#17
Let * be the binary operation on the rational numbers given by a * b = a + b + 2ab, which of the following are true?
I: * is commutative
II: There is a rational number that is a *indentity
III: Every rational number has a *inverse
The answer is C: I and II.
Obviously, * is commuatative, but I have problem finding the indentity. Can anyone show me?
#18
A group G in which (ab)^2 and a^2 b^2 for all a, b in G is necessarily
(A) finite
(B) cyclic
(C) of order two
(D) abelian
(E) none of the above.
The answer is D.
Can anyone explain this to me?
#35
The rank of the matrix is?
 1 2 3 4 5 
 6 7 8 9 10
 11 12 13 14 15
 16 17 18 19 20
 21 22 23 24 25
The answer is 2.
Can anyone show me?
Form 8767: #17, #18 and #35
Re: Form 8767: #17, #18 and #35
For #17, the identity cannot be zero right? since zero is not a rational number by defintion.
For #35, according to theorem, the dimension of the col space and row space is always equal. Meaning that 3 of the col vectors are linear combination of the other two. Can anyone show me how this linear combination works?
Thanks
For #35, according to theorem, the dimension of the col space and row space is always equal. Meaning that 3 of the col vectors are linear combination of the other two. Can anyone show me how this linear combination works?
Thanks

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Form 8767: #17, #18 and #35
For 17, I have no idea what definition you're looking at... zero is certainly a rational number. For III, try 1/2. That will screw things up

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Form 8767: #17, #18 and #35
For 18, just write out what it means for (ab)^2 = a^2b^2
abab = aabb
Hit both sides from the left with a^1 and from the right by b^1
ba=ab
abab = aabb
Hit both sides from the left with a^1 and from the right by b^1
ba=ab

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Form 8767: #17, #18 and #35
For 35, you can do row operations and it preserves the rank. For each row i , 1 <i </= 5, subtract row i1. This will give you a matrix:
1 2 3 4 5
5 5 5 5 5
5 5 5 5 5
5 5 5 5 5
5 5 5 5 5
And then the answer is obvious
1 2 3 4 5
5 5 5 5 5
5 5 5 5 5
5 5 5 5 5
5 5 5 5 5
And then the answer is obvious
Re: Form 8767: #17, #18 and #35
Thank you very much, got it !!