## 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

1. anonymous

If you help me out with what a Jacobi matrix is, I bet we can figure this out.

2. 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

3. anonymous

Ok, but that is essentially taken care of in the rest of the statement.

4. anonymous

right, what I wrote initially represents the matrix pretty well

5. anonymous

Have you tried expanding the determinant by minors?

6. anonymous

no I haven't yet

7. 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.

8. anonymous

I've been able to get the equation from solving detH3, but I cannot extend the equation for all n above 3

9. anonymous

if this is undoable, my next question asks me to find H6..

10. anonymous

sorry A6

11. anonymous

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

12. anonymous

$+\ldots\pm a_{1n}\left|\begin{array}{cccc}a_{21}&a_{22}&\ldots&a_{2(n-1)}\\a_{31}&a_{32}&\ldots&a_{3(n-1)}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\ldots&a_{n(n-1)}\end{array}\right|$

13. anonymous

Notice that lopping off the first row and first column just gives us $$A_{n-1}$$, so the first term will be the $$a\det(A_{n-1})$$

14. anonymous

right, and lopping b gives [c b 0], [0 a b], [0 c a]

15. anonymous

Is that for the second one?

16. anonymous

yeah, sorry my comp is not running too well

17. anonymous

how do you make matrices?

18. anonymous

that's not what I got for the matrix in the second term

19. anonymous

sorry, that was my matrix for the second term in row 1

20. anonymous

row 2, column 1 is b right? and row 2 column 3 is c?

21. anonymous

[a b 0 0], [c a b 0], [0 c a b], [0 0 c a]

22. anonymous

Oh, I had the b and c switched. Okay, we're good. You're right about the second term.

23. anonymous

Ok, now we are going to expand the second term determinant by minors again.

24. anonymous

the other property that this matrix has is that Hii=a, Hi, i+1=b Hi, i-1=c

25. 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

26. anonymous

$a_{12}\left|\begin{array}{cccc}a_{21}&a_{23}&\ldots&a_{2n}\\a_{31}&a_{33}&\ldots&a_{3n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n3}&\ldots&a_{nn}\end{array}\right|=b\left|\begin{array}{cccc}c&b&0&\ldots&0\\0&a&b&\ldots&0\\0&c&a&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&a\end{array}\right|$

27. anonymous

Does that look good?

28. anonymous

That property just says what you already said.

29. anonymous

yup, I keep forgetting that we have to deal with An

30. anonymous

Okay, so what do the rest of the terms after term 2 look like?

31. 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...]

32. 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?

33. 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)

34. anonymous

Right. What did we do to find the original determinant?

35. anonymous

minor a12

36. 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?

37. anonymous

one sec, I'm still absorbing how the first determinant is A(n-1) I think i got it

38. anonymous

Oh, ok, sure.

39. 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]

40. anonymous

But the determinant isn't $$A_{n-a}$$, it's $$\det(A_{n-1})$$

41. anonymous

ur right I copied it wrong

42. anonymous

so it's [c b 000] [0 c b 000] [ 0 0 c b 000]

43. anonymous

But you skip the second column because that is the column that $$a_{12}$$ is in.

44. anonymous

Right.

45. anonymous

what can we take by that? I get that there is c's down a diagonal and b's down a diagonal

46. anonymous

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

yup

48. anonymous

c b 0000, 0 c a b 00000, 0 0 c a b 0000?

49. anonymous

I think the second on is 0ab0...

50. anonymous

one

51. anonymous

yup, ive been doing math al day haha, not working at full potential, good catch

52. anonymous

I know how that goes. I usually top out after an hour at a time. Haha.

53. 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.

54. anonymous

we have to take out the c somehow, so -bc*detA(n-2), know how we could take out the c?

55. anonymous

Right, we take out the c by expanding along a row or column with the c in it.

56. anonymous

well every row has a c except the second one

57. anonymous

sorry how do you expand? is that taking out a row and column and then multiplying the determinant by the entry?

58. anonymous

mathworld.wolfram.com/Determinant.html

59. anonymous

if we take out the a11 we have Hn again don't we?

60. anonymous

Bingo!

61. 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?

62. 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

63. anonymous

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

64. anonymous

sorry i keep putting H, H or A same thing

65. anonymous

We expanded by minors, which means we were working with a smaller matrix.

66. anonymous

Yeah, I get it.

67. anonymous

To find that second determinant, we are going to expand again.

68. anonymous

oh ok, but we couldn't have said -b(detHn-1) hey?

69. 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.

70. anonymous

ok, I get it now, we expanded and ended up with Hn-2, cool

71. anonymous

Right. But it's important which row or column we expand along.

72. anonymous

for H6 I'm guessing it's just long but straightforward

73. anonymous

...oh ok, thx a lot you've been a really big help

74. 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.

75. anonymous

cool, that gives our equation

76. anonymous

thx, now on to a Calc assignment, going to be up all night haha

77. anonymous

Ha, fun.

78. anonymous

Good luck!

79. anonymous

thanks again have a good one!

80. anonymous

You too.