Krdytkyu

I have the vectors $A=ahat e_x$ and $B=ahat e_y$, so
$$
Atimes B =
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix}=hat za^2
$$
Q1: But why is the following wrong
begin{align}
Atimes B &=ahat e_xtimes ahat e_y\
&=(ahat e_x)times (ahat e_y) tag 1\
&=(ahat e_x)times (hat e_ya) tag 2\
&=a(hat e_x)times (hat e_y)a tag 3\
&=a((hat e_x)times (hat e_y)) tag 4\
&=a(hat e_xtimes hat e_y) tag 5\
&=ahat e_z quad ?
end{align}

Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}

$$=a(hat e_x)times (hat e_y)a tag 3$$
$$=a((hat e_x)times (hat e_y)) tag 4$$

What happens here...?

You may be mixing it up with the distributive property of the cross product over addition.

Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}

For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
$$begin{vmatrix}
a & b & c \
color{purple}{k}d& color{purple}{k}e & color{purple}{k}f \
g & h & i
end{vmatrix}=color{purple}{k}begin{vmatrix}
a & b & c \
d& e & f \
g & h & i
end{vmatrix}$$
This means that if you want to factor out the $a$, you do that for the second and the third row:
$$begin{vmatrix}
hat e_x &hat e_y & hat e_z \
color{blue}{a}& 0 & 0\
0 & color{red}{a} &0
end{vmatrix}=color{blue}{a}color{red}{a}begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\
0 & 1 &0
end{vmatrix}=a^2begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\
0 & 1 &0
end{vmatrix}$$

We have that by cross product definition

$$Atimes B =ahat e_xtimes ahat e_y=a^2( e_xtimes hat e_y)$$

and by the properties of the determinant

$$Atimes B =
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} =
a^2
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}$$

Your first argument is wrong because you lost an $a$ between lines (3) and (4).

Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.