The only point that remains to prove is that these values for the unknowns, the only possible ones, do indeed together form a solution. But if the matrix A is invertible with inverse A−1, then x = A−1b will be a solution, thus showing its existence. To see that A is invertible when det(A) is nonzero, consider the n × n matrix M obtained by stacking the one-line matrices L(j) on top of each other for j = 1, ..., n (this gives the adjugate matrix for A).