Before you read my question please consider that i HAVE to do this exercise as i did below.
So i am showing there exists no division algebra on $\mathbb R^3$.
To show that, i have to show that for $*: \mathbb R^3 \times \mathbb R^3 \to \mathbb R^3$ it exists $0 \ne x,y \in \mathbb R^3$ with $x*y = 0$ $(1)$
I first showed that there exists a linear matrix $A(x) \in \mathbb R^{3 \times 3}$ with $A: x \to A(x)$.
Then i did show that $det(A(x)) = 0$ for some $x \in \mathbb R^3 \setminus \{0\}$
At last i defined $L_x:\mathbb R^3 \to \mathbb R^3, y \to x*y$ and i showed it is linear and it satisfies $L_x(y) = A(x).y$.
Now to my problem:
I have to use what i did here and the fact that a Matrix $A$ is invertible if $Ay \ne 0$ for all $y \in \mathbb R^3 \setminus \{0\}$.
How and Why follows $(1)$ from what i did here.
I am unable to make a connection between what i did and what i shall show ( maybe because i am still newbie ).