## anonymous 4 years ago Turing????

1. anonymous

Hey

2. TuringTest

who wassup?

3. TuringTest

whoa*

4. anonymous

I am getting really confused

5. TuringTest

I hope I can help...

6. saifoo.khan

Chatting here? Use chat please. :D

7. anonymous

LOL I need to find an inverse of a matrix

8. TuringTest

I anticipate a question officer ;)

9. saifoo.khan

As soon as i came, he started studying!!!

10. anonymous

I learnt one way that you use kind of gauss elimination

11. anonymous

|dw:1327265979609:dw| like this

12. TuringTest

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

wait a sec hold ur horses

14. TuringTest

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

...for 2x2 at least

16. anonymous

ik but i gotta use that way and i am really solving a 3 by 3 matrix

17. TuringTest

what do you mean you are really solving a 3x3 ?

18. anonymous

the problem I am doing now is a 3by 3 matrix

19. TuringTest

oh, so you want to work out the inverse of the 3x3 with the same method?

20. anonymous

|dw:1327266165384:dw|

21. TuringTest

well post it, though it's a pain to write out...

22. anonymous

Well this is the method right?

23. TuringTest

yeah, that's the same method as before

24. anonymous

ok well now I have another method and i wld like to know th edifference

25. anonymous

http://stattrek.com/matrix-algebra/how-to-find-inverse.aspx This is the other method

26. anonymous

It is so similar to the first one. how is it different?

27. TuringTest

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

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

It is like bothering me and I cant figure it out

30. TuringTest

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

k wait let me ask u what is an e matrice?

32. anonymous

I have a sneaky feeling i am doing something wrong here

33. TuringTest

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

But we r doing the same thing with the other method IDK i am sooo confused

35. TuringTest

|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

do u have twidlla or aomething like taht?

37. TuringTest

|dw:1327267501628:dw|now you can check that multiplying\[E_1A\]will be the exact same effect as using -2R1+R2

38. TuringTest

yeah sure, this seems to be lagging, post a link to twiddla if you want

39. anonymous

lol i dont have one do u ?

40. TuringTest

no...

41. TuringTest

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

lol maybe i will have to reread it:D

43. anonymous

I have skype and tehn we cld share screens

44. anonymous

LOL U dont have too

45. TuringTest

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

ok is e1 only the right side or the whole matrix

47. TuringTest

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

typo* that should be 'check that (E1)A1=A2' above

49. anonymous

okkkk got that so far

50. TuringTest

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

|dw:1327268423853:dw|

52. anonymous

wtf omg u clarified it for me

53. TuringTest

hooray! that's all I could ask for :D

54. anonymous

LOL my book never explained that difference taht u sart a new matrix every time HAHA

55. anonymous

I hate my book LOL

56. anonymous

Oh man I am so excited Thanks

57. TuringTest

That's why you need people to communicate with. Books can't cater to people very well.

58. anonymous

I have been trying to tackle this since yesterday evening LOL and noon ecld understand me

59. anonymous

Ya i am studying online so I am having difficulties

60. anonymous

Thanks turing u r awesome :DDDDD

61. TuringTest

anytime, thanks for refreshing my memory on this method :)