Property of $10\times 10 $ matrix 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 $2^9$?
 A: Answer based on the comments by Ludolila and Erick Wong as an answer:
The answer follows from three easily proven rules:


*

*Adding or subtracting a row of a matrix from another does not change its determinant.

*Multiplying a line of the matrix by a constant $c$ multiplies the determinant by that constant.

*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.
A: 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:
$$A = {\left[ {\begin{array}{*{20}{c}}
  1&1& \ldots & \ldots &1 \\ 
  0&2& \ldots & \ldots &2 \\ 
  0&0& \ldots & \ldots &2 \\ 
   \vdots &{}&{}&{}&{} \\ 
  0&0& \ldots &0&2 
\end{array}} \right]_{10 \times 10}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\det (A) = {2^9}$$
