Let V denote the set of ordered triples (x, y, z) and define addition in V as inR3.For each of the following definitions of scalar multiplication, decide whether V is a vector space.(a) a(x,y,z)=(ax,y,az)(b) a(x,y,z)=(ax,0,az)(c) a(x,y,z)=(0,0,0)(d) a(x,y,z)=(2ax,2ay,2az)

Answer:a) Nob) Noc) Nod) NoStep-by-step explanation:Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:1. u+v∈V2. u+v=v+u3. (u+v)+w=u+(v+w).4. Exist 0∈V such that u+0=u5. For each u∈V exist −u∈V such that u+(−u)=0.6. if c is an escalar and u∈V, then cu∈V7. c(u+v)=cu+cv8. (c+d)u=cu+du9. c(du)=(cd)u10. 1u=ulet's check each of the properties for the respective operations:Let [tex]u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)[/tex]Observe that  1. u+v∈V2. u+v=v+u, because the adittion of reals is conmutative3. (u+v)+w=u+(v+w). because the adittion of reals is associative4. [tex](u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)[/tex]5. [tex](u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)[/tex]then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.a) 6. [tex]c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V[/tex]7. [tex]c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv[/tex]8.[tex](c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du[/tex]Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(ax,y,az)[/tex]b)  6. [tex]c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V[/tex]7. [tex]c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv[/tex]8. [tex](c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du[/tex]9.[tex]c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u[/tex]10[tex]1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)[/tex]Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(ax,0,az)[/tex]c) Observe that [tex]1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)[/tex]Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(0,0,0)[/tex].d)  Observe that [tex]1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u[/tex]Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(2ax,2ay,2az)[/tex].