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chrisplusian

  • one year ago

Another Linear algebra question.... see attachments please

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  1. chrisplusian
    • one year ago
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  2. chrisplusian
    • one year ago
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  3. chrisplusian
    • one year ago
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    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.

  4. anonymous
    • one year ago
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    it is always true (matrices or whatever) that \[(AB)^{-1}=B^{-1}A^{-1}\]

  5. chrisplusian
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    so \[B(AB)^{-1}=BB^{-1}A^{-1}=A^{-1}\]

  7. chrisplusian
    • one year ago
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    I have that on my work

  8. chrisplusian
    • one year ago
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    Did I do something wrong?

  9. anonymous
    • one year ago
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    ok what are you trying to check?

  10. chrisplusian
    • one year ago
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    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?

  11. anonymous
    • one year ago
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    you do not know what A is right? so there is no way to find \(A^{-1}\) other than to compute \[B(AB)^{-1}\]

  12. chrisplusian
    • one year ago
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    That's what I did

  13. anonymous
    • one year ago
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    i guess i am confused how are you trying to check your answer?

  14. chrisplusian
    • one year ago
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    Are you able to see the picture I uploaded?

  15. anonymous
    • one year ago
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    yeah i checked your first multiplication, it is correct

  16. anonymous
    • one year ago
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    OOOH i see the mistake

  17. chrisplusian
    • one year ago
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    I checked by multiplying B^(-1)A^(-1) to see if I got back (AB)^-1

  18. anonymous
    • one year ago
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    your determinant is wrong

  19. chrisplusian
    • one year ago
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    Which one?

  20. anonymous
    • one year ago
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    should be \(16-28\) not \(16-14\)

  21. chrisplusian
    • one year ago
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    wow

  22. chrisplusian
    • one year ago
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    Your right....

  23. anonymous
    • one year ago
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    actually no, i am wrong

  24. anonymous
    • one year ago
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    it is the determinant of the original matrix hold on

  25. anonymous
    • one year ago
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    lol no i am right

  26. anonymous
    • one year ago
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    that is the determinant of the original matrix sorry, still should be \(16-28\)

  27. chrisplusian
    • one year ago
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    Is what I did to check correct? (Not the numbers, I know they are wrong, but is the method correct?)

  28. anonymous
    • one year ago
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    yeah that should work you are checking that \[(AB)^{-1}=A^{-1}B^{-1}\] it should work out

  29. chrisplusian
    • one year ago
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    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?

  30. chrisplusian
    • one year ago
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    All I know is that if (AB)^-1 exists than B^-1 and A^-1 should exist right?

  31. anonymous
    • one year ago
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    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

  32. anonymous
    • one year ago
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    yes you are right

  33. chrisplusian
    • one year ago
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    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?

  34. anonymous
    • one year ago
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    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

  35. anonymous
    • one year ago
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    to answer your question, yes, but if you are told the product has an inverse, then so must the factors

  36. chrisplusian
    • one year ago
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    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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