\input{ccs_gyak_defs}
\author{Matrix algebra}

\begin{document}
\maketitle

\begin{cpropw}*{}\label{pr:1}
    If $A \in \mathbb{R}^{{n}\times{n}}$ and $D \in \mathbb{R}^{{m}\times{m}}$ are nonsingular, than the determinant of the following block-triangle matrix can be computed as follows:
    \begin{align}
        \begin{aligned}
            \vmqty{ A & 0 \\ C & D } = \det(A) \cdot \det(D)
        \end{aligned}
    \end{align}
\end{cpropw}
\begin{proof}
    \begin{align}
        \begin{aligned}
            \pmqty{I&0\\-CA^{-1}&I}\pmqty{A&0\\C&D}=\pmqty{A&0\\0&D}\;
        \end{aligned}
    \end{align}
\end{proof}

\begin{cpropw}*{}\label{pr:2}
    If $A \in \mathbb{R}^{{n}\times{n}}$ and $D \in \mathbb{R}^{{m}\times{m}}$ are nonsingular, than the determinant of the following block matrix can be computed as follows:
    \begin{align}
        \begin{aligned}
            M= \begin{pmatrix} A & B \\C & D \end{pmatrix}
            \Rightarrow
            \det(M)=\det(A-BD^{-1}C)\det(D).
        \end{aligned}
    \end{align}
    If $m = n$ and if $C$ and $D$ commute then ${\det(M)=\det(AD-BC)}$.
\end{cpropw}
\begin{proof}
    \begin{align}
        \begin{aligned}
            \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} I & 0 \\ -D^{-1}C & I \end{pmatrix} = \begin{pmatrix} A-BD^{-1}C & B \\ 0 & D \end{pmatrix}
        \end{aligned}
    \end{align}
\end{proof}

\end{document}
