# Inequality for invertible matrices with natural number entries.

Let $$A$$ be an invertible matrix with entries in the natural numbers (including 0).

Question: In case A has only positive diagonal entries and determinant 1, is the sum of all entries of A always larger than or equal to the sum of entries of the inverse of A?

Can we omit the condition that the diagonal entries are positive or that the determinatn is equal to 1? Also note that the sum of entries of the inverse matrix might be a negative number, so a possibly better question might be to ask whether the sum of all entries of A are always larger than or equal to the absolute value of the sum of entries of the inverse of A.

One can show this for 2x2 matrices easily with the explicit from for the inverse matrix, without assuming the condition that the diagonal entries are positive or that the determinatn is equal to 1.

I would think that one can also do the case of 3x3-matrices with an explicit form but this looks a little more complicated.

• Is $A$ invertible as a matrix with only natural number entries (i.e. the inverse has natural entries too), or is it invertible as a real matrix, and just happens to have natural entries? – Arthur Sep 29 at 13:19
• @Arthur invertible as a real matrix, as it probably rarely happens that the inverse has natural numbers as entries. – Mare Sep 29 at 13:19
• It happens quite often, namely for all such $A$ with $\det(A)=\pm 1$. So is $A\in GL_n(\Bbb Z)$ or not? – Dietrich Burde Sep 29 at 13:48
• @DietrichBurde yes, then the inverse has integer entries but in general not natural entires. – Mare Sep 29 at 13:55

Random counterexample: $$A=\pmatrix{1&2&5\\ 2&5&5\\ 5&5&51}, \quad A^{-1}=\pmatrix{ 230&-77&-15\\ -77& 26& 5\\ -15& 5& 1}.$$ $$A$$ has determinant $$1$$ but $$e^TAe=81<83=e^TA^{-1}e$$.