Let v be the set of all ordered pairs of real numbers and consider the scalar addition and multiplication operations defined by: u+v=(x,y)+(s,t)=(x+s+1,y+t -two) au=a.(x,y)=(ax+a-1,ay-2a+2) It is known that this set with the operations defined above is a vector space. A) calculate u+v is au for u=(-2,3),v=(1,-2) and a=2 B) show that (0,0) #0 Suggestion find a vector W such that u+w=u C) who is the vector -u D) show that axiom A4 holds:-u+u=0

A) To calculate u+v and a*u, we can substitute the values of u, v, and a into the given definitions of scalar addition and multiplication:

For u=(-2,3) and v=(1,-2), we have:
u+v = (-2,3) + (1,-2)
= (-2+1+1, 3+(-2)-2)
= (0,-1)

For a=2 and u=(-2,3), we have:
a*u = 2*(-2,3)
= (2*(-2)+2-1,2*3-2*2+2)
= (-3, 2)

Answer:
u+v = (0, -1)
a*u = (-3, 2)

B) To show that (0,0) ≠ 0, we need to find a vector w such that u+w = u for any u in the vector space.
By choosing w = (0,0), let's see what happens:
u + w = (-2,3) + (0,0)
= (-2+0+1, 3+0-2)
= (-1, 1)

Since (-1,1) ≠ (-2,3), we can conclude that (0,0) ≠ 0.

C) To find the vector -u, we can multiply u by -1:
-u = -1 * u = -1 * (-2,3)
= (2*(-2)+2-1, 2*3-2*2+2)
= (-3, 2)

Answer:
-u = (-3, 2)

D) To show that axiom A4 holds, we need to prove that -u + u = 0 for any u in the vector space.
Using the values of u=(-2,3), we can calculate -u:
-u = (-3, 2)

Now, let's calculate -u + u:
-u + u = (-3, 2) + (-2, 3)
= (-3+(-2)+1, 2+3-2)
= (-4, 3)

Since (-4, 3) ≠ 0, axiom A4 does not hold in this vector space.

Answer:
-u + u = (-4, 3) ≠ 0