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anonymous
 5 years ago
I have a Jacobi matrix (An) with all a's down the main diagonal, all b's in the diagonal above the a's and all c's in the diagonal below, everything else is zeroes...need help proving that detAn=a*detA(n1)bc*detA(n2) for all n
anonymous
 5 years ago
I have a Jacobi matrix (An) with all a's down the main diagonal, all b's in the diagonal above the a's and all c's in the diagonal below, everything else is zeroes...need help proving that detAn=a*detA(n1)bc*detA(n2) for all n

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you help me out with what a Jacobi matrix is, I bet we can figure this out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well it's actually not a jacobi, it's modified, jacobi means that whenever absolute(ij) is bigger or equal to 2, Aij=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, but that is essentially taken care of in the rest of the statement.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, what I wrote initially represents the matrix pretty well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you tried expanding the determinant by minors?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see how to get the first term this way. I'm having some trouble seeing the second term without writing it out, but I can see how would most likely fall out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've been able to get the equation from solving detH3, but I cannot extend the equation for all n above 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if this is undoable, my next question asks me to find H6..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, if we expand by minors in the first row, then we get \[\det(A_{n})=a_{11}\left\begin{array}{cccc}a_{22}&a_{23}&\ldots&a_{2n}\\a_{32}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n2}&a_{n3}&\ldots&a_{nn}\end{array}\righta_{12}\left\begin{array}{cccc}a_{21}&a_{23}&\ldots&a_{2n}\\a_{31}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n3}&\ldots&a_{nn}\end{array}\right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[+\ldots\pm a_{1n}\left\begin{array}{cccc}a_{21}&a_{22}&\ldots&a_{2(n1)}\\a_{31}&a_{32}&\ldots&a_{3(n1)}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\ldots&a_{n(n1)}\end{array}\right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Notice that lopping off the first row and first column just gives us \(A_{n1}\), so the first term will be the \(a\det(A_{n1})\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, and lopping b gives [c b 0], [0 a b], [0 c a]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that for the second one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, sorry my comp is not running too well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you make matrices?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's not what I got for the matrix in the second term

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, that was my matrix for the second term in row 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0row 2, column 1 is b right? and row 2 column 3 is c?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0[a b 0 0], [c a b 0], [0 c a b], [0 0 c a]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, I had the b and c switched. Okay, we're good. You're right about the second term.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, now we are going to expand the second term determinant by minors again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the other property that this matrix has is that Hii=a, Hi, i+1=b Hi, i1=c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so maybe the matrix is different for A5? not sure, they give us A4, so I don't exactly know what a5 looks like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a_{12}\left\begin{array}{cccc}a_{21}&a_{23}&\ldots&a_{2n}\\a_{31}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n3}&\ldots&a_{nn}\end{array}\right=b\left\begin{array}{cccc}c&b&0&\ldots&0\\0&a&b&\ldots&0\\0&c&a&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&a\end{array}\right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That property just says what you already said.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup, I keep forgetting that we have to deal with An

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so what do the rest of the terms after term 2 look like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they should all be zero because abs(ij) is bigger or equal to 2, in which case the term is zero...first row is always [a b 0 0 0 0 0...]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right. So, now all we have to do is show that the second term is what we want it to be. Any ideas on how to do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0second term is always b because the property says that Hi, i+1=b, which will always be for all n, the 2nd determinant is what I dont know how to express as H(n2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right. What did we do to find the original determinant?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, yeah, we expanded along the first row. We are going to expand by minors again. Which row or column are we going to use?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec, I'm still absorbing how the first determinant is A(n1) I think i got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the 2nd determinant the first 3 entries are [c a b 0 0 0 0 0], [0 c a b 0000], [0 0 c a b 00]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the determinant isn't \(A_{na}\), it's \(\det(A_{n1})\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ur right I copied it wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it's [c b 000] [0 c b 000] [ 0 0 c b 000]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you skip the second column because that is the column that \(a_{12}\) is in.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what can we take by that? I get that there is c's down a diagonal and b's down a diagonal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, no, it's what I have up there, you only lose the first \(a\). After that you start getting \(a\)'s again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c b 0000, 0 c a b 00000, 0 0 c a b 0000?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the second on is 0ab0...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup, ive been doing math al day haha, not working at full potential, good catch

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know how that goes. I usually top out after an hour at a time. Haha.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So take a look at \[\left[\begin{array}{ccccc}c&b&0&0&\ldots\\0&a&b&0&\ldots\\0&c&a&b&\ldots\\0&0&c&a&\ldots\\\vdots&\vdots&\vdots&\vdots&\ldots\end{array}\right]\] and compare it to the original matrix.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have to take out the c somehow, so bc*detA(n2), know how we could take out the c?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right, we take out the c by expanding along a row or column with the c in it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well every row has a c except the second one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry how do you expand? is that taking out a row and column and then multiplying the determinant by the entry?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mathworld.wolfram.com/Determinant.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we take out the a11 we have Hn again don't we?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But right now we are smaller by 1 row and 1 column. After we take out another, we are smaller by 2, which is?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's A(n2)...so let's add it all up....det(An)= a11*det(Hn1)bc(det(Hn1)) that's what i got so far

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that last part is wrong...I don't understand how we went from Hn1 to Hn2 in the 2nd determinant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i keep putting H, H or A same thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We expanded by minors, which means we were working with a smaller matrix.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find that second determinant, we are going to expand again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok, but we couldn't have said b(detHn1) hey?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But it's not \(\det(A_{n1})\), it was only like that the first time because of how the structure of the matrix and what happened after we remove the *first* row and *first* column.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I get it now, we expanded and ended up with Hn2, cool

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right. But it's important which row or column we expand along.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for H6 I'm guessing it's just long but straightforward

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...oh ok, thx a lot you've been a really big help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So for the second determinant, we expand down the first column. That way, we multiply the b by c and \(A_{n2}\). The rest of the terms b gets multiplied by are 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool, that gives our equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thx, now on to a Calc assignment, going to be up all night haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks again have a good one!
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