Finding the matrix $X$ in $AX$ 
Given that $A$ is a $3$ x $3$ matrix such that $$ A \begin{pmatrix}
        1 \\
        0 \\
        0 \\
        \end{pmatrix} = \begin{pmatrix}
        1 \\
        1 \\
        1 \\
        \end{pmatrix} $$ and $$ A \begin{pmatrix}
        0 \\
        0 \\
        1 \\
        \end{pmatrix} = \begin{pmatrix}
        0 \\
        0 \\
        1 \\
        \end{pmatrix} $$
Find a matrix $X$ such that 
$$A X = \begin{pmatrix}
        1 & 0 & 4 \\
        1 & 0 & 4 \\
        1 & 0 & 7 \\
        \end{pmatrix}$$

How do I approach this question? From the two matrix equations given above, I can find the first and the third columns, but I don't think I have the second column to find matrix $A$. Thanks. 
 A: You can take$$X=\begin{pmatrix}1&0&4\\0&0&0\\0&0&3\end{pmatrix}.$$
And how do I know that? Note that$$X.\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}\implies AX\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix},$$that$$X.\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\implies AX.\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$and that$$X.\begin{pmatrix}0\\0\\1\end{pmatrix}=4\begin{pmatrix}1\\0\\0\end{pmatrix}+3\begin{pmatrix}0\\0\\1\end{pmatrix}\implies AX.\begin{pmatrix}0\\0\\1\end{pmatrix}=4\begin{pmatrix}1\\1\\1\end{pmatrix}+3\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}4\\4\\7\end{pmatrix}.$$
A: The first column of your result matrix (let's note $AX=Y$, and $Y_i$ the columns of Y), will be $A X_1$, ($X_1$ is the first column of X), the second $Y_2  =A X_2$ and $Y_3 = A X_3$.
Then you want $Y_1 = [1 1 1]$, so you know that $X_1=[100]$ works. 
To get $Y_2=[000]$ will be easy, you just need $X_2$ to be null everywhere.
Then you have to find $X_3$ so that $Y_3=A X_3$, you know you can have $[444]=A[400]$.
Then you just need to add $[003]=3* A[001]$.
So finally $X_3=[403]$ works.
