Newton correction for a square vector residual

Let $n\ge 1$, let $A:{\mathbb{R}}^n\to {\mathbb{R}}^n$ be a linear map, and let $r\in {\mathbb{R}}^n$ be a residual vector. If $A$ is not invertible in the sense of Invertible Euclidean linear maps, return $(\mathtt{\text{singular-jacobian}},\mathrm{\varnothing })$. If $A$ is invertible, return $(\mathtt{\text{correction}},s)$, where $s$ is the unique vector solving $As=-r\mathrm{.}$

Use the following exact-arithmetic procedure, with exact real zero tests. Represent $A$ by its matrix in the standard basis, with indices $0\le i,j<n$.

1. To compute the determinant of any matrix $M=(m_{ij})$ below, initialize $d=0$. Enumerate the permutations $\sigma$ of $\{0,\dots ,n-1\}$ in lexicographic order. For each, count its inversions $\nu (\sigma )=\mathrm{\#}\{(i,j):i<j,\sigma (i)>\sigma (j)\}$, multiply $p={\prod }_{i=0}^{n-1}{m}_{\sigma (i),i}$ in increasing index order, and add $(-1{)}^{\nu (\sigma )}p$ to $d$. The resulting $d$ is the determinant in For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix.
2. Compute $d=\det (A)$ by step 1. If $d=0$, return $(\mathtt{\text{singular-jacobian}},\mathrm{\varnothing })$, using A finite square real matrix is invertible if and only if its determinant is nonzero for the invertibility test.
3. Otherwise, for $j=0,\dots ,n-1$, copy $A$ to $A_j$, replace column $j$ by $-r$, compute $d_j=\det (A_j)$ by step 1, and set $s_j=d_j\mathrm{/}d$. Return $(\mathtt{\text{correction}},s)$, the solution specified by Cramer's rule over a commutative ring: every solution satisfies $\det (A)x_j=\det (A_j(b))$, and a unit determinant gives the unique quotient formula over $\mathbb{R}$.

The output is a linear correction, not a claim that $r$ came from a nonlinear function or that applying the correction decreases a residual.