Determine the determinant!

Adding a multiple of a row to another doesn't change the determinant, so add -1 times the first row to each of the other rows:

$\left( \begin{matrix} 2017 & 1 & 1 & \ldots & 1 \\ -2016 & 2016 & 0 & \ldots & 0 \\ -2016 & 0 & 2016 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -2016 & 0 & 0 & \ldots & 2016 \end{matrix} \right)$

Except for top left element (which is 2017), the first row contains all 1s, the diagonal contains all 2016s, and the first column contains all -2016s.

Add $-\frac{1}{2016}$ of each row except the first to the first row. The first row vanishes, except for the first element which becomes 4033.

Since the matrix is now lower diagonal, the determinant is the product of the diagonal entries: $4033 \cdot 2016^{2016} = (37 \cdot 109) \cdot (2^5 \cdot 3^2 \cdot 7)^{2016}$ . It's well known that the number of positive divisors of an integer with factorization $p_1^{e_1} p_2^{e_2} p_3^{e_3} \ldots$ is $(1+e_1)(1+e_2)(1+e_3) \ldots$ , so applying to this, the factorization is $37^1 \cdot 109^1 \cdot 2^{10080} \cdot 3^{4032} \cdot 7^{2016}$ and the number of positive divisors is $2 \cdot 2 \cdot 10081 \cdot 4033 \cdot 2017 = \boxed{328018037764}$ .

Clearly, $2016$ is an eigenvalue of the matrix (call the determinant $A$ ).

Also, the matrix $A-2016I$ has all its elements $1$ , so its nullity is $2017-1 = 2016$ . Therefore, the geometric multiplicity of the eigenvalue, $2016$ is $2016$ , and hence its algebraic multiplicity is at least $2016$ (and at most $2017$ , because the order of the matrix is $2017$ ).

If the eigenvalue $2016$ would have had multiplicity $2017$ , then (trace $A$ ) $=$ sum of all eigenvalues of $A$ ,and so

$2017 \times 2017 = 2016 \times 2017$ , which is absurd.

It follows that the eigenvalue $2016$ has multiplicity $2016$ . Let $\lambda$ be the last eigenvalue. Then (trace $A$ ) $=$ sum of all eigenvalues of $A$ , so

$2017 \times 2017 = 2016 \times 2016 + \lambda$ , therefor $\lambda = 4033$ .

It follows that

$|A| = 2016^{2016} \times 4033 = (2^{5} \times 3^{2} \times 7)^{2016} \times 37 \times 109$

$= 2^{10080} \times 3^{4032} \times 7^{2016} \times 37 \times 109$ , which has

$10081 \times 4033 \times 2017 \times 2 \times 2 = 328018037764$ factors.