Turing????

- anonymous

Turing????

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- schrodinger

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- anonymous

Hey

- TuringTest

who wassup?

- TuringTest

whoa*

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## More answers

- anonymous

I am getting really confused

- TuringTest

I hope I can help...

- saifoo.khan

Chatting here?
Use chat please. :D

- anonymous

LOL I need to find an inverse of a matrix

- TuringTest

I anticipate a question officer ;)

- saifoo.khan

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

- anonymous

I learnt one way that you use kind of gauss elimination

- anonymous

|dw:1327265979609:dw|
like this

- 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

- anonymous

wait a sec hold ur horses

- TuringTest

do you want to do the one you have with the gauss-jordan thing?
cuz the other way is faster and more reliable...

- TuringTest

...for 2x2 at least

- anonymous

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

- TuringTest

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

- anonymous

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

- TuringTest

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

- anonymous

|dw:1327266165384:dw|

- TuringTest

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

- anonymous

Well this is the method right?

- TuringTest

yeah, that's the same method as before

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- 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

- anonymous

It is like bothering me and I cant figure it out

- 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

- anonymous

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

- anonymous

I have a sneaky feeling i am doing something wrong here

- 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...

- anonymous

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

- 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

- anonymous

do u have twidlla or aomething like taht?

- TuringTest

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

- TuringTest

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

- anonymous

lol i dont have one do u ?

- TuringTest

no...

- 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.

- anonymous

lol maybe i will have to reread it:D

- anonymous

I have skype and tehn we cld share screens

- anonymous

LOL U dont have too

- 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...

- anonymous

ok is e1 only the right side or the whole matrix

- 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.

- TuringTest

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

- anonymous

okkkk got that so far

- 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

- TuringTest

|dw:1327268423853:dw|

- anonymous

wtf omg u clarified it for me

- TuringTest

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

- anonymous

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

- anonymous

I hate my book LOL

- anonymous

Oh man I am so excited Thanks

- TuringTest

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

- anonymous

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

- anonymous

Ya i am studying online so I am having difficulties

- anonymous

Thanks turing u r awesome :DDDDD

- TuringTest

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

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