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?
 A: Another way to derive this result:
When the sum of all rows are zero, the row vectors are linearly dependent. Hence the determinant is zero. 
In fact, if any two or three or four rows add up to pure zero, the determinant is zero as well.
A: 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.
A: Let your matrix be called $A$.  Then set $x$ to be the column vector of all 1's.  You have $$Ax=0=0x$$
Hence $x$ is an eigenvector of $A$, while $0$ is an eigenvalue. Since the determinant is the product of the eigenvalues, and one of those is zero, the determinant must be zero.
A: It doesn't change the determinant of a matrix if we take a column and replace it by the sum of itself and (any scalar multiple of) another column in the matrix.
So take the last column and replace it by the sum of itself with the first column. Next, replace it by the sum of itself with the second column. Continue. Repeating this process till you get to the next-to-last-column, you see that you have replaced each entry the original last column with the sum of all the entries in that row, which is zero. So now the entire last column is zero, and the determinant must be zero.
A: Hint: The vector
$$\begin{bmatrix}1 \\  1 \\ \vdots  \\ 1  \end{bmatrix}$$
is an eigenvector with eigenvalue __?
