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]