That is a good way to go, but the variables may be confusing you.
(left*right) = left + 2right + 4
(x*y) = x + 2y + 4
(y*x) = y + 2x + 4 -- A little different from your result.
I think you're on the right track, but I don't know what that means.
It's not commutative because we demonstrated that it wasn't. (x*y) != (y*x)
Well, on second thought, it has nothing to do with our demonstration. It's not commutative because it isn't. We managed to demonstrate that this was so.
Without wanting to start a new thread, I'd like to ask something about commutatitve properties in the case of matrix multiplication.
I know, matrix multiplication is not commutative in general, that is, AB ≠ BA. Yet there are apparently cases when matrix multiplication is commutative. In which case is that?
Hello again, JacobE!
I know that matrix multiplication is not commutative in general, that is,
Yet there are apparently cases when matrix multiplication is commutative.
In which cases is that?
It is difficult to create a general rule for this,
but here is an example: .
One can show generally that two matrices are commutative if they have the same eigenvectors. Essentially the reason is that we can make them both diagonal (or at least in Jordan Normal Form) in the same basis- using their mutual eigenvectors as basis. The fact that operators having the same eigenvectors commute is important in Quanum Mechanics.