It is well known that for a matrix $A = (A_{i,j})$, the determinant of $A$ is
$$\det(A) = \sum_{\sigma \in S_n} \mbox{sgn} (\sigma)\prod_{i \in [n]} A_{i,\sigma(i)}.$$
Is there any way to manipulate the matrix to obtain, or any existing interpretation for, the "absolute" version of the determinant, which is
$$\det^+(A) = \sum_{\sigma \in S_n} \prod_{i \in [n]} A_{i,\sigma(i)}?$$
The quantity $$\operatorname{perm}(A) = \sum_{\sigma \in S_n} \prod_{i \in [n]} A_{i,\sigma(i)}$$ is called the permanent of matrix $A$.
Technically, the answer to "is there any way to manipulate the matrix to obtain" $\operatorname{perm}(A)$ is "yes": just go through all $\sigma \in S_n$, compute $\prod_{i \in [n]} A_{i,\sigma(i)}$ for each $\sigma$, and add these values up.
However, if you are asking if there is any reasonable way to do this, the answer is probably not. Computing $\operatorname{perm}(A)$ is $♯P$-complete, which means there is no known polynomial-time algorithm, and good reason to believe one does not exist.
This remains true if the entries of $A$ are all either $0$ or $1$, which also leads to one of the important interpetations of $\operatorname{perm}(A)$. Let $G$ be a bipartite graph with $n$ vertices on each side; let $A$ record which edges $G$ has by setting $A_{ij} = 1$ if there is an edge between the $i^{\text{th}}$ vertex on the left side and the $j^{\text{th}}$ vertex on the right side, and $0$ otherwise. Then $\operatorname{perm}(A)$ counts the number of perfect matchings of $G$: the number of bijections $\sigma$ between the two sides with an edge between $i$ and $\sigma(i)$ for all $i$.
