Linear Algebra scholarship exam

Maybe_Memorie
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Homework Statement



This is one of the three linear algebra scholarship questions given by my university last year. I've solved the other two, but this one is posing a bit of a problem. Question 1 on the file.


Given that for a nxn matrix A both matricies A and A-1 have integer entries, show that det(A) = +-1

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The Attempt at a Solution



I'm completely lost. I have a feeling co-factor expansion isn't the way to go, as that would be very messy, and the other two worked out fairly nicely when you know what you're doing.
 

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Can you show that det(A) and det(A-1) are integers??
 
I can reason it now, but not really mathematically show it.
All the entries of A are integers. So by cofactor expansion for det(A) along the first row every part will be a product of integers, so an integer. Same reasoning for det(A-1)

But det(A-1) = det(A)-1

So, an integer = 1/that integer, so det(A) =1
 
Maybe_Memorie said:
so det(A) =1

or -1.

That is indeed the correct reasoning!
 
I can't edit for some reason, but I meant +-1
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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