5
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We have $\det M=(a+b)(c+d)$ where $M=\begin{bmatrix} a& 0& -1& 0\\ 0& c& 0& -1\\ b& 0& 1& 0\\ 0& d& 0& 1 \end{bmatrix}$ and $\det M'=(a'+b')(c'+d')$ where $M'=\begin{bmatrix} a'& 0& -1& 0\\ 0& c'& 0& -1\\ b'& 0& 1& 0\\ 0& d'& 0& 1 \end{bmatrix}$.

Is there a matrix $A$ with linears forms in $a,b,c,d,a',b',c',d',\pm1,0$ as entries that somehow uses $M$ and $M'$ as building blocks that gives $$ \det A = (a+b)(c+d)+(a'+b')(c'+d') $$ (if needed we can use other $\Bbb Z$ entries but I would prefer not)?

It is ok to modify $M$ and $M'$ and use certain auxiliary supporting matrices as long as some structure of underlying matrices involved is evident.

Determinant of the following matrix suffices.

$$\begin{bmatrix} 1&0&0&0&0&0&0&0&c\\ 0&1&0&0&0&0&0&0&c\\ 0&0&1&0&0&0&0&0&d\\ 0&0&0&1&0&0&0&0&d\\ 0&0&0&0&1&0&0&0&c'\\ 0&0&0&0&0&1&0&0&c'\\ 0&0&0&0&0&0&1&0&d'\\ 0&0&0&0&0&0&0&1&d'\\ -a&-b&-a&-b&-a'&-b'&-a'&-b'&0 \end{bmatrix}$$

So we know such matrices exist. However there is a certain trouble here because it needs us to expand the polynomial fully and write down the determinant. There is no intuition of using $M$ and $M'$ as a building block.

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  • $\begingroup$ The query came from understanding formula complexity. Don't understand the meaning of this ridicule. $\endgroup$ – T.... Jul 24 '17 at 9:10
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    $\begingroup$ I didn't vote to close, so this is an honest question: do you require A to also be a 4 by 4 matrix, or can it be larger? $\endgroup$ – Yemon Choi Jul 24 '17 at 20:35
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    $\begingroup$ I actually suspect this is not possible, but I don't know how to prove that. It seems an interesting question to me, but maybe I missed something. I voted to reopen. $\endgroup$ – Christian Remling Jul 25 '17 at 4:09
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    $\begingroup$ The more general question is: Given matrices $M$ and $N$, when does there exist a matrix $P$ all of whose entries appear in $M$ and $N$ and such that $det(P)=det(M)+det(N)$ ? Can you settle any non-trivial cases of this? $\endgroup$ – Steven Landsburg Jul 25 '17 at 4:16
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    $\begingroup$ @StevenLandsburg I think if I am right any polynomial can be expressed as determinant (and the size is the central issue as in permanent determinant case) and so $det(M)+det(N)$ should be representable as a determinant. $\endgroup$ – T.... Jul 25 '17 at 7:39

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