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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 $n\neq m$?

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    $\begingroup$ $D$ should be an invertible matrix. $\endgroup$ – user296113 Sep 22 at 16:37
  • $\begingroup$ 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..... $\endgroup$ – Lord Shark the Unknown Sep 22 at 16:38
  • $\begingroup$ Oh yes of course, bad mistake from my side. Sorry! $\endgroup$ – vogs Sep 22 at 16:44
  • $\begingroup$ Related: math.stackexchange.com/q/1296257/339790 $\endgroup$ – Rodrigo de Azevedo Sep 23 at 16:23
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Since all four blocks are square and the bottom ones do commute (the zero matrix commutes with everyone),

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

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