I have an arbitrary $n \times n$ matrix with ones on the main diagonal and free elements everywhere else. I am trying to understand if I can write the determinant of this matrix in a "nice" way, as a function of its elements and $n$. For example I thought that I could rewrite the matrix as:
$det(A) = det (\tilde{L} + U)$
where $\tilde{L}$ is strictly lower triangular and $U$ is the upper triangular matrix with the ones on its main diagonal. My idea is to exploit the fact that $det(\tilde{L}) = 0$ and $det(U) = 1$ in the formula for the determinant of the product of two matrices look here for instance. With $n = 2$ for example I would have something like:
$det(\tilde{L} + U) = det(\tilde{L}) + det(U) + tr( adj(\tilde{L})U ) $
Then since $\tilde{L}$ is strictly singular, following this, $adj(\tilde{L})$ will be a matrix with zeros everywhere except on the bottom left corner element. Indeed I would have:
$det(\tilde{L} + U) = det(\tilde{L}) + det(U) + tr( adj(\tilde{L})U ) $ $ = 1 + tr( (-1)^{n+1}l_{21}l_{32}\ldots l_{n-1,n}e_{n1})U) $
where $e_{n1}$ is a an $n \times n$ matrix with all zeros except the element $(n,1)$. In terms of the elements of $A$ this would be:
$ det(A) = 1 + tr( (-1)^{n+1}a_{21}a_{32}\ldots a_{n-1,n}e_{n1})U) $
making the computation with n = 2 would be:
$det(A) = 1 + (-1)^{2+1}a_{21} a_{12}) + a_{21}((-1)^{2+1}a_{21}a_{12}) $
How could I generalize this approach to $n>2$? Do you now another, maybe simpler way, to write the determinant of $A$ as a function of the elements of $A$ and $n$ in my framework?
