= Choosing a different block pivot
An <invertible matrix> need not have an invertible chosen diagonal block. For example,
$$
M=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad M^{-1}=M,
$$
has <singular matrices> as both one-by-one diagonal blocks. Neither diagonal-block formula applies in that partition. Interchanging the two rows produces the <identity matrix> and a usable <pivot element>. This is why a failed block pivot is not proof that the full <matrix> is a <singular matrix>; <Pivoted LU decomposition> changes the order before eliminating.
Conversely, invertible diagonal blocks do not suffice: $M=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ has $A=D=1$ but <Schur complement> $S=0$, so it is a <singular matrix>. Both hypotheses in the <block matrix inverse> formula have work to do.
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