= Schur complement determinant identity
{c}
{title2=$\det M=\det A\,\det S$}
When $A$ is an <invertible matrix>, the factorization in <block elimination> has outer triangular factors with <determinant> one. Hence $\det M=\det A\det S$, without a symmetry assumption. For the original example under <block matrix inverse>, this gives $\det M=(2\cdot3)(19/6)=19$. An invertible block $A$ therefore reduces the question of whether the full <matrix> is an <invertible matrix> to the smaller <Schur complement>; a singular $S$ makes $M$ a <singular matrix>.
Back to article page
