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  1. Let $A$ be a $k \times k$ matrix and $B$ be a $\left(n-k\right) \times \left(n-k\right)$ matrix, and $Z$ be the $n \times n$ matrix $$ Z = \left( \begin{matrix} A & C \\ 0 & B \end{matrix} \right),$$ where $C$ is any $k \times \left(n-k\right)$ matrix. Show that $$\det Z = \det A \cdot \det B.$$

So I understand this in my head. I get the det(Z)=det(A)det(B)-det(C)(0). I am having trouble figuring out how to prove this though. There may be some theorem that I am missing or something. Any tips or leads would be helpful though.

darij grinberg
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user139985
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  • Would det(AB)=det(A)det(B) be related? – zahbaz Apr 15 '15 at 05:50
  • "det(Z)=det(A)det(B)-det(C)(0)" is not the right way to go. If there was a fourth matrix $D$ instead of $0$, then you would not generally have $\det Z = \det A \cdot \det B - \det C \cdot \det D$. Anyway this is a well-known result, and I hope someone answers it nicely to have a reference for the future generations... – darij grinberg Apr 15 '15 at 05:51
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    $\det \left(AB\right) = \det A \cdot \det B$ holds when $A$ and $B$ are of the same size, which they here are not necessarily. – darij grinberg Apr 15 '15 at 05:51
  • You can see some general rules about block matrices here: http://en.wikipedia.org/wiki/Determinant#Block_matrices. – Emilio Novati Apr 15 '15 at 05:56
  • And here: http://math.stackexchange.com/questions/75293/determinant-of-a-block-lower-triangular-matrix, there is a possible duplicate of your question. – Emilio Novati Apr 15 '15 at 05:57

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