Exercise with proposed solution
So for this question, the functions I am getting for the additive inverse are the following:
-u and -6-u
following the process of the answer in the following question:
Finding the additive inverse in a vector space with unusual operations
Which will give me -3 and -3 for the vector.
For some reason it is still wrong.
I think I am misunderstanding something
Update: (u1,u2) element of V, (v1,v2) element of V: Addition is defined as (u1+v1-3, u2+v2+3) and not as simply (u1+v1, u2+v2). So I can't just say that the additive inverse is -(u2,u2). If I set u1+v1-3=3 and u2+v2+3=-3, I get v1=-u1 and v2 = -6-u2 which would represent the i functions as mentioned above. However, they don't seem to work.
The first several are correct but you seem to have misunderstood the last question. You are asked for the additive inverse of (x, y) which will depend on x and y, not constants. (x, y)+ (p, q) is defined as (x+ p- 3, y+ q+ 3). Since the additive identity is (3, -3) we must have x+ P- 3= 3 and y+ q+ 3= -3 so p= 6- x and q= -y- 6.
The additive inverse of (x, y) is (6- x, -y- 6).