A determinant identity

Question

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?

Answer 1

I'll repeat the argument cited in the comments, adjusted to fit your situation.

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.

$$\pmatrix{0&1&0\cr A^{1,n}_{n-1,n}&0&0\cr 0&0&1\cr}$$ where the $1$s are $1\times 1$ 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.