Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

right, what I wrote initially represents the matrix pretty well

Answer 5

Have you tried expanding the determinant by minors?

Answer 6

no I haven't yet

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

sorry A6

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

Is that for the second one?

Answer 16

yeah, sorry my comp is not running too well

Answer 17

how do you make matrices?

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

so maybe the matrix is different for A5? not sure, they give us A4, so I don't exactly know what a5 looks like

Answer 26

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

Answer 27

Does that look good?

Answer 28

That property just says what you already said.

Answer 29

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

Answer 30

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

Answer 31

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...]

Answer 32

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?

Answer 33

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)

Answer 34

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

Answer 35

minor a12

Answer 36

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?

Answer 37

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

Answer 38

Oh, ok, sure.

Answer 39

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]

Answer 40

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

Answer 41

ur right I copied it wrong

Answer 42

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

Answer 43

But you skip the second column because that is the column that \(a_{12}\) is in.

Answer 44

Right.

Answer 45

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

Answer 46

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

Answer 47

yup

Answer 48

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

Answer 49

I think the second on is 0ab0...

Answer 50

one

Answer 51

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

Answer 52

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

Answer 53

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.

Answer 54

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

Answer 55

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

Answer 56

well every row has a c except the second one

Answer 57

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

Answer 58

mathworld.wolfram.com/Determinant.html

Answer 59

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

Answer 60

Bingo!

Answer 61

But right now we are smaller by 1 row and 1 column. After we take out another, we are smaller by 2, which is?

Answer 62

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

Answer 63

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

Answer 64

sorry i keep putting H, H or A same thing

Answer 65

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

Answer 66

Yeah, I get it.

Answer 67

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

Answer 68

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

Answer 69

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.

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

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.

Answer 75

cool, that gives our equation

Answer 76

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

Answer 77

Ha, fun.

Answer 78

Good luck!

Answer 79

thanks again have a good one!

Answer 80

You too.