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anonymous

  • 5 years ago

Turing????

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  1. anonymous
    • 5 years ago
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    Hey

  2. TuringTest
    • 5 years ago
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    who wassup?

  3. TuringTest
    • 5 years ago
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    whoa*

  4. anonymous
    • 5 years ago
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    I am getting really confused

  5. TuringTest
    • 5 years ago
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    I hope I can help...

  6. saifoo.khan
    • 5 years ago
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    Chatting here? Use chat please. :D

  7. anonymous
    • 5 years ago
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    LOL I need to find an inverse of a matrix

  8. TuringTest
    • 5 years ago
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    I anticipate a question officer ;)

  9. saifoo.khan
    • 5 years ago
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    As soon as i came, he started studying!!!

  10. anonymous
    • 5 years ago
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    I learnt one way that you use kind of gauss elimination

  11. anonymous
    • 5 years ago
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    |dw:1327265979609:dw| like this

  12. TuringTest
    • 5 years ago
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    Well there are quite a few ways to do that. I know you know at least one and I sent you a link on more. http://tutorial.math.lamar.edu/Classes/LinAlg/FindingInverseMatrices.aspx for instance theorem 4 on this page is what FoolForMath was going to use

  13. anonymous
    • 5 years ago
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    wait a sec hold ur horses

  14. TuringTest
    • 5 years ago
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    do you want to do the one you have with the gauss-jordan thing? cuz the other way is faster and more reliable...

  15. TuringTest
    • 5 years ago
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    ...for 2x2 at least

  16. anonymous
    • 5 years ago
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    ik but i gotta use that way and i am really solving a 3 by 3 matrix

  17. TuringTest
    • 5 years ago
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    what do you mean you are really solving a 3x3 ?

  18. anonymous
    • 5 years ago
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    the problem I am doing now is a 3by 3 matrix

  19. TuringTest
    • 5 years ago
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    oh, so you want to work out the inverse of the 3x3 with the same method?

  20. anonymous
    • 5 years ago
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    |dw:1327266165384:dw|

  21. TuringTest
    • 5 years ago
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    well post it, though it's a pain to write out...

  22. anonymous
    • 5 years ago
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    Well this is the method right?

  23. TuringTest
    • 5 years ago
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    yeah, that's the same method as before

  24. anonymous
    • 5 years ago
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    ok well now I have another method and i wld like to know th edifference

  25. anonymous
    • 5 years ago
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    http://stattrek.com/matrix-algebra/how-to-find-inverse.aspx This is the other method

  26. anonymous
    • 5 years ago
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    It is so similar to the first one. how is it different?

  27. TuringTest
    • 5 years ago
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    Oh I hate this stupid way, I think it works best for computer programs. Conceptually it winds up being the exact same thing: using transformations to an identity matrix to tell you how to find the inverse of the original. This one just formalizes all the steps at the end into a big matrix product (all the E's). Since each matrix in the product transforms a matrix just like a step in your method, this amounts to the same thing, but allows you to write each step as a separate matrix. Like I say, I think it works best for programming the inverse of large matrices.

  28. anonymous
    • 5 years ago
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    but wait isnt the last step the inverse. Like in the first method it was and in the seond methos it isnt. What are we doing different

  29. anonymous
    • 5 years ago
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    It is like bothering me and I cant figure it out

  30. TuringTest
    • 5 years ago
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    Imagine you are doing your inverse the old way each step is the same as multiplying by one of those E matrices One way of formalizing the first method you learned is as\[AA^{-1}=I\]now let each step be a matrix E_n. We have to multiply both sides\[(E_1E_2...E_nA)A^{-1}=I(E_nE_{n-1}...E_1)\]the first group of terms is Identity matrix, because that's what we are supposing we do in your method, so we have\[IA^{-1}=I(E_nE_{n-1}...E_1)\]\[A^{-1}=E_nE_{n-1}...E_1\]I hope that demonstrates that they are essentially the same method if you look closely to what we did. The only difference is representing each step in the transformation with a matrix E_n

  31. anonymous
    • 5 years ago
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    k wait let me ask u what is an e matrice?

