Linear Algebra scholarship exam

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Homework Help Overview

The problem involves a linear algebra question regarding an nxn matrix A with integer entries, specifically asking to show that the determinant of A is either +1 or -1. The original poster has attempted to solve similar problems but finds this one challenging.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the properties of determinants, questioning whether both det(A) and det(A-1) can be integers. There is mention of cofactor expansion as a potential method, though some express concerns about its complexity.

Discussion Status

Some participants have provided reasoning regarding the integer nature of the determinants, suggesting that since all entries of A are integers, the determinants must also be integers. There is an acknowledgment of the relationship between det(A) and det(A-1), with some participants indicating that the reasoning leads to the conclusion that det(A) could be +1 or -1.

Contextual Notes

There is a note about the original poster's difficulty with the problem, indicating that they have successfully solved other related questions but are struggling with this specific one. The discussion reflects a mix of reasoning and attempts to clarify the mathematical relationships involved.

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

Homework Equations





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
 

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