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Algebra Difficulty 4.8 AIME Prove it Brazil

Let nn be an integer and n1n_1 be one of its divisors. Let AA be a n×nn \times n symmetric matrix defined by ai,i=4a_{i,i} = 4, ai,i+1=ai+1,i=1a_{i,i+1} = a_{i+1,i} = -1 for all ii such that 1in11 \le i \le n-1 and i+1i+1 is not a multiple of n1n_1, ai,i+n1=ai+n1,i=1a_{i,i+n_1} = a_{i+n_1,i} = -1 and ai,j=0a_{i,j} = 0 otherwise.

Solution

See problem 3, grades 10–12.

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