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If $A$ and $B$ are $n\times n$ matrices and $C$ is defined to be

$$ C=\begin{pmatrix} O&A\\ B&O \end{pmatrix} $$ Where $O$ denotes the zero matrix.

Can I conclude that $O$ needs to be only square matrix of size $n\times n$ because if $O$ has any other size, then it does not make sense?

Can I also then say that $\det(C)=-\det(A)\det(B)$?

If I am wrong, then can you explain why?

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  • $\begingroup$ For the first question, yes, both $O$'s have to be $n\times n$. For the upper left one has to have the same number of rows as $A$ and the same number of columns as $B$. $\endgroup$ – Keivan Mar 24 '16 at 3:38
  • $\begingroup$ For the second question, also yes, the determinant of a block matrix is defines the same way. One good way to see it is by looking at diagonal products and noting that each term there is a product of a term from $\det A$ and a term from $\det B$. $\endgroup$ – Keivan Mar 24 '16 at 3:40
  • $\begingroup$ And have a look at the wikipedia page: en.wikipedia.org/wiki/Determinant#Block_matrices $\endgroup$ – Keivan Mar 24 '16 at 3:42
  • $\begingroup$ Oops I forgot the a $(-1)^n$ in front. $\endgroup$ – Keivan Mar 24 '16 at 3:47
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Consider the example where $A$ and $B$ are both the $2\times2$ identity matrix. Then you may calculate that $\det(C)$ is $1$, which is not $-\det(A)\det(B)$. In fact we have in general $$\det(C)=(-1)^n\det(A)\det(B)\ .$$ Reason: use row-reduction. If we interchange rows $n$ times we get $$\det(C)=(-1)^n\det\pmatrix{B&O\cr O&A\cr}\ ,$$ and it's not too hard to show that the matrix on the RHS has determinant $\det(A)\det(B)$.

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  • $\begingroup$ Thanks. That makes it very clear. So if $$C=\begin{pmatrix}A&B\\ C&D\end{pmatrix}$$then $$\det(C)\not=\det(A)\det(D)-\det(B)\det(C)$$ $\endgroup$ – user99885 Mar 24 '16 at 4:09
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Yes, in your situation $O$ has to be $n\times n$. As for the determinant, $$ \det C=(-1)^n\,\det(A)\,\det(B). $$ This is because $$ \begin{bmatrix}O&A\\ B& O\end{bmatrix}=\begin{bmatrix}O&I\\ I& O\end{bmatrix}\,\begin{bmatrix}B&O\\ O&A\end{bmatrix} $$

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  • $\begingroup$ Thanks for your answer. Your answer is also excellent but I had to choose one :) $\endgroup$ – user99885 Mar 24 '16 at 4:11

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