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

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${(2 k + 1) \frac{π}{2}, k \in ℤ}$

${k π + \frac{π}{4}, k \in ℤ}$

${k π, k \in ℤ}$

$ℝ$

answer is D.

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Detailed Solution

If its invertible, then determinant value $\neq 0$
So ,
$| \begin{matrix} e^{t} & e^{- t} (\sin t - 2 \cos t) & e^{- t} (- 2 \sin t - \cos t) \\ e^{t} & e^{- t} (2 \sin t + \cos t) & e^{- t} (\sin t - 2 \cos t) \\ e^{t} & e^{- t} \cos t & e^{- t} \sin t \end{matrix} | \neq 0$

$\Rightarrow e^{t} \cdot e^{- t} \cdot e^{- t} | \begin{matrix} 1 & \sin t - 2 \cos t & - 2 \sin t - \cos t \\ 1 & 2 \sin t + \cos t & \sin t - 2 \cos t \\ 1 & \cos t & \sin t \end{matrix} | \neq 0$

Applying, $R_{1} \to R_{1} - R_{2}$ then $R_{2} \to R_{2} - R_{3}$
We get
$| 0 - \sin t - \cos t - 3 \sin t + \cos t |$
$e^{- 1} | \begin{matrix} 0 & 2 \sin t & - 2 \cos t \\ 1 & \cos t & \sin t \end{matrix} | \neq 0$
By expanding we have,
$e^{- t} \times 1 (2 \sin t \cos t + 6 \cos^{2} t + 6 \sin^{2} t - 2 \sin t \cos t) \neq 0$
$\Rightarrow e^{- t} \times 6 \neq 0$ for $\forall t \in ℝ$