# Thread: Commutative and Associative properties.

1. ## Commutative and Associative properties.

* is any operation. (x*y)=x+2y+4 is an operation for all Real numbers.

To prove if it is commutative (x*y)=(y*x)
so (x*y)=x+2y+4; (y*x)=2y+x+4 ?

so (x*y)=(y*x)

Is this a correct way to show it is commutative?

2. ## Re: Commutative and Associative properties.

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.

3. ## Re: Commutative and Associative properties.

So this equation is not commutative because x and y can be different numbers therefore (x*y) does not equal (y*x)?

4. ## Re: Commutative and Associative properties.

because x and y can be different numbers
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.

5. ## Re: Commutative and Associative properties.

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?

6. ## Re: Commutative and Associative properties.

off the top of my head, the identity matrix, the zero matrix,...

7. ## Re: Commutative and Associative properties.

$\text{The operation }\star\text{ is such that: }\:x\,\star\,y\;=\;x+2y+4\,\text{ for all real numbers.}$

$\text{Is }\star\text{ commutative?}$

First, note what the definition says: . $x\,\star\,y \;=\;x+2y + 4$
. . the sum of the first number, twice the second number, and four.

So that: . $2\,\star\,5 \;=\;2 + 2(5) + 4 \;=\;16$

n While: . $5\,\star\,2 \;=\;5 + 2(2) + 4 \;=\;13$

The operation $\star$ is not commutative.

8. ## Re: Commutative and Associative properties.

Thanks for that!
I've also heard that it's true for diagonal matrices and if the matrices are inverses of each other. Is that so?

9. ## Re: Commutative and Associative properties.

Hello again, JacobE!

I know that matrix multiplication is not commutative in general, that is, $AB \ne BA.$
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: . $A \:=\:\begin{pmatrix}1 &2 \\ 3&4 \end{pmatrix}\qquad B \:=\:\begin{pmatrix}7&4 \\ 6&13 \end{pmatrix}$

10. ## Re: Commutative and Associative properties.

Appreciate it!

11. ## Re: Commutative and Associative properties.

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.