Sylvester's determinant identity

Remarks

First assume that $A_0$ is invertible, and write $A$ in blocks with $A_0$ in the top left corner. By the Schur complement formula, $\det A = \det A_0 \cdot \det S$, where $S = (s_{ij})$ is the Schur complement of $A_0$ in $A$. The same formula on each bordered submatrix gives $b_{ij} = \det A_0 \cdot s_{ij}$, so $B = (\det A_0) S$ and the identity follows. Both sides are polynomials in the entries of $A$ with integer coefficients, so the identity holds over every commutative ring.

For $k = n - 1$, the matrix $B$ is $1 \times 1$ with entry $\det A$, and the identity is trivial. For $k = n - 2$, it is a form of the Desnanot–Jacobi identity, which is the basis of Dodgson condensation. It should not be confused with Sylvester's determinant theorem, which says $\det(I_m + AB) = \det(I_n + BA)$.

Going deeper

For a student

The identity explains why fraction-free Gaussian elimination works over the integers. The intermediate entries of the Bareiss algorithm are such bordered minors, and the division by the previous pivot is exact. It sits next to the determinant via the Schur complement and to the determinant of a block triangular matrix.