Q:

Why can't we have these alternative cross product relationships?

I am currently studying Introduction to Electrodynamics, fourth edition, by David J. Griffiths. Chapter 1.1.2 Vector Algebra: Component Form says the following:

$$\mathbf{\hat{x}} \times \mathbf{\hat{x}} = \mathbf{\hat{y}} \times \mathbf{\hat{y}} = \mathbf{\hat{z}} \times \mathbf{\hat{z}} = \mathbf{0},$$
$$\mathbf{\hat{x}} \times \mathbf{\hat{y}} = - \mathbf{\hat{y}} \times \mathbf{\hat{x}} = \mathbf{\hat{z}},$$
$$\mathbf{\hat{y}} \times \mathbf{\hat{z}} = -\mathbf{\hat{z}} \times \mathbf{\hat{y}} = \mathbf{\hat{x}},$$
$$\mathbf{\hat{z}} \times \mathbf{\hat{x}} = -\mathbf{\hat{x}} \times \mathbf{\hat{z}} = \mathbf{\hat{y}}.$$
These signs pertain to a right-handed coordinate system ($x$-axis out of the page, $y$-axis to the right, $z$-axis up, or any rotated version thereof). In a left-handed system ($z$-axis down), the signs would be reversed: $\mathbf{\hat{x}} \times \mathbf{\hat{y}} = - \mathbf{\hat{z}}$, and so on. We shall use right-handed systems exclusively.

I don't understand why we can't also have that
$$\mathbf{\hat{x}} \times \mathbf{\hat{y}} = \mathbf{\hat{y}} \times - \mathbf{\hat{x}} = \mathbf{\hat{z}},$$
$$\mathbf{\hat{y}} \times \mathbf{\hat{z}} = \mathbf{\hat{z}} \times - \mathbf{\hat{y}} = \mathbf{\hat{x}},$$
$$\mathbf{\hat{z}} \times \mathbf{\hat{x}} = \mathbf{\hat{x}} \times - \mathbf{\hat{z}} = \mathbf{\hat{y}}.$$
I would greatly appreciate it if people would please take the time to explain this.

A:

We do also have those "alternative cross product relationships" as well. 
The cross-product is compatible with scalar multiplication 
(in particular, multiplication by $-1$),
so $\mathbf{-\hat y}\times\mathbf{\hat x}=\mathbf{\hat y}\times -\mathbf {\hat x}$, for example.

