The set of all values of $$t \in \mathbb{R}$$, for which the matrix $$\begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \\ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}$$ is invertible, is

We need to find all $$t \in \mathbb{R}$$ for which the given matrix is invertible (determinant $$\neq 0$$).

By factoring out $$e^t$$ from column 1, $$e^{-t}$$ from column 2, and $$e^{-t}$$ from column 3, the determinant becomes $$\det = e^{-t} \cdot \det \begin{bmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\[6pt] 1 & 2\sin t + \cos t & \sin t - 2\cos t \\[6pt] 1 & \cos t & \sin t \end{bmatrix}$$.

Applying the row operations $$R_1 \to R_1 - R_3$$ and $$R_2 \to R_2 - R_3$$ gives Row 1 as $$(0,\ \sin t - 2\cos t - \cos t,\ -2\sin t - \cos t - \sin t) = (0,\ \sin t - 3\cos t,\ -3\sin t - \cos t)$$ and Row 2 as $$(0,\ 2\sin t + \cos t - \cos t,\ \sin t - 2\cos t - \sin t) = (0,\ 2\sin t,\ -2\cos t)$$. Therefore the determinant is $$e^{-t} \det \begin{bmatrix} 0 & \sin t - 3\cos t & -3\sin t - \cos t \\[6pt] 0 & 2\sin t & -2\cos t \\[6pt] 1 & \cos t & \sin t \end{bmatrix}$$.

Expanding along the first column yields $$e^{-t}\Bigl[(\sin t - 3\cos t)(-2\cos t) - (-3\sin t - \cos t)(2\sin t)\Bigr],$$ which simplifies to $$e^{-t}\Bigl[-2\sin t\cos t + 6\cos^2 t + 6\sin^2 t + 2\sin t\cos t\Bigr] = e^{-t}\bigl[6\cos^2 t + 6\sin^2 t\bigr] = 6e^{-t}.$$

Since $$6e^{-t} \neq 0$$ for all $$t \in \mathbb{R}$$, the determinant never vanishes, and the matrix is invertible for all real $$t$$. Hence the correct answer is Option D: $$\mathbb{R}$$.