Solving $Ax=B$ where $x$ is a $3\times 3$ matrix I've often seen questions regarding how to find matrix $x$ in $Ax = B$, but almost all of these include $x$ as a $3\times 1$ matrix. 
I'm trying to solve this question:

Find a matrix x that satisfies the following equation: $x ⋅
\begin{bmatrix}2 \\ 0 \\2\end{bmatrix} = \begin{bmatrix}2 \\ 0.5 \\ 1.5\end{bmatrix}$

So my thought was that x is supposed to be a $3\times 3$ matrix, otherwise you won't get $B$. I am however at a loss on how to solve the equation for such a matrix.
I tried writing this as a system of equations, which gives me:
$$ \begin{bmatrix}a & b & c \\ d & e& f \\g & h & i\end{bmatrix} ⋅ \begin{bmatrix}2 \\ 0 \\2\end{bmatrix} =\begin{bmatrix}2 \\ 0.5 \\ 1.5\end{bmatrix}$$
$$2a + 2c = 2 \\
2d + 2f = 0.5 \\
2g + 2i = 1.5$$
but I really don't think this is right. I also wouldn't know how to go from here. I tried row reduction on the augmented matrix of this system but it gave me the wrong answer.
If anyone could guide me in te right way, I would really appreciate it!
 A: You have $9$ unknowns, and $3$ equations for the system:
$$
\begin{bmatrix}
2 \\
1/2 \\
3/2
\end{bmatrix}
=
x 
\begin{bmatrix}
2 \\
0 \\
2
\end{bmatrix}
=
\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
\begin{bmatrix}
2 \\
0 \\
2
\end{bmatrix}
=
\begin{bmatrix}
2 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 2 \\
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c \\
d \\
e \\
f \\
g \\
h \\
i
\end{bmatrix}
$$
This means you have $6$ free variables.
There are a lot of $3\times 3$ matrices which satisfy this.
So let us pick an easy one.
$$
x =
\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
=
\begin{bmatrix}
x_{11} & x_{12} & x_{13} \\
x_{21} & x_{22} & x_{23} \\
x_{31} & x_{32} & x_{33}
\end{bmatrix}
$$
The middle column unknowns $x_{i2}$ do not show up, this means we can choose any value for them, so we set them to zero. $3$ free variables of the $6$ original ones determined, $3$ free variables to go.
Otherwise we have equations
$$
2 x_{i1} + 2 x_{i3} = b_i \quad (i \in \{1,2,3\})
$$
E.g. if we wish $x_{i1} = x_{i3}$ we end up with
$$
x_{i1} = x_{i3} = b_i/4
$$
This gives:
$$
x = 
\begin{bmatrix}
1/2 & 0 & 1/2 \\
1/8 & 0 & 1/8 \\
3/8 & 0 & 3/8
\end{bmatrix}
$$
Or if we instead wish $x_{i3} = 0$ we end up with
$$
x_{i1} = b_i/2
$$
This gives:
$$
x = 
\begin{bmatrix}
1 & 0 & 0 \\
1/4 & 0 & 0 \\
3/4 & 0 & 0
\end{bmatrix}
$$
A: You are correct in finding $$2a + 2c = 2 \\
2d + 2f = 0.5 \\
2g + 2i = 1.5$$
Do not worry if the solutions are not unique.
pick some values for $a$,$d$,and $g$ and solve for $c$,$f$,and $i$.
You will find many marices to work for your equation.
