The set of all values of $\mathrm{t\in \mathbb{R}}$, for which the matrix

$$\left[ {\matrix{ {{e^t}} & {{e^{ - t}}(\sin t - 2\cos t)} & {{e^{ - t}}( - 2\sin t - \cos t)} \cr {{e^t}} & {{e^{ - t}}(2\sin t + \cos t)} & {{e^{ - t}}(\sin t - 2\cos t)} \cr {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr } } \right]$$ is invertible, is :

If the matrix is invertible then its determinant should not be zero. <br/><br/>So, $$ \left|\begin{array}{ccc} e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right| \neq 0 $$ <br/><br/>$$ \Rightarrow e^t \times e^{-t} \times e^{-t}\left|\begin{array}{ccc} 1 & \sin t-2 \cos t & -2 \sin t-\cos t \\ 1 & 2 \sin t+\cos t & \sin t-2 \cos t \\ 1 & \cos t & \sin t \end{array}\right| \neq 0 $$ <br/><br/>Applying, $R_1 \rightarrow R_1-R_2$ then $R_2 \rightarrow R_2-R_3$, <br/><br/>$$ \begin{aligned} & e^{-t}\left|\begin{array}{ccc} 0 & -\sin t-3 \cos t & -3 \sin t+\cos t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right| \neq 0 \\\\ & \Rightarrow e^{-t}\left(2 \sin t \cos t+6 \cos ^2 t+6 \sin ^2 t-2 \sin t \cos t\right) \neq 0 \\\\ & \Rightarrow 6e^{-t} \neq 0, \text { for } \forall t \in R . \end{aligned} $$