hello there!

I need some clarity on the following please:

The question asked that it be shown that the following set is not a vector space:

'The set of all triples of real numbers with addition defined by

(x, y, z) + (u, v, w) = (z+w, y+v, x+u)'

I worked out that the following axiom holds:

(k+m)u=ku +ku

however, the answer works it out tohold.not

It goes as follows:

Setu= (u1, u2, u3), then

(k+m)u

=(k+m)(u1, u2, u3)

= ((k+m)u1,(k+m)u2,(k+m)u3)

and

ku+ mu

=(ku1, ku2, ku3) + (mu1, mu2, mu3)

= (ku3 + mu3, ku2 + mu2, ku1+ mu1)

since these are not equal, this axiom does not hold.'

Now, I don't understand why these two are not equal?

Can we not further simplify:

(ku3 + mu3, ku2 + mu2, ku1+ mu1)

= ((k+m)u3,(k+m)u2,(k+m)u1)?

That way, does it not equal = ((k+m)u1,(k+m)u2,(k+m)u3))?

or is it not equal because the order of the triple is different?

I'm quite confused and any help would be much appreciated!

Have a great day all