The set of all values of t , for which the 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 ] is invertible, is [2023]

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Solution

Q.
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 [2023]

1 ${(2 k + 1) \frac{π}{2}, k \in ℤ}$

2 $ℝ$

3 ${k π + \frac{π}{4}, k \in ℤ}$

4 ${k π, k \in ℤ}$

Ans.
(2)

$If the matrix is invertible, then its determinant should not be zero.$

So, $| \begin{matrix} e^{t} & e^{- t} (\sin t - 2 \cos t) & e^{- t} (- 2 \sin t - \cos t) \\ e^{t} & e^{- t} (2 \sin t + \cos t) & e^{- t} (\sin t - 2 \cos t) \\ e^{t} & e^{- t} \cos t & e^{- t} \sin t \end{matrix} |$ $\neq 0$

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

$Applying R_{1} \to R_{1} - R_{2} then R_{2} \to R_{2} - R_{3}, we get:$

$e^{- t}$ $| \begin{matrix} 0 & - \sin t - 3 \cos t & - 3 \sin t + \cos t \\ 0 & 2 \sin t & - 2 \cos t \\ 1 & \cos t & \sin t \end{matrix} |$ $\neq 0$

$\Rightarrow e^{- t} (2 \sin t \cos t + 6 \cos^{2} t + 6 \sin^{2} t - 2 \sin t \cos t) \neq 0$

$\Rightarrow e^{- t} \times 6 \neq 0, \forall t \in ℝ .$

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