Question

Please help me solve the following :

Answer 1

Let A be a cube matrix.
Compute the following Determinant :

Answer 2

the determinant of a triangular matrix is the product of the diagonal entries.
you can use elimination to make this matrix upper triangular.

It looks like you get a "telescoping" product, where the successive entries cancel quite a bit, and you can find a nice closed form solution as a function of the number of rows

Answer 3

First of all thank you for your comment.
I was thinking to use elimination to make the matrix upper triangular and then compute the determinant as the product of the diagonal entries.

There are two problems I couldn't solve:
1. How do I show (in this generic matrix) that it is upper triangular using elimination?
2. Can you explain me a little more about the "telescoping" product I'm supposed to get - wouldn't I finish with fraction in the end of the day?

Answer 4

this is tridiagonal, with 1 diagonal below the main diagonal. if you get rid of the lower diagonal, then you have an upper triangular matrix.
do elimination on a 3 x 3 or a 4x4
what do you get?

Answer 5

Let me think about it, just a second.

Answer 6

Lets take a look at a simple example of this matrix in 3x3:
\[\left[\begin{matrix}3 & 2 & 0\\ 1 & 3 & 2\\ 0 & 1 & 3 \end{matrix}\right]\]

It seems impossible to get rid of these ones without getting tons of fractions...

Answer 7

It seems really complicated to me. Am I right?

Answer 8

don't be a wuss...

Answer 9

Just never had to deal with this kind of problem before. I'll try to work on it as you instructed hoping I will get the idea.

Answer 10

Thank you very much for your help!

Answer 11

Did you get an answer?

Answer 12

I understood what you meant. I understand the logic.
Actually If I do the eliminate you just told me, I get a const set of numbers:
3 * 7/3 * 15/7 .... and so on.... ( that what you meant regarding the telescope series)?
I just have to think how to prove it now. How to explain my calculations.
Thank you very much for the response.

Answer 13

yes, looks good. for an n x n, the determinant becomes
\[ 2^{n+1} -1 \]

maybe prove using induction? I haven't thought about it, but that might work.

Answer 14

I didn't think about induction but It might be a good idea considering the fact I have no clue how to start this proof.

I'm trying to solve it now and in case I won't make it I will try again tomorrow and close this thread.

Again I really appriciate your assistance.

Answer 15

In the inudction step I get :
what can I do with this?