Form 8767: #17, #18 and #35

Forum for the GRE subject test in mathematics.
YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Form 8767: #17, #18 and #35

Postby YKM » Mon Aug 29, 2011 2:22 am

#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?

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: Form 8767: #17, #18 and #35

Postby YKM » Mon Aug 29, 2011 2:27 am

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

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #17, #18 and #35

Postby blitzer6266 » Mon Aug 29, 2011 3:33 am

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

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #17, #18 and #35

Postby blitzer6266 » Mon Aug 29, 2011 3:48 am

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

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #17, #18 and #35

Postby blitzer6266 » Mon Aug 29, 2011 3:54 am

For 35, you can do row operations and it preserves the rank. For each row i , 1 <i </= 5, subtract row i-1. 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

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: Form 8767: #17, #18 and #35

Postby YKM » Mon Aug 29, 2011 3:58 am

Thank you very much, got it !!




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