Matrices determinants problem Could you please give at least a hint to prove this? I really don't know how to start.
Let $A$ be the matrix
$A= \left( \begin{array}{ccc}
A_{11} & A_{12} \\
A_{21} & A_{22} \end{array} \right)$,
where $A_{11}$ and $A_{22}$ are square matrices of order $k$ and $m$ respectively. Prove that for any two matrices $D$ of order $k \times k$ and $B$ of order $m \times k$ we have
$\left| \begin{array}{ccc}
D \cdot A_{11} & D \cdot A_{12} \\
A_{21} & A_{22} \end{array} \right| = |D|\cdot |A|$,
and
$\left| \begin{array}{ccc}
A_{11} & A_{12} \\
A_{21}+B \cdot A_{11} & A_{22}+B \cdot A_{12} \end{array} \right|=|A|.$
 A: Note that
\begin{align*}
\left(\begin{array}{cc}D\cdot A_{11}&D\cdot A_{12}\\A_{21}&A_{22}\end{array}\right)=\left(\begin{array}{cc}D&0\\0&I\end{array}\right)\cdot\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}&A_{22}\end{array}\right),
\end{align*}
where $I$ is the identity matrix of order $m$. The first matrix in the second expression is a “block diagonal matrix,” whose determinant is just $|D|\cdot|I|=|D|$. Now use the product rule for determinants.
As for your second question (in which I'm pretty sure the $D$ should be $B$):
\begin{align*}
\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}+B\cdot A_{11}&A_{22}+B\cdot A_{12}\end{array}\right)=\left(\begin{array}{cc}\underset{k\times k}{I}&0\\B&\underset{m\times m}{I}\end{array}\right)\cdot\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}&A_{22}\end{array}\right).
\end{align*}
The first matrix in the second expression is a triangular matrix, whose determinant is equal to the product of the main diagonal elements, which is $1$. Now, use again the product rule.
