What is the value of this determinant? I started learning linear algebra few weeks ago, and the teacher asked us to evaluate the following determinant:
$$\begin{vmatrix} \cos 2019^{\circ} & \cos 1131^{\circ}  & \cos 1869^{\circ}  & \cos  1101^{\circ}\\ -\cos 1131^{\circ} & \cos 2019^{\circ}  & -\cos 1101^{\circ}  & \cos  1869^{\circ}\\  -\cos 1869^{\circ} & \cos 1101^{\circ}  & \cos 2019^{\circ}  & -\cos  1131^{\circ} \\  -\cos 1101^{\circ} & -\cos 1869^{\circ}  & \cos 1131^{\circ}  & \cos  2019^{\circ}\end{vmatrix}.$$
The teacher said that we don't need to use brute force to evaluate the determinant. But how?
I tried to use brute force and trigonometry formulas, after an hour, I got the answer $4$.
Is $4$ correct? And how to get the answer without using brute force?
 A: Note that the columns of your matrix are orthogonal; same is true for the rows.  Note also that
$$
\cos(2019^\circ) = \cos(219^\circ) = -\sin(51^\circ)\\
\cos(1131^\circ) = \cos(51^\circ)\\
\cos(1869^\circ) = \cos(69^\circ)\\
\cos(1101^\circ) = \cos(21^\circ) = \sin(69^\circ)
$$
All this enough to deduce that the determinant of your matrix must either be $+4$ or $-4$.  I'm not sure how to rule out the possibility that the determinant of the matrix is $-4$ using only elementary techniques.

Another approach (interesting to me, but overkill if you're starting linear algebra): in terms of Kronecker products, we have
$$
M = I \otimes \pmatrix{\sin(-51) & \cos(-51)\\-\cos(-51) & \sin(-51)} + \pmatrix{0&1\\-1&0} \otimes \pmatrix{\cos(69) & \sin(69)\\ -\sin(69) & \cos(69)}
$$
By simultaneously diagonalizing the two matrices being added (which can be done by separately diagonalizing the left matrices in the product and the right matrices in the product), we can find the eigenvalues of $M$ and thus compute the determinant of $M$.

Another (more reasonable) approach: your matrix has the form
$$
M = \pmatrix{A & B\\C&D}
$$
where $C,D$ satisfy $CD = DC$.  It follows that $\det(M) = \det(AD - BC)$.  After computing the $2 \times 2$ matrix $AD - BC$, finding the determinant is easy.
A: I don' t see that suitable after some weeks taking linear algebra anyway your matrix $O$ is orthogonal which is not direct as said, now $O^TO=16I_4$ so $|\det (O)|=\pm 4$. If you separate the main diagonal from $O$ you obtain a skew hermitian matrix whose eigenvalues are pure imaginary and conjugate one another now you add the diagonal so  you translate these eigenvalues by $\cos(2019^\circ)$. It should be clear that the determinant is $>0$
