Can I transform a solution to the matrix equation (over the field $\mathbb{F}_2$) $A\vec{x} = \vec{y}$ to a solution of $A\vec{x} = \vec{0}$?

Question

I'm working in the feld $\mathbb{F}_2$, and I have a matrix equation of the form $A\vec{x} = \vec{y}$. I have two related questions:

1. If I find a solution $\vec{x}$ that solves this inhomogeneous equation (in the case $\vec{y} \ne \vec{0}$), is there a way to "rotate" or transform $\vec{x} \rightarrow \vec{k}$ such that $A\vec{k} = \vec{0}$? I'm looking for something other than the solution $\vec{k} = \vec{x} + \vec{x} = \vec{0}$.

2. Suppose I have a solution $\vec{x}$ to $A\vec{x} = \vec{y}$ and another solution $\vec{w}$ such that $A\vec{w} = \vec{z}$. Is there a way to produce a third solution $\vec{k}$ such that $A\vec{k} = \vec{0}$, with $\vec{k} = \vec{k}(A,w,x,y,z)$?

For question 2, if $\vec{y} = \vec{z}$, then the solution $\vec{k} = \vec{x} + \vec{w}$ will do the trick, since:

$$ A\vec{k} = A \left( \vec{x} + \vec{w} \right) = \vec{y} + \vec{z} = \vec{0}.$$

The last equality comes from the fact $\vec{y} = \vec{z}$. So I have that case covered, but if $\vec{y} \ne \vec{z}$, is this still possible? I'm looking for the case where the equation has finitely many solutions (it could have one, or in general many).

Edit: This is related to my question I posted on the Physics StackExchange.

Answer 1

If $A$ is an invertible matrix over the field ${\Bbb F}_2$, for instance if $A$ is the identity matrix, the unique solution of the equation $Ax = 0$ is $x = 0$. Furthermore, if $Ax = Aw = y$, then $x = w = A^{-1}y$ and hence $x + w = 0$.

Note. The term Boolean matrix is not really appropriate in this case. The entries of your matrix are indeed $0$ or $1$, but you are not using the Boolean operations. For instance, $1 + 1 = 1$ in the Boolean case.