Given that the element $(x, y)$ is mapped to $(x-y, x+y)$ under the mapping $f$, then the preimage of $(2,-4)$ under $f$ is 
(A) $(1,3)$
(H) $(\mathrm{I},-3)$
(C) $(-1,3)$
(D) $(-1,-3)$
1. (D).

From the problem, we know $x-y=2, x \rightarrow y=-4$, thus $x=-1, y=-3$. Therefore, the preimage of $(2,-4)$ under $f$ is $(-1,-3)$