Determinant of specific infinite matrix What is the limit, as n approaches infinity, of the determinant of an n x n matrix where each cell has the value $\cos(n * row + column)$? 
My friend and I believe the answer to be 0, but can't figure out a method of finding the solution without n! steps.
We are interested in how this can be solved as well as the answer.
 A: The matrix has determinant equal to $0$ for $n \ge 3$. It is enough to consider the first $3$ rows. After some explicit Gaussian elimination it turns our the following linear combination vanishes
$$r_1 - 2\cos(n) r_2 + r_3 = 0$$
and therefore the first three rows are linearly dependent, so the matrix has determinant $0$. Indeed, we take the $k$-th entry of this row vector
$$r_{1,k} - 2\cos(n) r_{2,k} + r_{3,k} = \cos(n+k) - 2\cos(n)\cos(2n+k) + \cos(3n+k)$$
By the product-to-sum identity we have
$$2\cos(n)\cos(2n+k) = \cos(n+k) + \cos(3n+k)$$
establishing the linear dependence. The same process will yield that the determinant will be zero in the case of a sine matrix of the same form as well (for $n \ge 3$). It's curious to note that if we took $C$ to be the cosine matrix and $S$ to be the sine matrix, together they give a matrix $E = C + iS$ of exponentials. It's quite easy to show that the first two rows of this matrix are linearly dependent and therefore $\det(E) = 0$.

Just for completeness here is the Gaussian elimination itself. We start with
$$r_2 \mapsto r_{1,1}r_2 - r_{2,1}r_1 = r_2'$$
$$r_3 \mapsto r_{1,1}r_3 - r_{3,1}r_1 = r_3'$$
Next we do this again for the 3rd row
$$r_3' \mapsto r_{2,2}'r_3' - r_{3,2}'r_2' = r_3''$$
where
$$r_{2,2}' = r_{1,1}r_{2,2} - r_{2,1}r_{1,2} = -\sin(1) \sin(n)$$
$$r_{3,2}' = r_{1,1}r_{3,2} - r_{3,1}r_{1,2} = -\sin(1) \sin(2n)$$
and I've used the product-to-sum identities and then the sum-to-product identities to simplify the expressions. Next we have
$$r_{2,2}'r_3' - r_{3,2}'r_2' = (r_{3,2}'r_{2,1}-r_{2,2}'r_{3,1})r_1 - r_{3,2}'r_{1,1}r_2 + r_{2,2}'r_{1,1}r_3$$
Again, simplifying these expressions we have
$$r_3'' = -\sin(1)\sin(n)\cos(n+1)(r_1 - 2\cos(n) r_2 +  r_3)$$
When performing the actual calculation we can ignore the $-\cos(n+1)\sin(1)\sin(n)$ factor so we really have
$$r_3'' = r_1 - 2\cos(n) r_2 +  r_3$$
