Let V = R². For (u₁, U2), (v₁, v₂) € V and a ER define vector addition by (u₁, U₂) (V1₁, V₂) = = (U₁ + V₁ − 1, U₂ + V2 − 3) and scalar multiplication by a (u₁, ₂) = (au₁ — a +1, au2 − 3a + 3). It can be shown that (V, , ) is a vector space over the scalar field R. Find the following: the sum: (-6, 0) (9,2)=(2 the scalar multiple: 40 (-6,0) =( the zero vector: Ov 1 the additive inverse of (x, y): B(x, y) =( 3 -1

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