Answer
$T(4,−1)=(3,-2)$
The pre-image is $(6,2)$.
Work Step by Step
We know that $T(v_1,v_2)=(v_1+v_2,2v_2)$
Thus $T(4,−1)=(4+(-1),2(-1))=(3,-2)$
Let $(x,y)$ be the pre-image of $(8,4)$. Then $T(x,y)=(x+y,2y)=(8,4)$. Thus $2y=4\\y=2$, $x+y=8\\x+2=8\\x=6$.
From this we get $x=6,y=2.$ So the pre-image is $(6,2)$.
