Assignment 8: Problem 22 (1 point) Let V=R2. For (1, 2), (v1, v2) EV and a ER define vector addition by (u1, 2) (v1, v2)=(1+v1 - 1,u2+2-2) and scalar multiplication by a (u, u2)=(au-a+1, au2-2a + 2). It can be shown that (V, BB, D) is a vector space over the scalar field R. Find the following: the sum: (-8,-7) (-1,-3)=-10-12 the scalar multiple: 0(-8,-7)=(1,2) the zero vector: 012 the additive inverse of (x, y): B(x, y) = -2x-2y Note: You can earn partial credit on this problem.