Let V be the set of vectors in R2 with the following definition of addition and scalar multiplication:

Addition: [x1x2]?[y1y2]=[0x2+y2]
Scalar Multiplication: ??[x1x2]=[0?x2]

answer yes or no

x?y=y?x for any x and y in V

(x?y)?z=x?(y?z) for any x, y and z in V

There exists an element 0 in V such that x?0=x for each x?V

For each x?V, there exists an element ?x in V such that x?(?x)=0

??(x?y)=(??x)?(??y) for each scalar ? and any x and y V

(?+?)?x=(??x)?(??x) for any scalars ? and ? and any x?V

(??)?x=??(??x) for any scalars ? and ? and any x?V

1?x=x for all x?V

See below

Step-by-step explanation:

I will denote your new addition by +', and your new scalar multiplication by * to distinguish them from the usual operations.

1) x+'y=y+'x. Yes: Let x=[a,b] and y=[c,d].  Then x+'y=[0,b+d]=[0,d+b]=y+'x

2) (x+'y)+'z=x+'(y+'z). Yes: Let x=[a,b], y=[c,d], z=[e,f]

Then (x+'y)+'z=[0,b+d]+'[e,f]=[0,(b+d)+f]=[0,b+(d+f)]=[a,b]+'[0,d+f]=x+'(y+'z)

3) There exists an element 0 in V such that x+'0=x for each x in V: No

Take x=[1,0]. Then x+'[a,b]=[0,b]≠[1,0], so 0=[a,b] does not exist (no choice of 0 works for this x)

4) For each x in V, there exists an element -x in V such that x+'(-x)=0. Yes (if 0=[0,0])

Given x=[a,b], take -x=[a,-b]. Then x+'(-x)=[0,b-b]=[0,0]=0.

5) *k(x+y)=(*kx)+'(*ky) for each scalar k and any x and y in V. Yes

Let x=[a,b] and y=[c,d]. Then *k(x+'y)=*k[0,b+d]=[0,k(b+d)]=[0,kb]+[0,kd]=*kx+*ky.

6) (s+t)*x=(*sx)+(*tx) for any scalars s and t and any x in V. Yes

Let x=[a,b]. Then (s+t)*x=[0,(s+t)b]=[0,sb]+[0,tb]=*sx+*tx

7) (*s)(*tx)=*(st)x for any scalars s and t and any x in V. Yes

Let x=[a,b]. Then (*s)(*tx)=*s[0,tb]=[0,stb]=*(st)x

8) *1x=x for all x in V. No

*1[1,2]=[0,2]≠[1,2]

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

go on the tube you might find some information there