Why is $\mathbf{x}$ yielding a product of $\mathbf{y}$ and the right inverse for $\mathbf{A}$? When you solve the equation $\mathbf{Ax}=\mathbf{y}$ where $\mathbf{x} \in \mathbb{R}^{5}$, $\mathbf{y} \in \mathbb{R}^{4}$ and $\mathbf{A}$ is a $4 \times 5$ matrix with a right inverse, I understand that the $\mathbf{x}$ is $\mathbf{A}^{-1}\mathbf{y}$ but how come it can happen that $\mathbf{x}$ is not $\mathbf{A}^{-1}\mathbf{y}$ but some right inverse times $\mathbf{y}$? So if $\mathbf{X}$ were a right inverse to $\mathbf{A}$, $\mathbf{x}$ is $\mathbf{X}\mathbf{y}$?
I would like some algebraic proof such as for the fact that $\mathbf{x}$ is $\mathbf{H}^{-1}\mathbf{y}$ for some square matrix $\mathbf{H}$
$$
\mathbf{H}^{-1}\mathbf{Hx} = (\mathbf{H}^{-1}\mathbf{H})\mathbf{x} = \mathbf{I}_4 \mathbf{x} = \mathbf{x} = \mathbf{H}^{-1}\mathbf{y}
$$
I do not think this is valid statement
$$
\mathbf{AXx} = (\mathbf{AX})\mathbf{x} = \mathbf{I}_4 \mathbf{x} = \mathbf{x} = \mathbf{X}\mathbf{y}
$$
so that does probably not that why $\mathbf{x}$ in my case is a the result of post-multiplying a right inverse of $\mathbf{A}$ by $\mathbf{y}$
To give details about my case where $\mathbf{x}$ is $\mathbf{X}\mathbf{y}$?, I have 
$$
\mathbf{A} =
\left[\begin{array}{rrrrr}
2&-4&-1&-3&2\\
-1&2&1&0&1\\
1&-2&-1&-3&-1\\
-1&4&-1&0&5
\end{array}\right]
$$ and $\mathbf{y}$ is a column vector of four unknowns. Then I have the solution for $\mathbf{x}$ as
$$
\mathbf{x} = 
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]
=
\left[\begin{array}{r}
3y_1+y_4-3y_3+y_2-15t\\
\frac{1}{2}y_4+y_1-y_3+\frac{1}{2}y_2-6t\\
-y_3+y_1+y_2-4t\\
-\frac{1}{3}y_2-\frac{1}{3}y_3\\
t
\end{array}\right]
\quad
\text{where }
t \in \mathbb{R}
$$
For exapmle when $t = 0$, we have that 
$$
\mathbf{x} = 
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]
=
\left[\begin{array}{r}
3y_1+y_4-3y_3+y_2-15 \cdot 0\\
\frac{1}{2}y_4+y_1-y_3+\frac{1}{2}y_2-6 \cdot 0\\
-y_3+y_1+y_2-4 \cdot 0\\
-\frac{1}{3}y_2-\frac{1}{3}y_3\\
0
\end{array}\right]
=
\left[\begin{array}{rrrr}
3&1&-3&1\\
1&\frac{1}{2}&-1&\frac{1}{2}\\
1&1&-1&0\\
0&-\frac{1}{3}&-\frac{1}{3}&0\\
0&0&0&0
\end{array}\right]
\left[\begin{array}{r}
y_1\\
y_2\\
y_3\\
y_4
\end{array}\right]
$$
We now have $\mathbf{x} = \mathbf{Xy}$ where $\mathbf{X}$ is a right inverse to $\mathbf{A}$ - so why this relationship ($\mathbf{x} = \mathbf{Xy}$ with $\mathbf{X}$ being a right inverse)?

Claim. Let $V,W$ be vector spaces, $A\colon V\to W$ and $B\colon W\to V$ be linear maps with $AB=1_W$. Also suppose that $Ax=y$ for certain $x\in V$, $y\in W$ with $y\ne 0$.
Then there exists a linear map $B'\colon W\to V$ such that $AB'=1_W$ and $x=B'y$.
Proof: It suffices to find $C\colon W\to V$ such that $AC=0$ and $Cy=x-By$ as we then can take $B'=B+C$.
Let $v:=x- By$. Note that $v\in\ker A$. By extending $y$ to a to a basis of $W$, we find a linear map $W\to \ker A$ that maps $y\mapsto v$. Composed with the inclusion $\ker A\to V$, we obtain a linear map $C\colon W\to V$ such that $AC=0$ and $Cy=x-By$. $\square$
Remark. If we allow $y=0$, the claim is no longer true because there may exist non-zero $x$ with $Ax=0$.
