$\mathrm{det}(A-B)\neq0$ if and only if $\mathrm{det}(A+B)\neq0$? [closed]

Is it true that $\mathrm{det}(A-B)\neq0$ if and only if $\mathrm{det}(A+B)\neq0$?

(For $A,B$ real $n\times n$ matrices, say.)

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closed as off-topic by Jonas Meyer, Davide Giraudo, Daniel Robert-Nicoud, TravisJ, John MaMay 27 at 0:06

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No, let $A = B = I$ –  Vectk Jul 26 '12 at 3:45
@Brian: That's more of an answer than a comment. –  user26872 Jul 26 '12 at 3:46
I am curious to know where this bizarre conjecture comes from. The implication doesn't even hold for $1\times 1$ matrices, i.e. real numbers. –  Erick Wong Jul 26 '12 at 4:10
This result would be true in characteristic $2$. Otherwise the guess is without a chance: you can easily make $A+B=P$ and $A-B=Q$ for any square maretices $P,Q$ of your choosing. –  Marc van Leeuwen Jul 26 '12 at 4:44
-1 This is the kind of questions I can't understand as it takes about 3 minutes of serious work, try-and-fail, examples with 1x1, 2x2 matrices, etc. to realize what's going on. Besides all this, the question has no background, what's it related to, no own effort... –  DonAntonio Jul 26 '12 at 6:09

It is not true, even if you require $A\ne B$. For instance, take $$A=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Then $\det(A-B)=0$, $\det(A+B)=2$.