exterior product definition i have question from  vector mathematics,i know that if there is given two vector,   for instance $a=\{a_1,a_2,a_3\}$,$b={b_1,b_2,b_3}$; then  so called exterior product is  determined as $a\wedge b=\frac{1}{2}(ab-ba)$, but I am confused. My question is what is $ab$ or $ba$? is it scalar? in this case $ab=ba$  and $a\wedge b=0$   which does not have any meaning  or $ab=(b_1-a_1,b_2-a_2,b_3-a_3)$ or vector?  thanks
Update:
consider for example following equation
$$[a,b][X,Y]=a*x+b*y=c$$
then solution is given by   following    equalities
$$[ X,Y]=\frac{1}{a\wedge b}[c\wedge b,a\wedge c]$$
 A: I suspect that what you mean is $a\wedge b = 1/2(a\otimes b-b\otimes a)$, in which case $a\otimes b$ is the tensor product of $a$ and $b$.
$a\otimes b$ is an element of a new vector space that is higher-dimensional than the vector space containing $a$ and $b$. You can think about it as just a formal construction (basically it is mostly like the pair $(a,b)$), it cannot be simplified further, and satisfies some linearity properties like $(a+b)\otimes c = a\otimes c + b \otimes c$.
A: I'm struggling to interpret your equation, especially as you wrote down two three-dimensional vectors. Here's what I think is meant: you are working in a two-dimensional space. You have unknown scalars $x, y$, and column vectors $a, b, c$. You want to solve $A\begin{pmatrix}x \\ y\end{pmatrix} = c$, where the $2 \times 2$ matrix $A$ has columns $a$ and $b$. All you're doing is inverting $A$ (when that is possible) to get $\begin{pmatrix}x \\ y\end{pmatrix} = A^{-1}c$. If $a = \begin{pmatrix}a_1 \\ a_2\end{pmatrix}$ and $b = \begin{pmatrix}b_1 \\ b_2\end{pmatrix}$, then defining a scalar (see my comment above)
$$
a \wedge b = \frac{1}{2}(a_1b_2 - b_1a_2) = \frac{1}{2}\det A
$$
makes this work. You'll also see that the $\frac{1}{2}$ plays no role in solving the system. Your formula is essentially Cramer's rule, using the fact that $\bigwedge^n (\mathbf{R}^n)$ is $1$-dimensional.
If solving matrix equations is your goal then thinking about the exterior product may just confuse things.
