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I need your help finding where is my mistake while calculating 5x5 matrix determinant using block matrices.

My calculations are enter image description here

I tripple checked everything and the online calculation of matrix A gives determinant of 4. And I get -8 using block matrices.

I just can't see why am I getting wrong answer.

Please help!

EDIT: The formula I am using is: det(AD-ACA^(-1)B) A being

0 1
1 0

B is

0 -2 1
3 1 1

C

1 -1
2 2
3 1

D

1 1 1
1 0 1
1 1 2

Det of part A is -1.

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  • $\begingroup$ Can you describe your approach? What formula are you using? $\endgroup$ – Exodd Jan 4 '17 at 16:19
  • $\begingroup$ I have inserted your calculations as an inline image. $\endgroup$ – Andreas Caranti Jan 4 '17 at 16:23
  • $\begingroup$ You can't replace $A$ with $-1$ to end up with $-D\pm CB$. $\endgroup$ – LinAlg Jan 4 '17 at 16:37
  • $\begingroup$ Yes, I am trying to get this correctly this time. $\endgroup$ – Deramite Jan 4 '17 at 16:42
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The error is in the first step. If you write your matrix as $\begin{pmatrix}A & B \\ C & D\end{pmatrix}$, you claim that the determinant equals $\textrm{det}(-D-CB)$. However, the correct formula is $\textrm{det}(A)\textrm{det}(D-CA^{-1}B)$.

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  • $\begingroup$ Yep, I understood it as well while editing more information. Thanks! $\endgroup$ – Deramite Jan 4 '17 at 16:34
  • $\begingroup$ With $-D+CB$ you end up with the wrong answer though. $\endgroup$ – LinAlg Jan 4 '17 at 16:36
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You're using the formula $$\det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(AD-BC)$$ where $A,B,C,D$ are not square matrices.

This is false in general. The formula holds if they're square matrices(in particular) and there is a commutation condition between $C$ and $D$.

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