Why the determinant of a matrix with the sum of each row's elements equal 0 is 0? I'm trying to understand the proof of a problem, but I'm stuck. In my book they consider that if all lines of a matrix has sum 0 then it's determinant is also 0. I checked some random examples and it's true, but I couldn't proof it. Could you help me?

Geometrically speaking, the determinate calculates the area, volume, or hyper-volume of a trapezoid, parallelepiped, of hyper-parallelepiped in $\mathbb{R}^2$, $\mathbb{R}^2$, or $\mathbb{R}^n$ (respectively). A row full of zeros in a $n\times n$ matrix implies that the hyper-parallelepiped is "missing" a dimension that is needed in order to compute the hyper-volume; thereby causing the determinate to become zero. 
To better visualize this explanation, visualize how a trapezoid (a two dimensional object) can be represented by a $3\times 3$ matrix. Since a trapezoid does not have a volume, then the determinate must be zero. 
Algebraically speaking, apply the technique of co-factor expansion.
