Let $$M = \begin{pmatrix}
A & B \\
C & D \\
\end{pmatrix}$$
and
$$M^{-1} = \begin{pmatrix}
P & Q \\
R & S \\
\end{pmatrix},$$
Where $A, \dots,S$ are $k \times k$ matrices. Show that $$\det M \det S = \det A.$$
This is an exam practice problem. Honestly not sure where to start with this one. Is there some way I can use the identity
$$\det M = \det (A – BD^{-1}C) \det (D)$$
for the determinant of block matrices?
Best Answer
You may need to find $S$ in terms of $A,B,C,D$. First assume that $A$ is invertible. Use elementary row operations to $$\begin{pmatrix}A & B & I & 0 \\ C & D & 0 & I\end{pmatrix}\to\begin{pmatrix}A & B & I & 0 \\ 0 & D-CA^{-1}B & -CA^{-1} & I\end{pmatrix}\to\begin{pmatrix}I & 0 & * & *\\ 0 & I & * & (D-CA^{-1}B)^{-1}\end{pmatrix}$$ (I only calculated the entry $S$.)
Hence, $$\det S=\det(D-CA^{-1}B)^{-1},$$ and also $$\det M=\det(A)\det(D-CA^{-1}B).$$ So we have $$\det S\det M=\det A.$$
This is also true even if $A$ is not invertible. One way to see this is to note that determinant is a continuous function. A slight perturbation will make $A$ invertible.