<s>[INST] 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?
 [/INST] 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 $0$.
</s>