How to solve Ax = b for A
Here you actually do not have a system of linear equations that needs to be seen at a whole and solved together. Here you have 3 independent equations, each of them with infinite valid answers. So:

$\begin{bmatrix}
a_{1} &amp; a_{2} &amp; a_{3}\\
a_{4} &amp; a_{5} &amp; a_{6}\\
a_{7} &amp; a_{8} &amp; a_{9}
\end{bmatrix}\times \begin{bmatrix} x_{1}\\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} b_{1}\\ b_{2} \\ b_{3} \end{bmatrix}$ 

equals solving 

$a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3} = b_{1}$

and

$a_{4}x_{1} + a_{5}x_{2} + a_{6}x_{3} = b_{2}$

and

$a_{7}x_{1} + a_{8}x_{2} + a_{9}x_{3} = b_{3}$

independently.

Each of them in for example in $\mathbb{R}$ has infinite right answers. For instance in $a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3} = b_{1}$ you just need to randomly choose two of $a_{i}$s and simply get the last one. For example for known $x$ vector $\begin{bmatrix} 1\\ 2 \\ 3 \end{bmatrix}$, $b$ vector $\begin{bmatrix} 4\\ 5 \\ 6 \end{bmatrix}$ and $a_{1}=1$ and $a_{2}=2$, we get $a_{3}$:

$5 + 3\times a_{3} = 4$ 

so 

$a_{3}=-\frac{1}{3}$!

Same procedure applies to other two equations as well. Randomly choose two $a_{i}$s for each equation and get the last one. 

PS: I feel either something is missing in your question or I did not understand well. In either cases, please drop a comment here and I will update the answer.