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Assume that $X$ is a vector space over $K$ and $y_i\in X$ ($i=1,...,n$) $a_{ij}\in K$ ( $i,j=1,...,n$).

Let's consider the system of linear equations:

$$ a_{11}x_1+a_{12}x_2+...+a_{1n}x_n=y_1 $$ $$ .... $$ $$ a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n=y_n $$ with vector variables $x_1,...,x_n\in X$ and deteminant $det[a_{ij}]\neq0$.

Motivated by a classical Cramer's theorem : does this system has solution and if this solution exists is it unique?

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Yes and yes. Choose a basis of $X$ and write each vector in your system as a linear combination of elements of the basis. Then your system will break down to one ordinary (i.e. involving scalar variables, not vector variables) linear system for each coordinate/element of the basis. Even if $\dim X = \infty$, your original system will have a unique solution since each $y_i$ will have non-zero coefficients only at finitely many coordinates.

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Yes, that's correct. All you need to show the system has a unique solution is $[a_{ij}]$ is an invertible matrix. You can derive a similar Cramer's formula if you use $y_1,\dots,y_n$ as a generating system and write $x_i$ as a linear combination of $y_j$'s. But that's really just to find the inverse of $[a_{ij}]$.

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