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?
 A: The rows of the LHS will be given by the rows of $A$, multiplied by $X$.  Since the first and third rows of $A$ are the same, the first and third rows of the product will be the same.  Therefore the product cannot equal the RHS.
A: You can argue by contradiction, using the relation 
$$\text{rank}(AB) \le \text{min}(\text{rang}(A),\text{rang}(B))$$
with $A= \begin{pmatrix}1 & 1 & 2 & -1 \\
        0 & 2 & 1 & 3 \\
        1 & 1 & 2 & -1 \\
        \end{pmatrix}$ and $B:=X$ a $4\times 3$ matrix with $\text{rank}(X)\le 3$.
Since then you would have
$$\text{rank}(AX) \le \text{min}(\text{rang}(A),\text{rank}(X)) \le \text{min}(2,3)=2,$$
while $\text{rank}\left(\begin{pmatrix}
        1 & 1 & 1 \\
        0 & 2 & 0 \\
        2 & 1 & 1 \\
        \end{pmatrix} \right)=3,$ since the determinant of the last matrix is different from $0$.
A: Hint : let there exist $X_{4 \times 3}$ matrix satisfiy the equaltiy then make a contradiction whit entries  of $X$
A: 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?
A: You have shown that the rank of your $3\times 4$ matrix is only (at most)$~2$ (since only two nonzero rows are left). On the other hand your $3\times 3$ matrix is invertible (easy computation), so it has rank$~3$. A matrix product $AB$ cannot have rank larger than the rank of either $A$ or $B$, which shows that no such matrix $X$ exists.
