Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
To solve this problem, you can’t let the strangeness of the jargon intimidate you. Rather, just accept the definition of the commutator without trying to understand it too much right away in the abstract; then plug in specific operations one at a time.
For instance, in the first statement, the operation x?y is defined as xy (the product of x and y). Since the commutator is generally defined as x?y – y?x, we can write an equation:
Commutator = x?y – y?x
Substitute the first statement’s definition. Note that y?x = yx (we switch the positions of the variables).
Commutator = xy – yx
Since xy equals yx, we get the following:
Commutator = xy – xy = 0
This means that for the operation of multiplication, the commutator always equals zero, for all values of x and y. As a result, the first statement does not fit the question, which asks for cases in which the commutator does not equal zero for some values of x and y. We can now rule out answer choices (A) and (E).
Next, we consider statement II. We write the definition of the commutator again, but we substitute the second definition of ?.
Commutator = x?y – y?x
We substitute the second statement’s definition, switching the x and y to replace y?x.
Commutator = (x – y)2 – (y – x)2
Commutator = x2 – 2xy + y2 – (y2 – 2yx + x2)
Commutator = x2 – 2xy + y2 – y2 + 2yx – x2
Commutator = 0
By now, we might notice that the commutator is always zero when x?y = y?x, or in other words, when switching x and y in the operation gives you the same result. For instance, with multiplication (in the first statement), you can write xy or yx to represent the same thing. In the second case, (x – y)2 always equals (y – x)2, since both expressions expand to equal x2 – 2xy + y2. Asking whether the commutator is zero is a fancy way of asking whether switching the variables gives you the same result.
Armed with this insight, we can examine the third statement’s definition:
x?y = x3 – 3x2y + 3xy2 – y3
Although this expression is complicated, we can see that if we switch x and y, we do NOT get the same expression, after rearrangement:
y?x = y3 – 3y2x + 3yx2 – x3
Notice, for instance, that in x?y, the x3 term is positive, but in y?x, that same term is negative. The overall expressions will only be equal when x = y.
Thus, for this operation, the commutator x?y – y?x will NOT always be zero, and this definition fits the question.
The correct answer is (C): III only.
