# Determinant of block matrix where all blocks are $n \times n$

I found the following formula on Wikipedia:

Suppose $$A$$, $$B$$, $$C$$, and $$D$$ are matrices of dimension $$n \times n$$, $$n \times m$$, $$m \times n$$, and $$m \times m$$, respectively.

$$\det \begin{pmatrix} A & B \\ C & D\end{pmatrix} = \det(D) \det \left(A - B D^{-1} C \right)$$

Hence, if $$D$$ is a zero matrix, the determinant should be zero. But when I calculate the determinant of the following matrix

$$\det \left[ \begin {array}{ccc|ccc} 0&0&0&-4&0&0\\ 0&-1&0 &0&-5&0\\ 0&0&-2&0&0&-6\\ \hline 7&0&0&0&0 &0\\ 0&8&0&0&0&0\\ 0&0&9&0&0&0 \end {array} \right] = 60480$$

Is this formula above really only applicable, when $$n\neq m$$?

• $D$ should be an invertible matrix. – user296113 Sep 22 at 16:37
• No, since $D^{-1}$ is then ill-defined. Taking your argument to its limit would mean that each determinant with lower right entry zero would vanish..... – Lord Shark the Unknown Sep 22 at 16:38
• Oh yes of course, bad mistake from my side. Sorry! – vogs Sep 22 at 16:44
• – Rodrigo de Azevedo Sep 23 at 16:23

## 1 Answer

Since all four blocks are square and the bottom ones do commute (the zero matrix commutes with everyone),

$$\det \left[ \begin {array}{ccc|ccc} 0&0&0&-4&0&0\\ 0&-1&0 &0&-5&0\\ 0&0&-2&0&0&-6\\ \hline 7&0&0&0&0 &0\\ 0&8&0&0&0&0\\ 0&0&9&0&0&0 \end {array} \right] = \det \left( \mathrm O_3 + \mbox{diag} (4 \cdot 7, 5 \cdot 8, 6 \cdot 9) \right) = \prod_{i=4}^9 i = \frac{9!}{3!} = \color{blue}{60480}$$