The set of all values of $t \in \mathbb{R}$, 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, is

(A) $\left\{(2 \mathrm{k}+1) \frac{\pi}{2}, \mathrm{k} \in \mathbb{Z}\right\}$

(B) $\left\{\mathrm{k} \pi+\frac{\pi}{4}, \mathrm{k} \in \mathbb{Z}\right\}$

(C) $\{\mathrm{k} \pi, \mathrm{k} \in \mathbb{Z}\}$

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Matrices

The set of all values of

t \in R

, for which the matrix

e^{t} e^{t} e^{t} e^{- t} (sin t - 2 cos t) e^{- t} (2 sin t + cos t) e^{- t} cos t e^{- t} (- 2 sin t - cos t) e^{- t} (sin t - 2 cos t) e^{- t} sin t

is invertible, is

Answer: (D)

Take

e^{t}

out of the first column and

e^{- t}

out of each of the other two; the common factor is

e^{t} \cdot e^{- t} \cdot e^{- t} = e^{- t}

. In the reduced determinant the operations

R_{1} \to R_{1} - R_{3}

and

R_{2} \to R_{2} - R_{3}

clear the first column except in the last row, and expanding the remaining

2 \times 2

block gives

6 sin^{2} t + 6 cos^{2} t = 6

. Hence the determinant equals

6 e^{- t}

, which is nonzero for every real

t

. The matrix is invertible for all

t \in R

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option