 1 Attachment

- 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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## More answers

- anonymous

it is always true (matrices or whatever) that \[(AB)^{-1}=B^{-1}A^{-1}\]

- chrisplusian

ok

- anonymous

so \[B(AB)^{-1}=BB^{-1}A^{-1}=A^{-1}\]

- chrisplusian

I have that on my work

- chrisplusian

Did I do something wrong?

- anonymous

ok what are you trying to check?

- 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

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

That's what I did

- anonymous

i guess i am confused
how are you trying to check your answer?

- chrisplusian

Are you able to see the picture I uploaded?

- anonymous

yeah
i checked your first multiplication, it is correct

- anonymous

OOOH i see the mistake

- chrisplusian

I checked by multiplying B^(-1)A^(-1) to see if I got back (AB)^-1

- anonymous

your determinant is wrong

- chrisplusian

Which one?

- anonymous

should be \(16-28\) not \(16-14\)

- chrisplusian

wow

- chrisplusian

Your right....

- anonymous

actually no, i am wrong

- anonymous

it is the determinant of the original matrix hold on

- anonymous

lol no i am right

- anonymous

that is the determinant of the original matrix sorry, still should be \(16-28\)

- chrisplusian

Is what I did to check correct? (Not the numbers, I know they are wrong, but is the method correct?)

- anonymous

yeah that should work
you are checking that
\[(AB)^{-1}=A^{-1}B^{-1}\] it should work out

- 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

All I know is that if (AB)^-1 exists than B^-1 and A^-1 should exist right?

- 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

yes you are right

- 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

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

to answer your question, yes, but if you are told the product has an inverse, then so must the factors

- 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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