I have an implementation that produces a matrix $A$. I am working on the trace of it's inverse $tr(A^{-1})$. In my problem, there are just two states for the result. Sometimes the result is very high and sometimes is low.
For example, in an implementation it produced: $$A= \begin{bmatrix} .8& .2& 0& .8& .8& \\ .2& 1& 0& .5& .2& \\ 0& 0& 1& 0& 0& \\ .8& .5& 0& 1.3& .8& \\ .8& .2& 0& .8& .8& \\ \end{bmatrix}$$
which the trace of it's inverse is $10^{32}$, while in a other implementation it produced: $$A= \begin{bmatrix} 1.4& .2& 0& .1& .7& \\ .2& 1& 0& .5& .2& \\ 0& 0& 1& 0& 0& \\ .1& .5& 0& .3& .1& \\ .7& .2& 0& .1& .4& \\ \end{bmatrix}$$
which $tr(A^{-1})=10^1$. Also, there are some same columns and rows in both of them (for example the second and third columns in the first matrix are the same as the second and third columns in the second matrix).
My question is, what parameters may have an effect on the trace of an inverse matrix ??
And my second question is what does it mean and is it possible to estimate that the $tr(A^{-1})$ is low or high from $A$ before calculating?
If some rows and columns are the same, then the matrix is singular and hence they are not invertible. That is $A^{-1}$ should not exists.
Notice that $trace(A)$ is the sum of the eigenvalues, if a matrix is singular, then the matrix has $0$ has an eigenvalue. Perhaps numerically you might get a small number but upon inverting it, you get a very huge number.
If your interest is just computing the trace of the inverse, you can compute the eigenvalues and check if they are zero, suppose not, you can invert the eigenvalues and sum them up.