# How can we solve a system Ax=y over finite field?

Suppose $x\in\mathbb{F}_{q}^{N}$ is a $N\times 1$ vector, $A\in\mathbb{F}_{q}^{N\times N}$ is a $N\times N$ matrix. Let $Ax=y$. The question is: how can we design a matrix $A$ to guarantee that we can uniquely determine $x$ with $y$? What's the requirements imposed on the matrix $A$?

In a linear system(not in finite field), we need $rank(A)=N$ to make sure we have a unique solution. So I guess: If $A$ has full rank over finite field(like the function gfrank() in matlab), is it certain that we can find a unique solution $x$?

For example, we can set $x\in\left\{ 0,1\right\} ^{N}$, $A\in\left\{ 0,1\right\} ^{N\times N}$. And then the problem becomes how to design a matrix $A$ so that we can solve $Ax=y \mod2$ uniquely.

Any one knows anything about it? Thanks.

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Yes, almost all the linear algebra you have learned (certain the parts about solving systems of linear equations) do not care about the field at all. – Tobias Kildetoft Dec 16 '13 at 9:19
Here's an exercise. Go back to linear algebra notes, where solvability of linear systems was studied. Check that at no point did you use properties of the reals other than those shared by all fields. Summary: Maximum rank will do. So will $\det A\neq0$. – Jyrki Lahtonen Dec 16 '13 at 14:14