# A determinant identity

The following identity involving determinants essentially appears in E.L. Ince’s book on Ordinary Differential Equations:

Let $$A$$ be an $$n times n$$ matrix, $$n geq 3$$. Denote by $$A_{j_1,ldots,j_r}^{k_1,ldots,k_r}$$ the $$(n-r) times (n-r)$$ matrix obtained from $$A$$ by
erasing the $$j_1$$-th, …, $$j_r$$-th row and the $$k_1$$-th, …, $$k_r$$-th column. Then,
$$left|A right| left|A_{n-1,n}^{1,n} right| = left|A_{n-1}^1 right|left|A_n^n right| – left|A_{n-1}^n right|left|A_n^1 right|.$$
Any ideas of how to prove it? Is this a special case of a more general identity?

MathOverflow Asked by Vassilis Papanicolaou on January 4, 2021

A polynomial identity holds in general if it holds on an open set. So it's enough to prove the identity for matrices $$A$$ in the open set where $$A^{1,n}_{n-1,n}$$ is invertible. That is, we can assume $$A^{1,n}_{n-1,n}$$ is invertible.

Now premultiply $$A$$ by the inverse of the matrix.

$$pmatrix{0&1&0cr A^{1,n}_{n-1,n}&0&0cr 0&0&1cr}$$ where the $$1$$s are $$1times 1$$ matrices with entry $$1$$ and the $$0's$$ are row or column matrices of the appropriate sizes, filled with zeros.

This does not affect the truth of the theorem and allows us to assume $$A^{1,n}_{n-1,n}$$ is the identity, after which row and column operations render the desired equality trivial.

Answered by Steven Landsburg on January 4, 2021

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