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

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?

• We have det($c\cdot A$)$=c^n$$\cdot$ det($A$), for any $A \in \mathbb{R}^{n \times n}$. – Olivier Oloa Mar 11 '17 at 14:31

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 $2$.