Determinant of a special $n\times n$ matrix [duplicate]
Solution 1:
A standard result (http://en.wikipedia.org/wiki/Matrix_determinant_lemma) is $\det(I+AB) = \det(I+BA)$.
Since the matrix above can be written as $I+ e e^T$, where $e$ is a vector of ones, we have $\det(I+ e e^T) = \det(1+ e^T e) = 1+e^Te = n+1$.
Solution 2:
Let $$v=(1,1,1,1...1)^T$$
Your matrix is $$ I + v v^T$$ This has $n-1$ eigenvalues equal to $1$ and one with value $n+1$. Hence the result.