Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
        \begin{pmatrix}
        1 & 1 & 2 & -1 \\
        0 & 2 & 1 & 3 \\
        1 & 1 & 2 & -1 \\
        \end{pmatrix}X = \begin{pmatrix}
        1 & 1 & 1 \\
        0 & 2 & 0 \\
        2 & 1 & 1 \\
        \end{pmatrix}
$$
When I try to peform Gaussian elimination to get the reduced form of A, I always get a row of zeroes, e.g:
\begin{pmatrix}
        1 & 1 & 2 & -1 \\
        0 & 2 & 1 & 3 \\
        1 & 1 & 2 & -1 \\
        \end{pmatrix}
$$ R_3 - R_1 \to R_3 $$
I get
\begin{pmatrix}
        1 & 1 & 2 & -1 \\
        0 & 2 & 1 & 3 \\
        0 & 0 & 0 & 0 \\
        \end{pmatrix}
What can I conclude from the fact that I got a zeroes row? 
Does this help solving the problem?

In this particular case it'd perhaps be easier to understand what's going on performing column elementary operations:
$$\begin{pmatrix}
        1 & 1 & 2 & -1 \\
        0 & 2 & 1 & 3 \\
        1 & 1 & 2 & -1 \\
        \end{pmatrix}\stackrel{C_2-C_1\,,\,\,C_3-2C_1\,,\,C_4+C_1}\longrightarrow\begin{pmatrix}
        1 & 0 & 0 & 0 \\
        0 & 2 & 1 & 3 \\
        1 & 0 & 0 & 0 \\
        \end{pmatrix}\stackrel{C_3-\frac12C_2\,,\,C_4-\frac32C_2}\longrightarrow$$
$$\begin{pmatrix}
        1 & 0 & 0 & 0 \\
        0 & 2 & 0 & 0 \\
        1 & 0 & 0 & 0 \\
        \end{pmatrix}$$
The above means that
$$\begin{align}&C_3-2C_1-\frac12(C_2-C_1)=0\implies &C_3=\frac32C_1+\frac12C_2\\
&C_4+C_1-\frac32(C_2-C_1)=0\implies &C_4=-\frac52C_1+\frac32C_2\end{align}$$
and the above means that any vector $\;(x_\;x_2\;x_3\;x_4)^t\;$in the image of $\;A\;$ has to fulfill the same relations as shown, meaning: 
$$\begin{align}&x_3=\;\;\;\,\frac32x_1+\frac12x_2\\&x_4=-\frac52x_1+\frac34x_2\end{align}$$
Now, does you right side matrix (formed with three row (column) vectors) fulfill this?
