Le Monde puzzle [29]
Although I am not receiving Le Monde magazine by mail any longer (maybe a consequence of highly critical posts?!), I bought it at a newsagent last Saturday and found a fairly easy mathematical puzzle. Or is it that easy? The question is to decide about the unicity of the solution to the equation
when the matrix is (p,p). When p=9,10,11, the matrix is invertible, so there is no problem. But, looking at p=12, it is no longer invertible! When running a quick check like
for (p in 3:30){ A=diag(p) A[(1:p)+p*(((1:p)+1)%%(p))]=1 print(p) b=try(solve(A))}
we actually spot that p=4,8,12,16,20,24,28,…, i.e. all multiples of 4, lead to a non-invertible matrix. It is easy to see why for the (4,4) matrix
since the first and second lines are repeated. The (8,8) matrix is equivalent to the bloc matrix
hence with the same determinant as … In the general case, the determinant of the matrix is a circular determinant. Because most entries (but two) are zero, the determinant can easily be written as
where ι is the basic complex number. Therefore, if, and only if, p is a multiple of 4, j in the above product can take the value n/4, leading to a zero in the product… I can see why this makes a good puzzle, because this case of the multiples of 4 is quite counterintuitive!
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