Prove determinator of matrix by induction.

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hlu4ff
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Hi, I have the matrix:
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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.
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ghostwalker
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(Original post by hlu4ff)
Hi, I have the matrix:
Name:  Matrix.PNG
Views: 124
Size:  7.4 KB
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.
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hlu4ff
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Yes, I have tried this, but how can I prove this statement for dn to be true?
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ghostwalker
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(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.
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hlu4ff
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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).
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ghostwalker
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(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}
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hlu4ff
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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))
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ghostwalker
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(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?
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hlu4ff
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Well, it worked, so does dn = d(n-1) +1, but I have no clue how to prove any of these formulas
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ghostwalker
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(Original post by hlu4ff)
Well, it worked, so does dn = d(n-1) +1, but I have no clue how to prove any of these formulas
"Well it worked" - Hum! Not the formula that would first come to mind if you were dreaming something up.

Go back to my first post - 7 hours ago! Expand the determinant - initially by the first row.
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