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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(n-1)-bc*detA(n-2) for all n

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  1. anonymous
    • 5 years ago
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    If you help me out with what a Jacobi matrix is, I bet we can figure this out.

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    Ok, but that is essentially taken care of in the rest of the statement.

  4. anonymous
    • 5 years ago
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    right, what I wrote initially represents the matrix pretty well

  5. anonymous
    • 5 years ago
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    Have you tried expanding the determinant by minors?

  6. anonymous
    • 5 years ago
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    no I haven't yet

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    I've been able to get the equation from solving detH3, but I cannot extend the equation for all n above 3

  9. anonymous
    • 5 years ago
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    if this is undoable, my next question asks me to find H6..

  10. anonymous
    • 5 years ago
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    sorry A6

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

  12. anonymous
    • 5 years ago
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    \[+\ldots\pm a_{1n}\left|\begin{array}{cccc}a_{21}&a_{22}&\ldots&a_{2(n-1)}\\a_{31}&a_{32}&\ldots&a_{3(n-1)}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\ldots&a_{n(n-1)}\end{array}\right|\]

  13. anonymous
    • 5 years ago
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    Notice that lopping off the first row and first column just gives us \(A_{n-1}\), so the first term will be the \(a\det(A_{n-1})\)

  14. anonymous
    • 5 years ago
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    right, and lopping b gives [c b 0], [0 a b], [0 c a]

  15. anonymous
    • 5 years ago
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    Is that for the second one?

  16. anonymous
    • 5 years ago
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    yeah, sorry my comp is not running too well

  17. anonymous
    • 5 years ago
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    how do you make matrices?

  18. anonymous
    • 5 years ago
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    that's not what I got for the matrix in the second term

  19. anonymous
    • 5 years ago
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    sorry, that was my matrix for the second term in row 1

  20. anonymous
    • 5 years ago
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    row 2, column 1 is b right? and row 2 column 3 is c?

  21. anonymous
    • 5 years ago
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    [a b 0 0], [c a b 0], [0 c a b], [0 0 c a]

  22. anonymous
    • 5 years ago
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    Oh, I had the b and c switched. Okay, we're good. You're right about the second term.

  23. anonymous
    • 5 years ago
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    Ok, now we are going to expand the second term determinant by minors again.

  24. anonymous
    • 5 years ago
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    the other property that this matrix has is that Hii=a, Hi, i+1=b Hi, i-1=c

  25. anonymous
    • 5 years ago
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    so maybe the matrix is different for A5? not sure, they give us A4, so I don't exactly know what a5 looks like

  26. anonymous
    • 5 years ago
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    \[a_{12}\left|\begin{array}{cccc}a_{21}&a_{23}&\ldots&a_{2n}\\a_{31}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n3}&\ldots&a_{nn}\end{array}\right|=b\left|\begin{array}{cccc}c&b&0&\ldots&0\\0&a&b&\ldots&0\\0&c&a&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&a\end{array}\right|\]

  27. anonymous
    • 5 years ago
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    Does that look good?

  28. anonymous
    • 5 years ago
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    That property just says what you already said.

  29. anonymous
    • 5 years ago
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    yup, I keep forgetting that we have to deal with An

  30. anonymous
    • 5 years ago
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    Okay, so what do the rest of the terms after term 2 look like?

  31. anonymous
    • 5 years ago
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    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...]

  32. anonymous
    • 5 years ago
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    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?

  33. anonymous
    • 5 years ago
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    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)

  34. anonymous
    • 5 years ago
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    Right. What did we do to find the original determinant?

  35. anonymous
    • 5 years ago
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    minor a12

  36. anonymous
    • 5 years ago
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    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?

  37. anonymous
    • 5 years ago
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    one sec, I'm still absorbing how the first determinant is A(n-1) I think i got it

  38. anonymous
    • 5 years ago
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    Oh, ok, sure.

