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

$\left| \begin{array}{ccc} D \cdot A_{11} & D \cdot A_{12} \\ A_{21} & A_{22} \end{array} \right| = |D|\cdot |A|$,


$\left| \begin{array}{ccc} A_{11} & A_{12} \\ A_{21}+B \cdot A_{11} & A_{22}+B \cdot A_{12} \end{array} \right|=|A|.$

  • 1
    $\begingroup$ Hint (for the first part): make a block diagonal matrix with $D$ and $I_m$. $\endgroup$
    – Maesumi
    Sep 2 '13 at 4:10

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.


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