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I'm asked to prove the following statement:

Let $A$ be an $n\times n$ matrix ($n\geq 2$) with all its elements being either $+1$ or $-1$. Check that its determinant is an even number (i.e. $det(A)=\pm 2k$, with $k$ being an integer number).

I'm not too sure how should I approach this problem. I don't even know where to start the proof, but a hint or two would be greatly appreciated.


Note: My professor showed how this property holds for several matrices $A$ she wrote down in the blackboard, but never gave a full mathematical demonstration. This is not an exercise or something, it's simply curiosity. I've used all my (very limited) Algebra knowledge to no avail. No demonstrations found on the internet. I asked for some help here, but I think I'll just go and ask her during tutoring hours.

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  • $\begingroup$ Corrected the question $\endgroup$ – Jose Lopez Garcia Jan 8 '18 at 8:41
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    $\begingroup$ Is it $2^k$ or $2k$? $\endgroup$ – Saad Jan 8 '18 at 8:42
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    $\begingroup$ Have you tried induction? $\endgroup$ – B. Pasternak Jan 8 '18 at 8:42
  • $\begingroup$ It is definitely 2k - not only because of the even in the title but also because of easy examples like the 'all 1' matrix with determinant 0 $\endgroup$ – Vincent Jan 8 '18 at 8:43
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    $\begingroup$ Also, the last example shows that the condition $k \neq 0$ should be omited (of course it made some sense when the $k$ was still in the exponent) $\endgroup$ – Vincent Jan 8 '18 at 8:44
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Using the definition, we see that \begin{align} \det A \mod 2\equiv \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}\mod 2 \equiv \sum_{\sigma \in S_n} 1 \mod 2 \equiv 0 \mod 2. \end{align}

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  • $\begingroup$ Thanks for your answer, what does the symbol $\sigma$ represent? $\endgroup$ – Jose Lopez Garcia Jan 8 '18 at 8:57
  • $\begingroup$ $S_n$ is the set of permutation of $n$ elements. For example $n = 3$, the function $\sigma(1) = 2, \sigma(2) = 3, \sigma(3) = 1$ is an example of a permutation. $\endgroup$ – Jacky Chong Jan 8 '18 at 8:59
  • $\begingroup$ Oh I see. I didn't know I could apply the $mod$ operand on matrices too, I found this answer very interesting. Thank you for your time ! $\endgroup$ – Jose Lopez Garcia Jan 8 '18 at 9:04
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    $\begingroup$ Just to make it clear, I applied mod to a number $\det A$. $\endgroup$ – Jacky Chong Jan 8 '18 at 9:06
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If one add the $1^{st}$ column of $A$ to the $2^{nd}$, $3^{th}, \cdots, n^{th}$ columns, we obtain a determinant $A'$ whose entries in $2^{nd}$ to $n^{th}$ columns are all even integers. Extract the $n-1$ copies of $2$ from these columns, we can express $\det(A)$ as $2^{n-1} \det(A'')$ for another matrix $A''$ with integer coefficients.

As a result, $\det(A)$ is not only even, it is divisible by $2^{n-1}$.

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