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  "body": "\n## Definition\n\nLet $F$ be a [[field|04C5ZBG1WT]], let $n \\ge 2$, and let $A$ be an $n \\times n$ matrix over $F$.\nLet $A^{(-1)}$ be the $(n+1) \\times (n+1)$ matrix with all entries $1$, and let $A^{(0)} = A$.\nFor $k = 1, \\dots, n-1$, let $A^{(k)}$ be the $(n-k) \\times (n-k)$ matrix with entries\n$$a^{(k)}_{ij} = \\frac{a^{(k-1)}_{ij}\\, a^{(k-1)}_{i+1,j+1} - a^{(k-1)}_{i,j+1}\\, a^{(k-1)}_{i+1,j}}{a^{(k-2)}_{i+1,j+1}}, \\qquad 1 \\le i, j \\le n-k.$$\nThis is defined when every divisor $a^{(k-2)}_{i+1,j+1}$ is nonzero.\n*Dodgson condensation* is the computation of the sequence $A^{(0)}, A^{(1)}, \\dots, A^{(n-1)}$.\nWhen it is defined, the single entry of $A^{(n-1)}$ equals the [[determinant|00KDZRW4C0]] of $A$.\n\n## Remarks\n\nThe 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.\nFor $k = 1$ all divisors are $1$, so $A^{(1)}$ is the matrix of connected $2 \\times 2$ [[minors|0BT9BYWVRY]] of $A$.\n\nExample: for $A = \\begin{pmatrix} 1 \u0026 2 \u0026 3 \\\\ 4 \u0026 5 \u0026 6 \\\\ 7 \u0026 8 \u0026 10 \\end{pmatrix}$, the first step gives $A^{(1)} = \\begin{pmatrix} -3 \u0026 -3 \\\\ -3 \u0026 2 \\end{pmatrix}$.\nThe only divisor in the second step is the central entry $5$ of $A$, so $A^{(2)} = \\big((-3)(2) - (-3)(-3)\\big)/5 = -3$.\nExpansion along the first row also gives $\\det A = 2 + 4 - 9 = -3$.\n\nNon-example: for $A = \\begin{pmatrix} 0 \u0026 1 \u0026 0 \\\\ 1 \u0026 0 \u0026 0 \\\\ 0 \u0026 0 \u0026 1 \\end{pmatrix}$, the central entry is $0$, so $A^{(2)}$ is not defined, although $\\det A = -1$.\nIn such a case one can first permute rows or columns, or use a different order of elimination.\nThe correctness of the method follows from the Desnanot–Jacobi identity, a form of the case $k = n - 2$ of [[Sylvester's determinant identity|7SSKBQYFW1]].\n\n## Explanations\n\n### L1 human\n\nThe method shrinks a matrix by one row and one column at each step, using only $2 \\times 2$ determinants and divisions.\nIt stops when one number is left, and that number is the determinant.\nIt fails when it must divide by zero.\n",
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