This question was motivated by a recent MO post. You know $n$ elements of the $N\times N$ matrix $M$ and you do not know $n$ elements of the inverse $M^{-1}$ (but you know the other $N^2-n$ elements of $M^{-1}$). Equating $(M^{-1})^{-1}=M$ gives $n$ nonlinear equations in $n$ unknowns, which in general will have multiple solutions. Under which additional condition can one reconstruct the matrix $M$ uniquely? Does it matter where in the matrix are the $n$ elements located?
Conjecture: For $n=N$ elements on the diagonal the reconstruction is unique if $M$ is positive definite.
Alternative formulation: there does not exist a pair of distinct positive definite $N\times N$ matrices with the same diagonal, such that their inverses differ only on the diagonal.
For $N=2$ it is true,$^\ast$ and some experimentation$^{\ast\ast}$ for larger $N$ suggests it is true for all $N$.
$^\ast$ For $N=2$ one has $M = \begin{pmatrix}a & b \\ b & c \end{pmatrix}$, $M^{-1} = \frac{1}{a c - b^2} \begin{pmatrix}c & -b \\ -b & a \end{pmatrix}$, we know $a,c$ and we know $\beta=b/(ac−b^2)$. There are two solutions for the unknown $b$, $b_\pm=(\pm\sqrt{4ac\beta^2+1}−1)/2\beta$, only $b_+$ gives a positive definite $M$.
$^{\ast\ast}$ Mathematica test for $N=3,4,5$, when there are, respectively, up to $5,14,22$ solutions for the unknown matrix elements, but only one of these gives a positive definite $M$.