The assertion is not true. Counter-example.
$$\begin{vmatrix}
0 & 2 & 1 & 1 \\
2 & 0 & 1 & 1 \\
1 & 1 & 0 & 2 \\
1 & 1 & 2 & 0
\end{vmatrix} $$
If entries of the matrix are distinct, then we need to show that rows (or columns) of this matrix are linearly independent. Let's do it with a $4 \times 4$ matrix and then we'll try to generalize.
$$ A = \begin{vmatrix}
0 & a & b & c \\
a & 0 & d & e \\
b & d & 0 & f \\
c & e & f & 0
\end{vmatrix}$$ where $ a \ne b \ne c \ne d \ne f \ne g$.
Assume that these rows are linearly dependent (i.e. we can write one of the rows in terms of other rows). Without loss of generality, we can write row 1 in terms of other rows. Then one must find $x_1, x_2, x_3$ (not all of them zero) such that,
$$ax_1 + bx_2 + cx_3 = 0 \\
0x_1 + dx_2 + ex_3 = a \\
dx_1 + 0x_2 + fx_3 = b \\
ex_1 + fx_2 + 0x_3 = c
$$
If we can show that there is a contradiction (i.e. there are no such $x_1, x_2, x_3$ which satisfy above system, we prove that rows of this matrix are linearly independent, thus determinant of this matrix in non-zero.