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# 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

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## Answer to a math question 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

Eliseo
4.6
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)
= $-3, 2$

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)
= $-3, 2$

-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.

-u + u = $-4, 3$ ≠ 0
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