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Let $n \in \mathbb{N}$

True or false: For all $A \in \mathbb{R}^{n \times n}$ we get that det($A+A$) $= 2^{n}$ det($A$)

I did several tests and they all worked fine so I'd say that the statement is true... I couldn't find a counter example either. But I'm not happy at all about my reasoning because there isn't really one :P

So how this could be solved correctly?

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    $\begingroup$ We have det($c\cdot A$)$=c^n$$\cdot$ det($A$), for any $A \in \mathbb{R}^{n \times n}$. $\endgroup$ – Olivier Oloa Mar 11 '17 at 14:31
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Are you aware that if you multiply a single row of $A$ by $\lambda\in\Bbb R$, and call the resulting matrix $A'$, then $\det(A')=\lambda\det(A)$? If so, you are done, because $A+A$ is just the matrix obtained by multiply every row by $2$.

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Yeah, it is true. It is det(A+A)=det(2A)=2^n det(A), since det is multilinear, which means, that it is linear in every row(#rows=n)

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