Is there any relation between determinant of a matrix with positive entries and its largest eigenvalue? Is there any relation between determinant of a matrix with positive entries and its largest eigenvalue ?
 A: Take the matrix $\begin{bmatrix} a & \epsilon \\ \epsilon & 1/a\end{bmatrix}$. 
The determinant is $1-\epsilon^2$. Taking $a$ to be large and $\epsilon$ to be small, you see that the largest eigenvalue behaves roughly like $a$. So, the determinant and largest eigenvalue can be arbitrarily off. 
As suggested in the comments by Evargalo, let $\lambda$ be the largest eigenvalue of $A$ in absolute value, and $A$ be a n x n matrix. Then, $det(A) \leq |\lambda|^n$. The proof is:Let $A$ have eigenvalues $\lambda_1,\ldots,\lambda_n$. Then, $det(A)= \lambda_1\ldots\lambda_n$, so $\det(A) \leq |\det(A)| = |\lambda_1|\ldots|\lambda_n| \leq |\lambda|\ldots|\lambda|=|\lambda|^n$. 
To see you can't do much better than the bound that Evargalo suggests, you can take matrices of the form 
$\begin{bmatrix} a & 1/a \\ 1/a & a\end{bmatrix}$
and look when $a$ is large. The determinant is $a^2 - \frac{1}{a^2} \approx a$ for large $a$, and the eigenvalues are $\frac{a^2 \pm 1}{a}\approx a$, so the determinant and square of the largest eigenvalue are close (both behave like $a^2$).  

Note that if you had non-negative entries (the case pointed out by John Hughes in the comments), the problem is even more straightforward -- take $\epsilon=0$ in the first example, and $aI$ for the second example, and you see matrices with determinant $1$ with largest eigenvalue $a$ and a matrix with determinant $a^2$ and largest eigenvalue $a$, respectively. 
