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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 is is devoted to the right vector space structures).

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).

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

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).

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 is is devoted to the right vector space 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 $$ \langle\ |\ \rangle\ :\ K^{1\times n}\otimes_K K^{n\times 1} $$ (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 $M$.