Every matrix $A$ with $\text{det}(A-A^T)=1$ can be realized as Seifert matrix

Question

Let $A\in\text{Mat}(2n\times2n;\mathbb{Z})$ be an integer matrix with $\text{det}(A-A^T)=1$. I got told that every such matrix can be realized as Seifert matrix of a Seifert surface $F$ of a knot $K$ with an appropriate basis of $H_1(F;\mathbb{Z})$. I would be thankful for any reference on this statement or an argument how such a matrix can be realized as a Seifert matrix of some knot.