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Let $A, B,C,D$ be square matrices, of same size $n\times n$.

Assume $A$ is invertible and $AC = CA.$

Prove that

$$ det \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix} = det[AD - BC] $$

I am getting my final answer as $det[AD - CB]$ instead, and my work looks fine -- just a simple factorization of the matrix, and using the multiplicativity of the determinant, $AA^{-1}=I$, and $AC=CA$.

Is the problem statement wrong? Or perhaps my answer of $det[AD - CB]$ is actually equal to $det[AD - BC]$, but I doubt it -- there are no extra assumptions on the relations of $B$ and $C$ other than that they are of the same size. So we can't assume that $B$ and $C$ commute.

Thanks,

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User001
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    The best way to sort this out is to test the formulas on some actual examples. (A computer algebra system is perfect for that). – Giuseppe Negro Oct 02 '15 at 01:56
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    isn't it $\det(AD - CB)$? – user251257 Oct 02 '15 at 01:59
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    To build non trivial commuting matrices $A$ and $C$, try taking diagonal matrices $D_A, D_C$ and conjugating it with the same invertible matrix $P$, that is: $A=PD_AP^{-1}, C=PD_CP^{-1}$. – Giuseppe Negro Oct 02 '15 at 01:59
  • Hi @user251257 - yes, my answer agrees with yours. I am betting at this point that the problem statement is wrong. But I am also considering what Guiseppe Negro is suggesting, too. Thanks... – User001 Oct 02 '15 at 02:00
  • Hi @GiuseppeNegro, why would we want to build commuting matrices A and C, when the issue is with the matrices B and C? Thanks, – User001 Oct 02 '15 at 02:03
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    @LebronJames: Your hypothesis is that $A,C$ commute, right? – hardmath Oct 02 '15 at 02:04
  • Hi @hardmath -- yes, absolutely. Hmm...can I deduce something about the relation between B and C from this... – User001 Oct 02 '15 at 02:06
  • Hi @hardmath, does the assumption that A and C commute tell us something about the relation between B and C? I don't see any, unfortunately... – User001 Oct 02 '15 at 02:16
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    While no special relationship between $B$ and $C$ can be deduced, it is true that $\det BC = \det CB$. – hardmath Oct 02 '15 at 02:36
  • Hi @hardmath, yes - I had thought of that, but the final answer is in the form of det( difference of matrices ), not det(product of matrices). So I still don't see the justification for swapping the order of CB. What do you think? Thanks, – User001 Oct 02 '15 at 02:44
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    You are right and your book is wrong. – user1551 Oct 02 '15 at 06:18
  • I mean, take some $2\times 2$ matrices and test both formulas, to see if one of them does not hold. – Giuseppe Negro Oct 02 '15 at 06:18
  • Ok, got it. Thanks so much, @user1551 - have a great day :-) – User001 Oct 02 '15 at 22:14

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