structural:reformatting, quality:correction
A proof by abstract linear algebra

This is a restatement of the proof above in abstract language.

Consider the map x = (x 1, …, x n) ↦ 1 det A (det (A 1), …, det (A n)), {\displaystyle \mathbf {x} =(x_{1},\ldots,x_{n})\mapsto {\frac {1}{\det A}}\left(\det(A_{1}),\ldots,\det(A_{n})\right),} where A i {\displaystyle A_{i}} is the matrix A {\displaystyle A} with x {\displaystyle \mathbf {x} } substituted in the i {\displaystyle i} th column, as in Cramer's rule. Because of linearity of determinant in every column, this map is linear. Observe that it sends the i {\displaystyle i} th column of A {\displaystyle A} to the i {\displaystyle i} th basis vector e i = (0, …, 1, …, 0) {\displaystyle \mathbf {e} _{i}=(0,\ldots,1,\ldots,0)} (with 1 in the i {\displaystyle i} th place), because determinant of a matrix with a repeated column is 0. So we have a linear map which agrees with the inverse of A {\displaystyle A} on the column space; hence it agrees with A − 1 {\displaystyle A^{-1}} on the span of the column space. Since A {\displaystyle A} is invertible, the column vectors span all of R n {\displaystyle \mathbb {R} ^{n}}, so our map really is the inverse of A {\displaystyle A}. Cramer's rule follows.