# Greatest natural number which divides the determinant of a matrix

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.

• See this MSE-question and do the same. – Dietrich Burde Dec 11 '15 at 15:19
• 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. – E.Lim Dec 11 '15 at 15:34
• @E.Lim could you please elaborate? – Synonym Dec 11 '15 at 16:10
• @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 – Synonym Dec 11 '15 at 16:12
• @Synonym For this you just give an example with determinant exactly equal to $2^{n-1}$ - this is not difficult. – Dietrich Burde Dec 11 '15 at 16:13

## 1 Answer

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.