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October 2nd, 2016, 08:11 AM  #1 
Newbie Joined: Oct 2016 From: Earth Posts: 3 Thanks: 0  Determinant of matrix with Aij = min(i, j)
Given a n×n matrix whose (i,j)th entry is i or j, whichever smaller, eg. [1, 1, 1, 1] [1, 2, 2, 2] [1, 2, 3, 3] [1, 2, 3, 4] The determinant of any such matrix is 1. How do I prove this? Tried induction but the assumption would only help me to compute the term for Ann mirror. 
October 2nd, 2016, 09:04 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,404 Thanks: 1306 
You'll always be able to reduce such a matrix to an upper triangular one with '1's appearing along the diagonal. The determinant of any upper triangular matrix is the product of the diagonal elements, in this case all '1's, and so the product will always be 1. You'll need to use induction to prove this. Last edited by skipjack; October 22nd, 2016 at 12:31 PM. 

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aij, determinant, matrix, mini 
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