JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The set of all values of $t \in \mathbb{R}$ , for which the matrix $e^{t} e^{t} e^{t} am p; e^{- t} (sin t - 2 cos t) am p; e^{- t} (2 sin t + cos t) am p; e^{- t} cos t am p; e^{- t} (- 2 sin t - cos t) am p; e^{- t} (sin t - 2 cos t) am p; e^{- t} sin t$ is invertible, is:

Choose the correct answer:

Explanation

Solution

Let $A = \begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \\ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}$

$\Rightarrow |A| = \begin{vmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \\ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{vmatrix}$

Taking $e^t$ common from $C_1$ and $e^{-t}$ from $C_2$ and $C_3$ :

$\Rightarrow |A| = e^t \cdot e^{-t} \cdot e^{-t} \begin{vmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\ 1 & 2\sin t + \cos t & \sin t - 2\cos t \\ 1 & \cos t & \sin t \end{vmatrix}$

Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$ , we get:

$|A| = e^{-t} \begin{vmatrix} 0 & -\sin t - 3\cos t & -3\sin t + \cos t \\ 0 & 2\sin t & -2\cos t \\ 1 & \cos t & \sin t \end{vmatrix}$

Expanding along $C_1$ :

$\Rightarrow |A| = e^{-t} [2\sin t \cos t + 6\cos^2 t - (-6\sin^2 t + 2\sin t \cos t)]$

$\Rightarrow |A| = e^{-t} [6(\sin^2 t + \cos^2 t)]$

$\Rightarrow |A| = 6e^{-t} > 0$

Since $|A| \neq 0$ for any real $t$ , $A$ is invertible for $t \in \mathbb{R}$ .