Block elimination
= Block elimination
{title2=$Sy=g-CA^{-1}f$}
Write a <linear system> as $Ax+By=f$ and $Cx+Dy=g$, with $A$ an <invertible matrix>. Solving the first equation for $x$ and substituting in the second gives
$$
x=A^{-1}(f-By),\qquad
(D-CA^{-1}B)y=g-CA^{-1}f.
$$
Thus the <Schur complement> is the coefficient <matrix> remaining after the $x$ variables have been eliminated. When it is an <invertible matrix>, solve the reduced <linear system> for $y$ and recover $x$. In computation, expressions such as $A^{-1}B$ mean solving with each column of $B$, using the same <LU decomposition> or <Cholesky decomposition> when applicable.
For the example under <block matrix inverse> and right-hand side $(3,1,4)^T$, take $f=(3,1)^T$ and $g=4$. Since $CA^{-1}f=13/6$, the reduced equation is $(19/6)y=11/6$. <Back substitution> gives
$$
\boxed{(x_1,x_2,y)=(23,-1,11)/19}.
$$
The original equations return $3$, $1$ and $4$, respectively. The reduction retained the influence of $x_1,x_2$ on $y$ through the correction $CA^{-1}B$.
A short proof outline also explains whether $M$ is an <invertible matrix>: the exact <matrix factorization>
$$
M=\begin{pmatrix}I&0\\CA^{-1}&I\end{pmatrix}
\begin{pmatrix}A&0\\0&S\end{pmatrix}
\begin{pmatrix}I&A^{-1}B\\0&I\end{pmatrix}
$$
has <invertible matrices> on the outside. Therefore $M$ is an <invertible matrix> precisely when the central <block diagonal matrix> is one. Reversing the factors gives the <block matrix inverse>, while taking <determinants> gives the <Schur complement determinant identity>.