  39. anonymous
    • 5 years ago
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    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]

  40. anonymous
    • 5 years ago
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    But the determinant isn't \(A_{n-a}\), it's \(\det(A_{n-1})\)

  41. anonymous
    • 5 years ago
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    ur right I copied it wrong

  42. anonymous
    • 5 years ago
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    so it's [c b 000] [0 c b 000] [ 0 0 c b 000]

  43. anonymous
    • 5 years ago
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    But you skip the second column because that is the column that \(a_{12}\) is in.

  44. anonymous
    • 5 years ago
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    Right.

  45. anonymous
    • 5 years ago
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    what can we take by that? I get that there is c's down a diagonal and b's down a diagonal

  46. anonymous
    • 5 years ago
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    Wait, no, it's what I have up there, you only lose the first \(a\). After that you start getting \(a\)'s again.

  47. anonymous
    • 5 years ago
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    yup

  48. anonymous
    • 5 years ago
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    c b 0000, 0 c a b 00000, 0 0 c a b 0000?

  49. anonymous
    • 5 years ago
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    I think the second on is 0ab0...

  50. anonymous
    • 5 years ago
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    one

  51. anonymous
    • 5 years ago
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    yup, ive been doing math al day haha, not working at full potential, good catch

  52. anonymous
    • 5 years ago
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    I know how that goes. I usually top out after an hour at a time. Haha.

  53. anonymous
    • 5 years ago
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    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.

  54. anonymous
    • 5 years ago
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    we have to take out the c somehow, so -bc*detA(n-2), know how we could take out the c?

  55. anonymous
    • 5 years ago
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    Right, we take out the c by expanding along a row or column with the c in it.

  56. anonymous
    • 5 years ago
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    well every row has a c except the second one

  57. anonymous
    • 5 years ago
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    sorry how do you expand? is that taking out a row and column and then multiplying the determinant by the entry?

  58. anonymous
    • 5 years ago
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    mathworld.wolfram.com/Determinant.html

  59. anonymous
    • 5 years ago
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    if we take out the a11 we have Hn again don't we?

  60. anonymous
    • 5 years ago
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    Bingo!

  61. anonymous
    • 5 years ago
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    But right now we are smaller by 1 row and 1 column. After we take out another, we are smaller by 2, which is?

  62. anonymous
    • 5 years ago
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    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

  63. anonymous
    • 5 years ago
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    that last part is wrong...I don't understand how we went from Hn-1 to Hn-2 in the 2nd determinant

  64. anonymous
    • 5 years ago
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    sorry i keep putting H, H or A same thing

  65. anonymous
    • 5 years ago
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    We expanded by minors, which means we were working with a smaller matrix.

  66. anonymous
    • 5 years ago
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    Yeah, I get it.

  67. anonymous
    • 5 years ago
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    To find that second determinant, we are going to expand again.

  68. anonymous
    • 5 years ago
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    oh ok, but we couldn't have said -b(detHn-1) hey?

  69. anonymous
    • 5 years ago
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    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.

  70. anonymous
    • 5 years ago
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    ok, I get it now, we expanded and ended up with Hn-2, cool

  71. anonymous
    • 5 years ago
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    Right. But it's important which row or column we expand along.

  72. anonymous
    • 5 years ago
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    for H6 I'm guessing it's just long but straightforward

  73. anonymous
    • 5 years ago
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    ...oh ok, thx a lot you've been a really big help

  74. anonymous
    • 5 years ago
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    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.

  75. anonymous
    • 5 years ago
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    cool, that gives our equation

  76. anonymous
    • 5 years ago
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    thx, now on to a Calc assignment, going to be up all night haha

  77. anonymous
    • 5 years ago
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    Ha, fun.

  78. anonymous
    • 5 years ago
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    Good luck!

  79. anonymous
    • 5 years ago
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    thanks again have a good one!

  80. anonymous
    • 5 years ago
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    You too.

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