Let $A$ be an $n\times n$ matrix and $Z$ be the $n\times n$ matrix, whose entries are all $m$. Let $S$ be the sum of all the adjoints of $A$. Then my conjecture is
$\det(A+Z)=\det(A)+Sm$ , in particular $\det(A+Z)=\det(A)$ holds if and only if $S=0$. If the conjecture is true, how can it be proven ?
For invertible $A$, sylvester's theorem can be used. $\det(A+Z)=\det(A)\det(I+A^{-1}Z)$ The matrix $A^{-1}Z$ is the product of the row vector containing the row sums of $A^{-1}$ and the column vector $(m,...,m)$. Sylvester's theorem states that the order of the vectors can be exchanged. The scalar product of the vectors is $0$ if and only if the sum of the row sums of $A^{-1}$ is $0$ and this is the same as the sum of the adjoints of $A$ divided by the determinant of $A$. So, for invertible matrices, my conjecture should hold. But how can I manage the case when $A$ is singular ?
You can use a continuity argument. The function $$A\longmapsto \det(A+Z)-\det(A)-Sm$$ is a polynomial function from the vector space $M_n(\mathbb C)$ of all $n\times n$ complex matrices to the complex numbers, so in particular it is continuous. You claim that it is zero on the set of matrices with non-zero determinant. Since this set is dense in $M_n(\mathbb C)$, the function is in fact identically zero.