Proving that there is no matrix $B$ for such $By = x$ Let $x = (2, 5, 4, 1, 3, 7), \;y = (6, 6, 7, 8)$
$$A = \begin{bmatrix}
1 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 
\end{bmatrix}$$
We know that for such defined $A$ we have $Ax = y$.
I'm looking for matrix $B$ which is associate with matrix $A$ that $By = x$.
My work so far
I want to prove that such matrix does not exist. When we take $Ax = y$ by multiplying this equation by $A^T$ we have:
$$A^TAx = A^Ty$$
Now if matrix B existed then it has to be in form of $B = (A^TA)^{-1}A^T$. However since $\det(A^TA) = 0$ inversion of $A^TA$ does not exists and so matrix B.
Is my justification correct?
 A: There are many possible matrices $B$ statisfying $By=x$. Note that $B$ represents a linear transformation that maps $y$ to $x$. To define such linear map and computing its matrix, extend $\{y\}$ to a basis $\{y, v_2,v_3,v_4\}$ of $\mathbb{R}^4$, which is possible since $y\neq 0$. Then define a map $f:\mathbb{R}^4\to\mathbb{R}^6$ by setting $f(y)=x$ and then decide where you want to send $f(v_i)$ (for instance $f(v_i)=0$, to make the matrix very easy to compute). After this, you just have to compute the matrix $B$ of $f$ and by definition it will satisfy $By=x$.
The matrix $B$ need not be of the form $(A^TA)^{-1}A^T$, that is where your argument fails.
Edit: I see that in your post you say "associate with $A$", I undertand that in this context (real vector spaces) this just mean $A^T$. In that case you are right, $B$ is not of the form $A^T$, but in this case since you know what $A$ is like, you can just check that $A^Ty\neq x$.
An extra comment on your justification is that $Mx=Ly$ does not imply $x=M^{-1}Ly$, as the inverse may not exist, as it is in your case. For example, take the vectors $x=(1,1)$ and $y=(1,0)\in\mathbb{R}^2$. If you take $M$ and $L$ to be both the matrix associated to the projection $p_1(a,b)=a$, you will get $Mx=Ly$, but $M^{-1}$ does not exist because the map is not injective. In particular this means that even if you were right, your proof was not.
