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All elements of a 100 x 100 matrix ,A are odd numbers. What is the greatest natural number that would always divide the determinant of A?

I have been able to show that it is always divisible by 2^99(y performing elementary row subtractions and additions) , but am stuck when it comes to showing that there is no natural number greater than this.

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  • $\begingroup$ See this MSE-question and do the same. $\endgroup$ – Dietrich Burde Dec 11 '15 at 15:19
  • $\begingroup$ To show there is no larger number, calculate the determinant of a matrix with diagonal entries $n$ and all other entries $1$ and choose $n$ odd. Then vary $n$ to disqualify other factors. $\endgroup$ – E.Lim Dec 11 '15 at 15:34
  • $\begingroup$ @E.Lim could you please elaborate? $\endgroup$ – Synonym Dec 11 '15 at 16:10
  • $\begingroup$ @DietrichBurde the mse question you directed to me seems to be asking to just prove that it is divisible by 2^n-1, i am more concerned wiht showing that ther can not be a greater integer than this $\endgroup$ – Synonym Dec 11 '15 at 16:12
  • $\begingroup$ @Synonym For this you just give an example with determinant exactly equal to $2^{n-1}$ - this is not difficult. $\endgroup$ – Dietrich Burde Dec 11 '15 at 16:13
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Consider the matrix,

$ M= \begin{bmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & 3 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 3 & 1 & \cdots & 1 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots\\ \end{bmatrix} $

This gives $2^{99}$ is the greatest natural number that would always divide the determinant.

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