determinant of this symmetric matrix I want to know whether the $\mbox{det}(A)\neq0$, 
$$A = \left( {\begin{array}{*{20}{c}}
{b_1^2 + {e_1}}&{{b_1}{b_2}{\rho _{12}}}&{{b_1}{b_3}{\rho _{13}}}\\
{{b_1}{b_2}{\rho _{12}}}&{b_2^2 + {e_2}}&{{b_2}{b_3}{\rho _{23}}}\\
{{b_1}{b_3}{\rho _{13}}}&{{b_2}{b_3}{\rho _{23}}}&{b_3^2 + {e_3}}
\end{array}} \right)$$
where $b_i,e_i>0$ for $i=1,2,3$, and $\rho_{ij}\in (-1,1)$.
From my tests with some values for the variables, det(A) is not zero. But I do not know whether it holds. Appreciate any suggestion!

One simple sufficient condition for $\det(A) > 0$ is the coefficients $\rho_{ij}$ satisfy an extra inequality:
$$1 - \rho_{12}^2 - \rho_{23}^2 - \rho_{13}^2 + 2\rho_{12}\rho_{23}\rho_{13} \ge 0\tag{*1}$$
If this is satisfied, the part of matrix excluding the $e_k$ pieces will be positive semi-definite. Together with the $e_k$ pieces, the matrix become positive definite and hence $\det(A) > 0$. The condition on $\rho_{ij}$ can be rewritten as
$$| \rho_{13} - \rho_{12}\rho_{23} | \le \sqrt{1 - \rho_{12}^2} \sqrt{1 - \rho_{23}^2}$$
If $\rho_{ij}$ are correlations among random variables $X_i$ and $X_j$, above condition can be interpreted as:

If $X_1$ is strongly correlated/anti-correlated to $X_2$ and $X_2$ is strongly correlated/anti-correlated to $X_3$, then $X_1$ is strongly correlated/anti-correlated to $X_3$. Furthermore, the correlation $\rho_{13}$ is roughly equal to $\rho_{12}\rho_{23}$, the product of the other two correlations.

To construct a counter-example for the problem, one start from a configuration
where $(*1)$ is violated the most.
For example, I start with the configuration
$$\rho_{12} = \rho_{23} = 1,\quad\rho_{13} = 0$$
perturb it a little bit to get
$$\begin{cases}
e_1 = e_2 = e_3 = t,\\
b_1 = b_2 = b_3 = 1,\\
\rho_{12} = \rho_{23} = 1-t,\\
\rho_{13} = 0
\end{cases}
\quad\iff\quad
A(t) = \begin{bmatrix}
1 + t & 1-t & 0\\
1 - t & 1+t & 1-t\\
0 & 1 -t & 1 + t
\end{bmatrix}
$$
It is easy to check 
$$\det(A(t)) = -(t+1)(t^2 - 6t + 1)
\quad\text{ and }\quad \det(A(0)) < 0
$$
Solving the quadratic equation $t^2 - 6t + 1 = 0$ gives us $t = 3 - \sqrt{8}$. 
For this particular $t$, the corresponding $A(t)$ has the form specified in question and yet $\det(A(t)) = 0$.
