anonymous
  • anonymous
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(n-1)-bc*detA(n-2) for all n
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
If you help me out with what a Jacobi matrix is, I bet we can figure this out.
anonymous
  • anonymous
well it's actually not a jacobi, it's modified, jacobi means that whenever absolute(i-j) is bigger or equal to 2, Aij=0
anonymous
  • anonymous
Ok, but that is essentially taken care of in the rest of the statement.

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anonymous
  • anonymous
right, what I wrote initially represents the matrix pretty well
anonymous
  • anonymous
Have you tried expanding the determinant by minors?
anonymous
  • anonymous
no I haven't yet
anonymous
  • anonymous
I 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
  • anonymous
I've been able to get the equation from solving detH3, but I cannot extend the equation for all n above 3
anonymous
  • anonymous
if this is undoable, my next question asks me to find H6..
anonymous
  • anonymous
sorry A6
anonymous
  • anonymous
Well, if we expand by minors in the first row, then we get \[\det(A_{n})=a_{11}\left|\begin{array}{cccc}a_{22}&a_{23}&\ldots&a_{2n}\\a_{32}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n2}&a_{n3}&\ldots&a_{nn}\end{array}\right|-a_{12}\left|\begin{array}{cccc}a_{21}&a_{23}&\ldots&a_{2n}\\a_{31}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n3}&\ldots&a_{nn}\end{array}\right|\]
anonymous
  • anonymous
\[+\ldots\pm a_{1n}\left|\begin{array}{cccc}a_{21}&a_{22}&\ldots&a_{2(n-1)}\\a_{31}&a_{32}&\ldots&a_{3(n-1)}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\ldots&a_{n(n-1)}\end{array}\right|\]
anonymous
  • anonymous
Notice that lopping off the first row and first column just gives us \(A_{n-1}\), so the first term will be the \(a\det(A_{n-1})\)
anonymous
  • anonymous
right, and lopping b gives [c b 0], [0 a b], [0 c a]
anonymous
  • anonymous
Is that for the second one?
anonymous
  • anonymous
yeah, sorry my comp is not running too well
anonymous
  • anonymous
how do you make matrices?
anonymous
  • anonymous
that's not what I got for the matrix in the second term
anonymous
  • anonymous
sorry, that was my matrix for the second term in row 1
anonymous
  • anonymous
row 2, column 1 is b right? and row 2 column 3 is c?
anonymous
  • anonymous
[a b 0 0], [c a b 0], [0 c a b], [0 0 c a]
anonymous
  • anonymous
Oh, I had the b and c switched. Okay, we're good. You're right about the second term.
anonymous
  • anonymous
Ok, now we are going to expand the second term determinant by minors again.
anonymous
  • anonymous
the other property that this matrix has is that Hii=a, Hi, i+1=b Hi, i-1=c
anonymous
  • anonymous
so maybe the matrix is different for A5? not sure, they give us A4, so I don't exactly know what a5 looks like
anonymous
  • anonymous
\[a_{12}\left|\begin{array}{cccc}a_{21}&a_{23}&\ldots&a_{2n}\\a_{31}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n3}&\ldots&a_{nn}\end{array}\right|=b\left|\begin{array}{cccc}c&b&0&\ldots&0\\0&a&b&\ldots&0\\0&c&a&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&a\end{array}\right|\]
anonymous
  • anonymous
Does that look good?
anonymous
  • anonymous
That property just says what you already said.
anonymous
  • anonymous
yup, I keep forgetting that we have to deal with An
anonymous
  • anonymous
Okay, so what do the rest of the terms after term 2 look like?
anonymous
  • anonymous
they should all be zero because abs(i-j) 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
  • anonymous
Right. 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
  • anonymous
second 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(n-2)
anonymous
  • anonymous
Right. What did we do to find the original determinant?
anonymous
  • anonymous
minor a12
anonymous
  • anonymous
Well, 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
  • anonymous
one sec, I'm still absorbing how the first determinant is A(n-1) I think i got it
anonymous
  • anonymous
Oh, ok, sure.
anonymous
  • anonymous
for 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
  • anonymous
But the determinant isn't \(A_{n-a}\), it's \(\det(A_{n-1})\)
anonymous
  • anonymous
ur right I copied it wrong
anonymous
  • anonymous
so it's [c b 000] [0 c b 000] [ 0 0 c b 000]
anonymous
  • anonymous
But you skip the second column because that is the column that \(a_{12}\) is in.
anonymous
  • anonymous
Right.
anonymous
  • anonymous
what can we take by that? I get that there is c's down a diagonal and b's down a diagonal
anonymous
  • anonymous
Wait, no, it's what I have up there, you only lose the first \(a\). After that you start getting \(a\)'s again.
anonymous
  • anonymous
yup
anonymous
  • anonymous
c b 0000, 0 c a b 00000, 0 0 c a b 0000?
anonymous
  • anonymous
I think the second on is 0ab0...
anonymous
  • anonymous
one
anonymous
  • anonymous
yup, ive been doing math al day haha, not working at full potential, good catch
anonymous
  • anonymous
I know how that goes. I usually top out after an hour at a time. Haha.
anonymous
  • anonymous
So 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
  • anonymous
we have to take out the c somehow, so -bc*detA(n-2), know how we could take out the c?
anonymous
  • anonymous
Right, we take out the c by expanding along a row or column with the c in it.
anonymous
  • anonymous
well every row has a c except the second one
anonymous
  • anonymous
sorry how do you expand? is that taking out a row and column and then multiplying the determinant by the entry?
anonymous
  • anonymous
mathworld.wolfram.com/Determinant.html
anonymous
  • anonymous
if we take out the a11 we have Hn again don't we?
anonymous
  • anonymous
Bingo!
anonymous
  • anonymous
But right now we are smaller by 1 row and 1 column. After we take out another, we are smaller by 2, which is?
anonymous
  • anonymous
it's A(n-2)...so let's add it all up....det(An)= a11*det(Hn-1)-bc(det(Hn-1)) that's what i got so far
anonymous
  • anonymous
that last part is wrong...I don't understand how we went from Hn-1 to Hn-2 in the 2nd determinant
anonymous
  • anonymous
sorry i keep putting H, H or A same thing
anonymous
  • anonymous
We expanded by minors, which means we were working with a smaller matrix.
anonymous
  • anonymous
Yeah, I get it.
anonymous
  • anonymous
To find that second determinant, we are going to expand again.
anonymous
  • anonymous
oh ok, but we couldn't have said -b(detHn-1) hey?
anonymous
  • anonymous
But it's not \(\det(A_{n-1})\), 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
  • anonymous
ok, I get it now, we expanded and ended up with Hn-2, cool
anonymous
  • anonymous
Right. But it's important which row or column we expand along.
anonymous
  • anonymous
for H6 I'm guessing it's just long but straightforward
anonymous
  • anonymous
...oh ok, thx a lot you've been a really big help
anonymous
  • anonymous
So for the second determinant, we expand down the first column. That way, we multiply the b by c and \(A_{n-2}\). The rest of the terms b gets multiplied by are 0.
anonymous
  • anonymous
cool, that gives our equation
anonymous
  • anonymous
thx, now on to a Calc assignment, going to be up all night haha
anonymous
  • anonymous
Ha, fun.
anonymous
  • anonymous
Good luck!
anonymous
  • anonymous
thanks again have a good one!
anonymous
  • anonymous
You too.

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