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chrisplusian
 one year ago
Another Linear algebra question.... see attachments please
chrisplusian
 one year ago
Another Linear algebra question.... see attachments please

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chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0First 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is always true (matrices or whatever) that \[(AB)^{1}=B^{1}A^{1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[B(AB)^{1}=BB^{1}A^{1}=A^{1}\]

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0I have that on my work

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0Did I do something wrong?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok what are you trying to check?

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0you do not know what A is right? so there is no way to find \(A^{1}\) other than to compute \[B(AB)^{1}\]

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0That's what I did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess i am confused how are you trying to check your answer?

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0Are you able to see the picture I uploaded?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i checked your first multiplication, it is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OOOH i see the mistake

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0I checked by multiplying B^(1)A^(1) to see if I got back (AB)^1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your determinant is wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should be \(1628\) not \(1614\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually no, i am wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is the determinant of the original matrix hold on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is the determinant of the original matrix sorry, still should be \(1628\)

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0Is what I did to check correct? (Not the numbers, I know they are wrong, but is the method correct?)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that should work you are checking that \[(AB)^{1}=A^{1}B^{1}\] it should work out

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0And 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
 one year ago
Best ResponseYou've already chosen the best response.0All I know is that if (AB)^1 exists than B^1 and A^1 should exist right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0Let 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
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0to answer your question, yes, but if you are told the product has an inverse, then so must the factors

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0I get that, but I just wanted to know if the concept of a nonzero determinant implying the inverse exists is true in general for any size matrix/
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