---

Solution

Let $m$ and $M$ be respectively the minimum and maximum values of \[ \left|\begin{matrix} \cos^2 x & 1+\sin^2 x & \sin 2x\\ 1+\cos^2 x & \sin^2 x & \sin 2x\\ \cos^2 x & \sin^2 x & 1+\sin 2x \end{matrix}\right|. \] Then the ordered pair $(m, M)$ is equal to : Solution: Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ \[ \Delta = \left|\begin{matrix} \cos^2 x & 1+\sin^2 x & \sin 2x\\ 1 & -1 & 0\\ 0 & -1 & 1 \end{matrix}\right| \] Expanding along the first row, \[ \Delta = \cos^2 x(-1) - (1+\sin^2 x)(1) - \sin 2x(1) \] \[ \Delta = -(\cos^2 x + \sin^2 x + 1 + \sin 2x) \] \[ \Delta = -2 - \sin 2x \] Since $\sin 2x \in [-1,1]$, \[ \Delta \in [-3, -1] \] Hence, $m = -3$, $M = -1$ Therefore, the ordered pair is $\boxed{(-3, -1)}$.