obtain matrix $ A$ if $X$ and $b$ given 
for the right and detailed answer refer to user9077 answer
 A: Yes, your work is correct, although you might want to explain why you can write
$$
\pmatrix{a & b \\ c & d}\pmatrix{0 \\ 1} = \pmatrix{0\\0},
$$
even if it seems obvious to you. 
Also: the "MathJax" that I typed to produce this answer was exactly this:
$$
\pmatrix{a & b \\ c & d}\pmatrix{0 \\ 1} = 
\pmatrix{0\\0},
$$

You should, if you're going to stick around here, learn how to format things like this. 
A: Since $\begin{pmatrix}1\\0\end{pmatrix}+c\begin{pmatrix}0\\1\end{pmatrix}$ are all solutions to $Ax=\begin{pmatrix}1\\3\end{pmatrix}$, then we have
$$
\begin{pmatrix}1\\3\end{pmatrix}=Ax=A\begin{pmatrix}1\\0\end{pmatrix}+cA\begin{pmatrix}0\\1\end{pmatrix}.
$$
The above is true for any $c$. This can only happen when $A\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$. From this we know that the second column of $A$ is $\begin{pmatrix}0\\0\end{pmatrix}$.
We also have $A\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}$. So the first column of $A$ is $\begin{pmatrix}1\\3\end{pmatrix}$. Therefore
$$A=\begin{pmatrix}1&0\\3&0\end{pmatrix}.$$
A: It is basically correct, but you can do it more easily.
If $a_1$ and $a_2$ are the columns of $A$, then you know that
$$
1a_1+ca_2=\begin{bmatrix} 1 \\ 3 \end{bmatrix}
$$
for every $c$; in particular this holds for $c=0$ and $c=1$, so
\begin{cases}
a_1=\begin{bmatrix} 1 \\ 3 \end{bmatrix} \\[4px]
a_1+a_2=\begin{bmatrix} 1 \\ 3 \end{bmatrix}
\end{cases}
