For approaches of (2) i.e. resolution of systems with coefficients
in a division ring (a skew field), you can consult

[1] A. Heyting, Die Theorie der linearen Gleichungen in einer Zahlen-
spezies mit nichtkommutativer Multiplikation, Math. Ann., 98, 465-
490 (1927).

[2] A.R. Richardson Simultaneous linear equations over a division
ring. Proc. Lond. Math. Soc., 28, 395-420 (1928).

Now, you have also the non-commutative determinants of
Gelfand and Retakh

arXiv:math/0208146

you will find in there a nice historical introduction on the subject of noncommutative determinants as well as a construction of the *free division ring.*

and in their book. Applied later to the theory of noncommutative
symmetric functions started by Gelfand, Krob, Lascoux, Leclerc, Retakh, Thibon (up to my knowledge continued through seven papers).

**A warning about the rank** Let $\Gamma=F(a,b)$ be the free group on two letters and order it with a total (group) ordering as it can be done this though series, see e.g.

*G. Duchamp, J.-Y. Thibon, Simple orderings for free partially commutative groups , International Journal of Algebra and Computation 2 No.3 (1992).*

Then consider the skew field $\mathbb{Q}((a,b))$ (Malcev Neumann series, for example as in arXiv:math/0405133) which is the set of functions $\Gamma \rightarrow \mathbb{Q}$ with well-ordered supports (and usual operations). Then, the matrix
$$
\begin{pmatrix}
ba & a\\
b^2 & b
\end{pmatrix}
$$
has its columns left proportional but not right proportional. So, the vector space generated by the columns on the left has dimension one and on the right has dimension 2. In fact $M$ is (two-sided) invertible. One has
$$
M^{-1}=
\begin{pmatrix}
[b,a]^{-1} & [a,b]^{-1}ab^{-1}\\
-b[b,a]^{-1} & -b[a,b]^{-1}ab^{-1}+b^{-1}
\end{pmatrix}
$$
and $M$ is a (two-sided) a zero divisor for the opposite field $K^{op}$
Concerning **question 1** using elementary operations, one can prove

**Proposition**: Let $K$ be a (skew) field and $M\in K^{n\times n}$ (square matrix of dimension $n$). The following are equivalent

- $M$ is right invertible
- $M$ is left invertible
- $M$ is not a right zero divisor
- $M$ is not a left zero divisor
- For $\lambda\in K^{1\times n}$ (a row) one has
$$
\lambda M=0\Longrightarrow \lambda=0
$$
- For $\gamma\in K^{n\times 1}$ (a column) one has
$$
M\gamma=0\Longrightarrow \gamma=0
$$

As regards **question 2**, in the general case the adapted notion is that of non-commutative (or quasi-)determinants of Gelfand and Retakh (see above) for example $M$ has four quasi-determinants given by its inverse.

**A bit on the relation between right-left kernels-images** I pursue a bit for the sake of completeness.

As remarked by Sebastian Goette and ACL, a matrix $M\in K^{n\times n}$ acts on the left on the space of columns $K^{n\times 1}$
(considered as a right $K$-vector space), defining then an element of $\mathrm{End}_K(K^{n\times 1})$, this correspondence is an
isomorphism and allows to speak of $ker(M)$ and $Im(M)$ which will be denoted $rker(M)$ and $rIm(M)$ (to express that it is devoted
to the right structures).

Likewise, $M$ acts on the right on the space of rows $K^{1\times n}$
(considered as a left $K$-vector space), and the correspondence $\mathrm{End}_K(K^{1\times n})$, is also an
isomorphism. Hence the notations $lker(M)$ and $lIm(M)$ (for the same reason).

Now, you have the non-degenerate pairing (still by matrix multiplication)
$$
\langle\ |\ \rangle\ :\ K^{1\times n}\otimes_K K^{n\times 1}\rightarrow K^{1\times 1}\simeq K
$$
(this time, the two spaces are considered as $K-K$-bimodules).

One can check easily that $lker(M)=(rIm(M))^\perp$ and $rker(M)=(lIm(M))^\perp$. This, with the classic
$$
dim(xker(M))+dim(xIm(M))=n
$$

where $x$ is one of the symbols $\{l,r\}$ allows to see geometrically that $dim(lker(M))=dim(rker(M))$ and

$dim(lIm(M))=dim(rIm(M))$, this last quantity should be considered as the rank of the matrix $M$.

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