I have taken reference for this question from On the determinant of a block matrix
Let
$$M:= \begin{pmatrix} A & B \\ C & D \end{pmatrix} $$
where $A$ and $D$ are singular, $\det A = 0$. What is $\det M$? Can one show that $\det M=0$?
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Consider a simple example where the sub-matrices are just $1\times1$: $A = (0), B = (1), C = (1), D = (0) \rightarrow |M| = -1$. So $|M|$ is not necessarily $0$. Whether there is a simple formula to calculate from the sub-matrices, I don't know. – Tony Mathew Jul 20 '23 at 09:49
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Yes agree with your answer Thanks for the cleanup No I'm not aware of it – Phalaksha C G Jul 21 '23 at 11:50
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Take a look at this – Rodrigo de Azevedo Jul 21 '23 at 11:54
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So in the formula using schur component either A or D have to have non zero determinant right? But as @TonyMathew pointed our with A and B having zero determinant we can have non zero determinant for M. Am I missing something? – Phalaksha C G Jul 21 '23 at 11:59
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1Sorry I meant A and D having zero determinant and still M has non zero det – Phalaksha C G Jul 21 '23 at 14:39
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Take a look at this – Rodrigo de Azevedo Jul 21 '23 at 14:46
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1Thanks! that's what I was looking for – Phalaksha C G Jul 22 '23 at 01:19