## 每週問題 May 25, 2015

Let $M=\begin{bmatrix} A&B\\ C&D \end{bmatrix}$, where $A$ and $D$ are square matrices of order $m$ and $n$, respectively. Let $E$ be an $m\times m$ matrix and $F$ be an $n\times m$ matrix. Prove the following identities.

(a) $\begin{vmatrix} EA&EB\\ C&D \end{vmatrix}=(\det E)(\det M)$.
(b) $\begin{vmatrix} A&B\\ C+FA&D+FB \end{vmatrix}=\det M$.

(a) 考慮分塊矩陣乘法

$\displaystyle \begin{bmatrix} EA&EB\\ C&D \end{bmatrix}=\begin{bmatrix} E&0\\ 0&I \end{bmatrix}\begin{bmatrix} A&B\\ C&D \end{bmatrix}$

$\displaystyle \begin{vmatrix} EA&EB\\ C&D \end{vmatrix}=\begin{vmatrix} E&0\\ 0&I \end{vmatrix}\cdot\begin{vmatrix} A&B\\ C&D \end{vmatrix}=(\det E)(\det I)(\det M)=(\det E)(\det M)$

(b) 考慮分塊矩陣乘法

$\displaystyle \begin{bmatrix} A&B\\ C+FA&D+FB \end{bmatrix}=\begin{bmatrix} I&0\\ F&I \end{bmatrix}\begin{bmatrix} A&B\\ C&D \end{bmatrix}$

$\displaystyle \begin{vmatrix} A&B\\ C+FA&D+FB \end{vmatrix}=\begin{vmatrix} I&0\\ F&I \end{vmatrix}\cdot\begin{vmatrix} A&B\\ C&D \end{vmatrix}=(\det I)(\det I)(\det M)=\det M$

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