Determinant via the Schur complement

Statement

Let $K$ be a commutative ring and let $A$ be an invertible $p \times p$ matrix over $K$. Let $B$, $C$ and $D$ be matrices over $K$ of sizes $p \times q$, $q \times p$ and $q \times q$. Then $$\det \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \det A \cdot \det(D - C A^{-1} B),$$ where $\det$ is the determinant. The matrix $D - C A^{-1} B$ is the Schur complement of $A$ in the block matrix.

Remarks

The proof factors the block matrix as $\begin{pmatrix} I & 0 \\ C A^{-1} & I \end{pmatrix} \begin{pmatrix} A & B \\ 0 & D - C A^{-1} B \end{pmatrix}$. Then it uses the rule for block triangular matrices. For $p = q = 1$ and $a \ne 0$, it gives $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = a(d - cb/a) = ad - bc$.

Explanations

L1 human

Block elimination removes the block $C$, in the same way as Gaussian elimination removes an entry. What remains in the lower right corner is the Schur complement. The determinant is then the product of two smaller determinants.