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A block matrix is block diagonal if all off-diagonal blocks vanish. It represents independent actions on a direct sum of vector spaces. Its determinant is the product of the diagonal-block determinants.

A block upper triangular matrix is a block matrix with square diagonal blocks and zero blocks below them. Its determinant is the product of the determinants of its diagonal blocks.

Use block matrix elimination with the invertible square block $A$:The left multiplier is block triangular with identity diagonal blocks, so its determinant is $1$. Taking the determinant of the block-triangular product provesThe matrix $D-C{A}^{-1}B$ is the Schur complement of $A$. The proof needs $A$ invertible, but does not assume the Schur complement or the full matrix is invertible.

For a block matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ with $A$ an invertible matrix, the Schur complement of $A$ in $M$ is $S=D-C{A}^{-1}B$. It is the effective coefficient matrix for the remaining variables after block elimination. The definition requires invertibility of $A$, but not of $S$ or $M$. Its size is that of $D$, even when $B$ and $C$ are rectangular matrices.

In the original example under block matrix inverse, $D=5$ becomes $S=19/6$ because changing the remaining variable also changes the two variables solved through $A$. This is what makes a Schur complement more informative than simply deleting rows and columns. It retains their mediated effect, and its matrix inverse is the corresponding block of $M^{-1}$ when the block matrix inverse exists.

The same elimination appears when minimizing a quadratic form. The Schur complement as a minimized quadratic form gives a geometric example; the Schur complement determinant identity gives an exact test for an invertible matrix.