Can vectors be inverted? I wish to enquire if it is possible to solve the below for $C$.
$$B^{-1}(x-\mu) = xc $$
Here obviously $B$ is an invertible matrix and both $c$ and $\mu$ are column vectors.
Would the solution be $$x^{-1}B^{-1}(x-\mu) = c $$ is it possible to invert vectors ?
How about if it was the other way
$$B^{-1}(x-\mu) = cx $$
Is there any other way to do this ?
Thanks in advance. 
 A: $$B^{-1}(x-\mu) = xc $$
If the matrix $B$ is known, why not write
$$
(B^{-1} - cI)x = B^{-1}\mu
$$
and then worry about whether $B^{-1}-cI$ is invertible?
A: Vectors, in general, can't be inverted under matrix multiplication, as only square matricies can inverses. However, in the situation you've described, it's possible to compute $c$ anyway, assuming the equation is satisfied for some $c$. If we multiply both sides by $X^T$, the result is $x^T B^{-1} (x-\mu) = x^T x c = |x|^2 c$. Assuming x is not the zero vector (in which case any $c$ will still have $xc=0$ so any choice of $c$ should work), we just get $c= \frac{1}{|x|^2} x^T B^{-1} (x-\mu)$.
I must caution that the expression above for $c$ is defined even when there is no solution of the original equation, which will be almost all of the time for randomly generated vectors and matrices. Hence, if you are going to use it, you should check that this works by plugging what you get for $c$ back into the original expression and see if it works. Also, the choice of $x^T \over |x|^2$ is not unique; any row vector $v$ such that $vx=1$ will work equally well in the above expression.
A: There's no such thing as an inverse of a vector (unless the vector is actually a $1\times 1$ vector, of course).
Otherwise, there would be a solution $C$ for any $B,X,\mu$ (or at least any $X$ "invertible"), but that is obviously not the case (e.g. for any $X$ if we put $B=I$, $\mu$ linearly independent from $X$, there is no $C$).
A: Perhaps this definition of the inverse vector will help you:
An inverse rectilinear vector   ā'  is a vector which is co-directed (in the same direction as) a vector ā and differs from it in magnitude according to: 
$$ |\bar{a'}|=\dfrac{1}{|\bar{a}|} $$
Projections on the coordinate axes of inverse rectilinear vectors are equal according to:
$$\quad a'_{x}=\dfrac{a_{x}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}};\quad a'_{y}=\dfrac{a_{y}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}\quad a'_{z}=\dfrac{a_{z}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}} $$
An example of solving problems with this vector can be found here https://en.wikipedia.org/wiki/Talk:Cross_product#Cross_product_does_not_exist
https://doi.org/10.5539/jmr.v9n5p71
A: In a quantum mechanics book, I've seen it defined this way:
If we have: $V= [v_1, v_2  ... v_n],$ then:
$$V^{-1} = [v_1^\dagger, v_2^\dagger, ... v_n^\dagger].$$
