Let V be the set of vectors in R2 with the following definition of addition and scalar multiplication: Addition: [x1x2]?[y1y2]=[

Question

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

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Madeline 3 weeks 2021-12-30T23:16:51+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-12-30T23:18:00+00:00

    Answer:

    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]

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