Inverting a vector If I have $Ax=b$ where $A$ is $n$ by $n$ while $x$ and $b$ are $n$ by 1, is it possible to find $A$ given $x$ and $b$. The idea would be some sort of $x^{-1}$ operation on the right of both equations but I'm not sure how to go about it. If necessary, we can consider a specific case of $n=2$.
Also, is the $A$ unique or not?
 A: There is no unicity: Let $y$ be a vector orthogonal to $x$, i.e. $x^\mathrm{T} y = 0$. For any $A^\ast$ solution, look at $A\stackrel{\rm{}def}{=} A^\ast+ yy^\mathrm{T}$. Then $Ax=b$ as well.
To find a solution $A$, you can write the problem as an (underspecified) linear system of $n$ equations with $n^2$ unknowns. See e.g. this.
A: No, inverse in this case is impossible, because you have a lot of solutions, for example consider:
$A\begin{bmatrix} 1 \\ 0\end{bmatrix}=\begin{bmatrix} 1 \\ 0\end{bmatrix}$
$A=\begin{bmatrix}1 && 0 \\ 0 && 1\end{bmatrix}$is one solution, but for example $A=\begin{bmatrix}1 && 1 \\ 0 && 1\end{bmatrix}$ is solution, too.
A: No, you need $n$ independent vectors $x$ to find $A$, in which case you would have one expression of $A$ in the base of you vectors $x_1,...x_n$.
A: An easy answer is pick a component of $x$, say $j$, that is not equal to zero (If $x = 0$, then $b$ better be zero; otherwise no solution). Then set the $j$th column of A, $A_j = b/x_j$. Choose the rest of $A$ matrix to be zero. 
