Question

show that the set x with the given operations fails to be a vector space by identifying all axioms that hold and fail
the set X=R with vector addition + defined by (a,b,c)=(x,y,z)=(1,y,c+z) and a scalar multiplecation . defined by k.(a,b,c)=(ka,kb ,kc)

Answer 1

I assume you mean vector addition is defined by\[(a,b,c)+(x,y,z)=(1,y,c+z)\]and scalar multiplication is defined by\[k(a,b,c)=(ka,kb,kc)\]now we have ten properties to check to see why this is not a vector space.
they are at the to of this page
http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx
so let's do a few interesting ones and I'll leave the rest to you to check

Answer 2

let\[\vec u=(a,b,c)\text{ and }\vec v=(x,y,z)\]right away we can see that\[\vec u+\vec r\in\mathbb R\]and\[ku\in\mathbb R\]so these are closed under addition and multiplication

so far it looks like a vector space...

Answer 3

It's not very clear from your post, but it looks like the rule for vector addition leads to this:\[ u+ v=(1,y,c+z)\]\[v+u=(1,b,z+c)\]which means that\[ u+ v\neq v+ u\]So we have now shown it's not a vector space. It also looks to me like\[(u+v)+w\neq u+(v+w)\]for a similar reason, but It's hard to say with just what we're given.

Answer 4

The last one I will demonstrate is\[u-u=u+(-u)=(1,-b,c-c)=(1,-b,0)\neq\vec 0\]because we can't get rid of the 1 by our definition of vector addition.

I'll leave it to you to verify that the following properties are or are not fulfilled:

e) no
g) no
h) no
i) yes
j) yes

Answer 5

thank you very much