I'd like to prove following:
$\det(M+U)=m·s $ where $m$ is a cofactor of $M$ and s is the sum of all entries in $U$ where $M$ is square matrix and sum of each row in $M$ is zero and $U$ is a square matrix with rank $1$
It would've been better if the $\det $ operation is multiplicative, but it looks not.
I think this proof start from the analogy of the case where we usually add the matrix $J$ where its all entries are $1$. $J$ also has rank 1. I'd like to use this fact so first prove $\det(M+J)=m\cdot n^2$ and then generalize it to show $\det(M+U)=m·s$.
- Is it true that $\det(M+J)=m\cdot n^2?$
- If not, any other approach to prove it?