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Matrix $A$ is a matrix of size $n×n$ with entries of even and odd natural numbers that differ from one another. In order for $A$ to be a non-singular matrix, the minimum number of odd number entries is · ·

GIve me a hint. We know non singular matrix is a matrix that have determinat is equal 0.

Nladougdiuew
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    Hint: Can you think of any particular nonsingular $n \times n$ matrix? – John Hughes Dec 21 '19 at 12:19
  • @JohnHughes i don't understand what did you mean.. – Nladougdiuew Dec 21 '19 at 12:30
  • give me other hint please – Nladougdiuew Dec 21 '19 at 12:31
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    To be honest, I misread "that differ from one another." Others take it to mean "all different" (which seems right), while I took it to mean "not all the same" (which seemed like a weird condition, because it's already implied by "nonsingular"). – John Hughes Dec 21 '19 at 12:41
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    "We know non singular matrix is a matrix that have determinat is equal 0." I thought the opposite, i.e. that a singular square matrix has determinant $0$ and a non-singular square matrix has a non-zero determinant – Henry Dec 21 '19 at 12:49

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We can easily construct a nonsingular matrix with distinct even positive integer entries. It is enough to make the diagonal elements larger than the sum of the other elements in the corresponding row.

Peter
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  • Do you have a theorem about that? – Nladougdiuew Dec 21 '19 at 12:31
  • and how to apply that to the problem? – Nladougdiuew Dec 21 '19 at 12:32
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    It is the Gershgorin circle theorem (you can look it up at wikipedia, for example). In our situation, start with the numbers $2,4,6,\cdots$ and make the diagonal larger than the sum. Than continue with the next even numbers in the second row, make the diagonal element larger than the sum and so on. – Peter Dec 21 '19 at 12:35
  • that theorem about eigenvalues.. – Nladougdiuew Dec 21 '19 at 12:36
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    Yes, in particular it guarantees that eigenvalue $\ 0\ $ is impossible , if we construct the matrix like described. – Peter Dec 21 '19 at 12:37
  • Do you have simple answer? – Nladougdiuew Dec 21 '19 at 12:46
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    You mean avoiding the "big guns" ? No, I have no easier approach in mind. – Peter Dec 21 '19 at 12:54
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    @Peter One may think of using the Vandermonde matrix in which every line is suitably multiplied by $\alpha_j$. For example:$\begin{pmatrix}&2&\dots&2^n\&4^n&\dots&4^{2n}\&\vdots&\ &\vdots\ &{2^n}^n&\dots&{2^n}^{2n}\end{pmatrix}$ –  Dec 21 '19 at 14:22