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 {\vec {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} .