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anonymous
 5 years ago
Turing????
anonymous
 5 years ago
Turing????

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am getting really confused

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3I hope I can help...

saifoo.khan
 5 years ago
Best ResponseYou've already chosen the best response.0Chatting here? Use chat please. :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL I need to find an inverse of a matrix

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3I anticipate a question officer ;)

saifoo.khan
 5 years ago
Best ResponseYou've already chosen the best response.0As soon as i came, he started studying!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I learnt one way that you use kind of gauss elimination

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327265979609:dw like this

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3Well 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait a sec hold ur horses

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3do you want to do the one you have with the gaussjordan thing? cuz the other way is faster and more reliable...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ik but i gotta use that way and i am really solving a 3 by 3 matrix

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3what do you mean you are really solving a 3x3 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the problem I am doing now is a 3by 3 matrix

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3oh, so you want to work out the inverse of the 3x3 with the same method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327266165384:dw

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3well post it, though it's a pain to write out...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well this is the method right?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3yeah, that's the same method as before

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok well now I have another method and i wld like to know th edifference

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://stattrek.com/matrixalgebra/howtofindinverse.aspx This is the other method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is so similar to the first one. how is it different?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3Oh 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is like bothering me and I cant figure it out

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3Imagine 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\[(E_1E_2...E_nA)A^{1}=I(E_nE_{n1}...E_1)\]the first group of terms is Identity matrix, because that's what we are supposing we do in your method, so we have\[IA^{1}=I(E_nE_{n1}...E_1)\]\[A^{1}=E_nE_{n1}...E_1\]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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k wait let me ask u what is an e matrice?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have a sneaky feeling i am doing something wrong here

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3it just represents one of the transformations that you do in order to get roweschelon form. Look at the matrix in the exampledw:1327267298132:dwnow we add 2R1 to R2 for both this and the identity matrix, just as we would do with your earlier method...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But we r doing the same thing with the other method IDK i am sooo confused

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3dw:1327267397287:dwthe only difference here is that now we are going to stop and take away our altered identity matrix and call it E1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do u have twidlla or aomething like taht?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3dw:1327267501628:dwnow you can check that multiplying\[E_1A\]will be the exact same effect as using 2R1+R2

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3yeah sure, this seems to be lagging, post a link to twiddla if you want

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol i dont have one do u ?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3Do 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol maybe i will have to reread it:D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have skype and tehn we cld share screens

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3Yeah, I did explain it rather clearly I think.. Look above, the first step in your old way produces a matrixdw:1327267836351:dwnow stop and call this new matrix E1...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok is e1 only the right side or the whole matrix

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3Each row operation makes a new matrix we call E from the identity matrix on the rightdw:1327267931741:dwcheck that matrix multiplication of (E1)A=A2 in other words multiplying by the changed identity matrix is the same as the row operation.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3typo* that should be 'check that (E1)A1=A2' above

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okkkk got that so far

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3now we stop and write a new identity matrix instead of continuing on the same one as before that is the main difference heredw:1327268217059:dwand 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

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3dw:1327268423853:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wtf omg u clarified it for me

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3hooray! that's all I could ask for :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL my book never explained that difference taht u sart a new matrix every time HAHA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh man I am so excited Thanks

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3That's why you need people to communicate with. Books can't cater to people very well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have been trying to tackle this since yesterday evening LOL and noon ecld understand me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ya i am studying online so I am having difficulties

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks turing u r awesome :DDDDD

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.3anytime, thanks for refreshing my memory on this method :)
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