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

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

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Hint: If $C$ has more rows than columns, then its rows are linearly dependent. – Harald Hanche-Olsen Jan 24 '13 at 19:16

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

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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 $B$

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$\det (-C\cdot B) = \det (C) \det(B)$ ?? – aiki93 Jan 24 '13 at 20:11
Determinant of a product of square matrices is the product of the determinants. But $\det(-C) = (-1)^n \det(C)$ for an $n \times n$ matrix. – Robert Israel Jan 24 '13 at 22:09