Given an invertible integer matrix $X \in \mathbb{Z}^n_+$, we set $A = X^T X + 1$ with $1$ denoting the identity matrix. Then, with the help of the Hadamard inequality, it is easy to find an upper bound on the determinant. But what is the lower bound of the determinant?
Since the matrix is positive definite (because $X$ is invertible), the matrix $A$ is invertible, hence, $\det A \neq 0$. It is integer and therefore we obtain $\det A \geq 1$. But so far I did not find a matrix $X$ such that $\det A = 1$. Does there exist a better bound on the determinant or does somebody know an example in order to verify that the bound is strict?
The matrix $X^tX$ is positive semidefinite so all it's eigenvalues $\lambda_i$, $1\leq i\leq n$ are nonnegative. Moreover since $X$ is invertible with integer entries we have $\det(X^tX)=1$, thus $\lambda_1\lambda_2\ldots \lambda_n=1$. We have that $\det(A)=(1+\lambda_1)(1+\lambda_2)\ldots (1+\lambda_n)$. For any nonnegative $x$ the inequality $1+x\geq 2\sqrt{x}$ holds so $\det(A)\geq 2^n$. There is no equality since this would imply $X=I_n$ and you assumed that $X$ has positive entries.