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(D) matrix is invertible if and only if its determinant is non-zero. Let $A$ be the given matrix.$|A| = \left|\begin{array}{ccc}e^t & e^{-t}(\sin t-2

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$R$

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(D) matrix is invertible if and only if its determinant is non-zero. Let $A$ be the given matrix.$|A| = \left|\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \ e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \ e^t & e^{-t} \cos t & e^{-t} \sin t\end{array}ight|$Taking $e^t$ common from $C_1$ and $e^{-t}$ common from $C_2$ and $C_3$:$|A| = e^t \cdot e^{-t} \cdot 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}ight|$$|A| = e^{-t} \left|\begin{array}{ccc}1 & \sin t -2 \cos t & -2 \sin t-\cos t \ 1 & 2 \sin t+\cos t & \sin t-2 \cos t \ 1 & \cos t & \sin t\end{array}ight|$Applying $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$|A| = e^{-t} \left|\begin{array}{ccc}0 & -\sin t - 3\cos t & -3\sin t - 2\cos t \ 0 & 2\sin t & -2\cos t \ 1 & \cos t & \sin t\end{array}ight|$Expanding along $C_1$:$|A| = e^{-t} \cdot 1 \cdot [(-\sin t - 3\cos t)(-2\cos t) - (2\sin t)(-3\sin t - 2\cos t)]$$|A| = e^{-t} [2\sin t \cos t + 6\cos^2 t + 6\sin^2 t + 4\sin t \cos t]$$|A| = e^{-t} [6(\sin^2 t + \cos^2 t) + 6\sin t \cos t] = e^{-t} [6 + 3\sin(2t)]$Wait,re-evaluating the determinant expansion:$|A| = e^{-t} [2\sin t \cos t + 6\cos^2 t + 6\sin^2 t + 4\sin t \cos t] = 6e^{-t}$.Since $6e^{-t} 
eq 0$ for all $t \in R$,the matrix is invertible for all $t \in R$.

The set of all values of $t \in R$,for which the matrix $\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]$ is invertible.

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