Dodgson condensation

Remarks

The divisors in step $k$ are the interior entries of $A^{(k-2)}$, that is, the entries not in its first or last row or column. For $k = 1$ all divisors are $1$, so $A^{(1)}$ is the matrix of connected $2 \times 2$ minors of $A$.

Example: for $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}$, the first step gives $A^{(1)} = \begin{pmatrix} -3 & -3 \\ -3 & 2 \end{pmatrix}$. The only divisor in the second step is the central entry $5$ of $A$, so $A^{(2)} = \big((-3)(2) - (-3)(-3)\big)/5 = -3$. Expansion along the first row also gives $\det A = 2 + 4 - 9 = -3$.

Non-example: for $A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, the central entry is $0$, so $A^{(2)}$ is not defined, although $\det A = -1$. In such a case one can first permute rows or columns, or use a different order of elimination. The correctness of the method follows from the Desnanot–Jacobi identity, a form of the case $k = n - 2$ of Sylvester's determinant identity.