Math  /  Algebra

QuestionLet V=R2V=\mathbb{R}^{2}. For (u1,u2),(v1,v2)V\left(u_{1}, u_{2}\right),\left(v_{1}, v_{2}\right) \in V and aRa \in \mathbb{R} define vector addition by (u1,u2)(v1,v2):=(u1+v1+3,u2+v21)\left(u_{1}, u_{2}\right) \boxplus\left(v_{1}, v_{2}\right):=\left(u_{1}+v_{1}+3, u_{2}+v_{2}-1\right) and scalar multiplication by a(u1,u2):=(au1+3a3,au2a+1)a \square\left(u_{1}, u_{2}\right):=\left(a u_{1}+3 a-3, a u_{2}-a+1\right). It can be shown that (V,,)(V, \boxplus, \square) is a vector space. Find the following: the sum: (5,5)(1,8)=(,)(-5,5) \boxplus(-1,8)=(\square, \square) the scalar multiple: 4-4 \square (5,5)=(-5,5)= \square \square ) the zero vector: \square 0=(,)\|0\|=(\square, \square) the additive inverse " v-v " of v=(x,y)v=(x, y) : v=(\prime \prime-v \prime \prime=( \square \square ) (Must be in terms of xx and yy )

Studdy Solution
To find the additive inverse "v""-v" of v=(x,y)v = (x, y), we need a vector (x,y)(-x', -y') such that:
(x,y)(x,y)=(0,0)(x, y) \boxplus (-x', -y') = (0, 0)
Using the vector addition operation:
(x,y)(x,y)=(xx+3,yy1)=(0,0)(x, y) \boxplus (-x', -y') = (x - x' + 3, y - y' - 1) = (0, 0)
Equating components:
xx+3=0x=x+3x - x' + 3 = 0 \quad \Rightarrow \quad x' = x + 3 yy1=0y=y1y - y' - 1 = 0 \quad \Rightarrow \quad y' = y - 1
Thus, the additive inverse is:
"v"=(x+3,y1)"-v" = (x + 3, y - 1)
The solutions are:
1. (5,5)(1,8)=(3,12)(-5, 5) \boxplus (-1, 8) = (-3, 12)
2. 4(5,5)=(5,23)-4 \square (-5, 5) = (5, -23)
3. Zero vector 0=(3,1)\|0\| = (-3, 1)
4. Additive inverse "v"=(x+3,y1)"-v" = (x + 3, y - 1)

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