Determinant of a block matrix with $\mathrm{Id}$ and $0$ in the diagonal

Question

How to compute the determinant $\det A$ depending on $B$ and $C$, where

$$ A = \left(\begin{matrix}\mathrm{Id} & B \\ C & 0 \end{matrix} \right), $$

a) when $C$ is square,
b) $C$ has more rows than columns

Answer 1

Hints:

• note that $$\begin{pmatrix} \text{Id} & B \\ C & 0 \end{pmatrix} = \begin{pmatrix} \text{Id} & 0 \\ C & \text{Id} \end{pmatrix} \cdot\begin{pmatrix} \text{Id} & B \\ 0 & - C\cdot B\end{pmatrix}.$$

• furthermore, in general $$\det\begin{pmatrix}A& 0\\ C& D\end{pmatrix} = \det\begin{pmatrix}A& B\\ 0& D\end{pmatrix} = \det(A) \det(D)$$ and $\det (A\cdot B) = \det (A) \det(B)$ for square matrices $A$ and $B$

Answer 2

The block determinant formula: if $A$ is invertible, $\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(A) \det(D - C A^{-1} B)$