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?