Let m and M be respectively the minimum and maximum values of $$\begin{vmatrix} \cos^2 x & 1 + \sin^2 x & \sin 2x \\ 1 + \cos^2 x & \sin^2 x & \sin 2x \\ \cos^2 x & \sin^2 x & 1 + \sin 2x \end{vmatrix}$$. Then the ordered pair (m, M) is equal to:

Let us write $$\cos x = c \quad\text{and}\quad \sin x = s.$$

With this notation one has $$\cos^2 x = c^2,\; \sin^2 x = s^2,\; \sin 2x = 2sc.$$

So the given determinant becomes

$$\Delta \;=\; \begin{vmatrix} c^2 & 1+s^2 & 2sc\\[2pt] 1+c^2 & s^2 & 2sc\\[2pt] c^2 & s^2 & 1+2sc \end{vmatrix}.$$

We now perform the elementary row operations $$R_2 \longrightarrow R_2-R_1$$ and $$R_3 \longrightarrow R_3-R_1.$$ (Row operations of the type $$R_i\to R_i+kR_j$$ do not alter the value of a determinant.)

After applying the operations we get

$$\Delta \;=\; \begin{vmatrix} c^2 & 1+s^2 & 2sc\\[2pt] 1 & -1 & 0\\[2pt] 0 & -1 & 1 \end{vmatrix}.$$

We now expand this determinant along the first row. The cofactor of the element in the $$i^{\text{th}}$$ row and $$j^{\text{th}}$$ column is defined as $$C_{ij}=(-1)^{i+j}M_{ij},$$ where $$M_{ij}$$ is the minor obtained by deleting the $$i^{\text{th}}$$ row and $$j^{\text{th}}$$ column.

Hence

$$\Delta = c^2\,C_{11} + (1+s^2)\,C_{12} + 2sc\,C_{13}.$$

We evaluate each cofactor carefully:

Minor $$M_{11}=\begin{vmatrix}-1 & 0\\ -1 & 1\end{vmatrix}=(-1)(1)-(0)(-1)=-1.$$ Since $$(-1)^{1+1}=+1$$, we get $$C_{11}=+(-1)=-1.$$

Minor $$M_{12}=\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix}=1\cdot1-0\cdot0=1.$$ Here $$(-1)^{1+2}=-1$$, so $$C_{12}=-\,1=-1.$$

Minor $$M_{13}=\begin{vmatrix}1 & -1\\ 0 & -1\end{vmatrix}=1\cdot(-1)-(-1)\cdot0=-1.$$ Now $$(-1)^{1+3}=+1$$, giving $$C_{13}=+(-1)=-1.$$

Substituting these cofactors we obtain

$$\Delta = c^2(-1) + (1+s^2)(-1) + 2sc(-1) = -c^2 -1 -s^2 -2sc.$$

Using the Pythagorean identity $$c^2+s^2=1,$$ we simplify:

$$\Delta = -(1) -1 -2sc = -2 -2sc.$$

Recognising $$2sc=\sin 2x,$$ the determinant further reduces to

$$\boxed{\;\Delta = -2 - \sin 2x\;}.$$

The range of $$\sin 2x$$ is $$-1 \le \sin 2x \le 1.$$ Consequently

$$\begin{aligned} \text{Maximum of } \Delta &: -2 - (\sin 2x)_{\min} = -2 - (-1) = -1,\\[4pt] \text{Minimum of } \Delta &: -2 - (\sin 2x)_{\max} = -2 - 1 = -3. \end{aligned}$$

Thus $$m=-3 \quad\text{and}\quad M=-1,$$ giving the ordered pair $$(m,M)=(-3,-1).$$

Hence, the correct answer is Option B.