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Specifically, I want to find a matrix $A$ with the lowest dimension which its elements are in simplest decompositioned form and has a predefined determinant as follows:

$\det(A)=abcd-{\frac {a}{ \left( a+1 \right) \left( d+1 \right) \left( cd+1 \right) }}-{\frac {b}{ \left( a+1 \right) \left( b+1 \right) \left( ad+1 \right) }}-{\frac {c}{ \left( c+1 \right) \left( b+1 \right) \left( ab+1 \right) }}-{\frac {d}{ \left( c+1 \right) \left( d+1 \right) \left( bc+1 \right) }}$

where $abcd=1$.

Surely, the elements of matrix $A$ should be $0,1,-1,a,b,c,d,\frac {1}{ \left( a+1 \right)},\frac{1}{\left( b+1 \right)},\frac {1}{ \left( c+1 \right)},\frac{1}{\left( d+1 \right)},\frac{1}{\left( cd+1\right) },\frac{1}{\left( ad+1 \right) } ,\frac{1}{\left( ab+1 \right) } ,\frac{1}{ \left( bc+1 \right) } $.

I asked this question to checkout if it is possible to use hadamard's inequality theorem to show that $\det(A)\leq \frac{1}{2}$.

In general does there an algorithmic way to make such matrix from its given determinant?

Update: This inequality which is equivalent to this spatial geometric problem has been asked and solved here by carlson inequality which is a special case of holder inequality, but I'm still curious about using hadamard's inequality by resolving an algorithmic method for building a matrix from its parametric determinant, so would someone know about a computer program which carryout this?

MasM
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    For any given $x\neq1$, the matrix of lowest rank having determinant $x$ is the $1\times 1$ matrix whose single entry is $x$. The matrix of lowest rank having determinant $1$ is the $0\times 0$ matrix. You probably want to edit the question to specify what you really wanted. – Andreas Blass Feb 15 '21 at 00:32
  • Consider diagonal matrices. – leslie townes Feb 15 '21 at 00:39
  • @AndreasBlass thank for your hint, I took into account – MasM Feb 15 '21 at 00:40
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    Here's a related question: what's the smallest $n$ for which there is an $n\times n$ zero-one matrix with a unique expression as a sum of permutation matrices? – Gerry Myerson Feb 15 '21 at 01:03
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    @GerryMyerson sorry ,I couldn't find the question uou mentioned, would you please send its link. – MasM Feb 15 '21 at 01:20
  • It's not a question posted to this website, it's a question that occurred to me as I thought about your question. But I left out an important detail – I want a unique expression as a sum of five permutation matrices. – Gerry Myerson Feb 15 '21 at 02:32
  • @MasM In your $4\times 4$ matrix an element in the first row is missing. – Raffaele Feb 18 '21 at 14:42

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