  32. anonymous
    • 5 years ago
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    I have a sneaky feeling i am doing something wrong here

  33. TuringTest
    • 5 years ago
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    it just represents one of the transformations that you do in order to get row-eschelon form. Look at the matrix in the example|dw:1327267298132:dw|now we add -2R1 to R2 for both this and the identity matrix, just as we would do with your earlier method...

  34. anonymous
    • 5 years ago
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    But we r doing the same thing with the other method IDK i am sooo confused

  35. TuringTest
    • 5 years ago
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    |dw:1327267397287:dw|the only difference here is that now we are going to stop and take away our altered identity matrix and call it E1

  36. anonymous
    • 5 years ago
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    do u have twidlla or aomething like taht?

  37. TuringTest
    • 5 years ago
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    |dw:1327267501628:dw|now you can check that multiplying\[E_1A\]will be the exact same effect as using -2R1+R2

  38. TuringTest
    • 5 years ago
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    yeah sure, this seems to be lagging, post a link to twiddla if you want

  39. anonymous
    • 5 years ago
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    lol i dont have one do u ?

  40. TuringTest
    • 5 years ago
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    no...

  41. TuringTest
    • 5 years ago
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    Do you see what I'm saying though? Try multiplying the matrices (E_1)A and you will see it has the same effect as the row operation we did. So they are the exact same thing.

  42. anonymous
    • 5 years ago
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    lol maybe i will have to reread it:D

  43. anonymous
    • 5 years ago
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    I have skype and tehn we cld share screens

  44. anonymous
    • 5 years ago
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    LOL U dont have too

  45. TuringTest
    • 5 years ago
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    Yeah, I did explain it rather clearly I think.. Look above, the first step in your old way produces a matrix|dw:1327267836351:dw|now stop and call this new matrix E1...

  46. anonymous
    • 5 years ago
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    ok is e1 only the right side or the whole matrix

  47. TuringTest
    • 5 years ago
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    Each row operation makes a new matrix we call E from the identity matrix on the right|dw:1327267931741:dw|check that matrix multiplication of (E1)A=A2 in other words multiplying by the changed identity matrix is the same as the row operation.

  48. TuringTest
    • 5 years ago
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    typo* that should be 'check that (E1)A1=A2' above

  49. anonymous
    • 5 years ago
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    okkkk got that so far

  50. TuringTest
    • 5 years ago
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    now we stop and write a new identity matrix instead of continuing on the same one as before that is the main difference here|dw:1327268217059:dw|and surprise surprise the new altered identity matrix after the last step will be called E2 again check that multiplying E2(A2)=A3 we repeat this process until we have the identity matrix on the left, each time with a new identity matrix on the right that we set aside and name after each row operation

  51. TuringTest
    • 5 years ago
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    |dw:1327268423853:dw|

  52. anonymous
    • 5 years ago
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    wtf omg u clarified it for me

  53. TuringTest
    • 5 years ago
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    hooray! that's all I could ask for :D

  54. anonymous
    • 5 years ago
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    LOL my book never explained that difference taht u sart a new matrix every time HAHA

  55. anonymous
    • 5 years ago
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    I hate my book LOL

  56. anonymous
    • 5 years ago
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    Oh man I am so excited Thanks

  57. TuringTest
    • 5 years ago
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    That's why you need people to communicate with. Books can't cater to people very well.

  58. anonymous
    • 5 years ago
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    I have been trying to tackle this since yesterday evening LOL and noon ecld understand me

  59. anonymous
    • 5 years ago
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    Ya i am studying online so I am having difficulties

  60. anonymous
    • 5 years ago
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    Thanks turing u r awesome :DDDDD

  61. TuringTest
    • 5 years ago
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    anytime, thanks for refreshing my memory on this method :)

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