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Let $A$ be a $10 \times 10$ matrix such that each entry is either $1$ or $-1$. Is it true that $\det(A)$ is divisible by $2^9$?

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    $\begingroup$ You can just try to do Gaussian elimination, until you have an upper triangular matrix. Then the result will be evident.. Just take a $3 \times 3$ example and see how it works... $\endgroup$ – Ludolila Mar 9 '13 at 21:09
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    $\begingroup$ If all the entries are the same, what does this mean? In othe words, is this forcing them to alternate? $\endgroup$ – Amzoti Mar 9 '13 at 21:15
  • $\begingroup$ @Amzoti:No. 1 0r -1 $\endgroup$ – rese Mar 9 '13 at 21:16
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    $\begingroup$ @Ludolila No need to reduce to upper triangular, it is already evident after eliminating the column below the first entry. $\endgroup$ – Erick Wong Mar 9 '13 at 21:18
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    $\begingroup$ @Amzoti $2^9$ divides $0$ just fine, what is your concern? $\endgroup$ – Erick Wong Mar 9 '13 at 21:21
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Alternatively, we can choose $A$ as below: $$A = {\left[ {\begin{array}{*{20}{c}} 1&1& \ldots &1 \\ { - 1}&1& \ldots &1 \\ { - 1}&{ - 1}& \ldots &{ - 1} \\ \vdots &{}&{}&{} \\ { - 1}&{ - 1}& \ldots &{ - 1} \end{array}} \right]_{10 \times 10}}$$ If we add fist row to another rows, we get: $$A = {\left[ {\begin{array}{*{20}{c}} 1&1& \ldots & \ldots &1 \\ 0&2& \ldots & \ldots &2 \\ 0&0& \ldots & \ldots &2 \\ \vdots &{}&{}&{}&{} \\ 0&0& \ldots &0&2 \end{array}} \right]_{10 \times 10}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\det (A) = {2^9}$$

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Answer based on the comments by Ludolila and Erick Wong as an answer:

The answer follows from three easily proven rules:

  1. Adding or subtracting a row of a matrix from another does not change its determinant.
  2. Multiplying a line of the matrix by a constant $c$ multiplies the determinant by that constant.
  3. The determinant of a matrix with integer entries is an integer.

Take a matrix $A=(a_{ij})\in M_{10}(\mathbb{R})$ such that all its entries are either $1$ or $-1$. If $a_{11}=-1$, multiply the first line by $-1$. For $2\le i\le10$, subtract $a_{i1}(a_{1\to})$ (where $a_{1\to}$ is the first row of $A$) from $a_{i\to}$.

Now all rows consist only of $0$'s and $\pm2$'s. Divide each of these rows by $2$ to obtain a matrix $B$ that has entries only in $\{-1,0,1\}$.

Note that $\det B = \pm 2^{-9} \det A$ following rules 1 and 2.

Following rule 3, $\det B$ is an integer, so $\det A = 2^9 \cdot n$ where $n$ is an integer.

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