Prove determinator of matrix by induction.

Original post by hlu4ff

Hi, I have the matrix:

This is a nxn matrix, named An, it's determinator is called dn.
I have to find a determinator for every n element of N (including 0).
I have an idea, that the determinator dn = n + 1. So d0 = 1, d1 = 2, d2 = 3 and d3 = 4, but I have absolutely no clue how I can prove this.
I have to use induction to show it, but I don't know how to.

One way, there may be a quicker; to start you want to set up a recurrence relation on

d_n

. So expand the determinant, and see if you can express it in terms of the determinant of smaller matrices

A_{n-1}, A_{n-2}

, etc. as necessary.

(edited 6 years ago)

Reply 2

OP

Yes, I have tried this, but how can I prove this statement for dn to be true?

Reply 3

Original post by hlu4ff

Yes, I have tried this, but how can I prove this statement for dn to be true?

If you've tried it, I presume you now have a formula for

d_n

in terms of the previous/lower d values.

This is what you need for the induction.

Base cases n=1, n=2, you have shown.

Assume true for all values up to

d_k

Then show it's true for

d_{k+1}

using that formula, and what you know of the previous values.

Reply 4

OP

Yes, I have found them and I have tried to show that it's true for d(k+1) but it didn't work yet... My formula is that dn = 2d(n-1) - d(n-2).

Reply 5

Original post by hlu4ff

Yes, I have found them and I have tried to show that it's true for d(k+1) but it didn't work yet... My formula is that dn = 2d(n-1) - d(n-2).

Agree with your formula.

So, your base bases are

d_1=2, d_2=3

, which you've already worked out explicitly.

We now asssume it's true for all

d_i

where

i\in\{1,2,...,k\}

- note we're using strong induction, rather than the weak induction usually used at A-level.

in particular.

d_{k}=( k)+1= k+1

and

d_{k-1}=(k-1)+1=k

And apply the formula to determine

d_{k+1}

(edited 6 years ago)

Reply 6

OP

Yes, I know how to prove that with strong induction, but why can I use this formula for dn, don´t I have to prove it first? (I mean the formula dn = 2d(n-1) - d(n-2))

Reply 7

Original post by hlu4ff

Yes, I know how to prove that with strong induction, but why can I use this formula for dn, don´t I have to prove it first? (I mean the formula dn = 2d(n-1) - d(n-2))

I thought you had proved the formula; otherwise where did it come from?

Reply 8

OP

Well, it worked, so does dn = d(n-1) +1, but I have no clue how to prove any of these formulas

Reply 9