TITLE: How to compute $\det(A+J)$?
QUESTION [5 upvotes]: If $A$  be an $n\times n$ matrix and $J$ be a matrix of same order with all entries $1$ then 

Show that $\det( A + J)=\det A$ + sum of all cofactors of $A$.

I have tried using Laplace Expansion but I am not getting it.
Please give some hints at doing the same.

REPLY [2 votes]: Hint: Let $a_1,\dots,a_n$ be the columns of $A$ and let $v_0$ be the vector $(1,1,\dots,1)$. When you expand $\det(a_1+v_0,\dots,a_n+v_0)$ according to multilinearity of the determinant, then apart from $\det(A)$ you only get $n$ further summands which can be non-zero. The latter summands can be easily computed by expansion along their "special column". 
Edit: The $n$ potentially non-zero additional summands are of the form $\det(a_1,\dots,a_{i-1},v_0,a_{i+1},\dots,a_n)$. Computing this determinant by expansion with respect to the $i$th column (the one containing $v_0$) you get a sum of cofactors of $A$ (with appropriate signs).