If $\mathrm{A}=\left[\begin{array}{ccc}e^{t} & e^{-t} \cos t & e^{-t} \sin t \\ e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\ e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t\end{array}\right]$ Then $A$ is

(C) invertible for all $t \in R$

(A) Invertible only if $t=\frac{\pi}{2}$

(B) not invertible for any $\mathrm{t} \in \mathrm{R}$

(D) invertible only if $\mathrm{t}=\pi$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Matrices

A = e^{t} e^{t} e^{t} e^{- t} cos t - e^{- t} cos t - e^{- t} sin t 2 e^{- t} sin t e^{- t} sin t - e^{- t} sin t + e^{- t} cos t - 2 e^{- t} cos t

Then

A

(A)Invertible only if $t = \frac{π}{2}$
(B)not invertible for any $t \in R$
(C)invertible for all $t \in R$ ✓ Correct
(D)invertible only if $t = π$

Answer: (C)

Solution

Compute the determinant of

A

. Factoring the shared exponential from each row and reducing with row operations, every surviving term combines into a single non-zero constant multiple of

e^{- t}

. Since

e^{- t} > 0

for every real

t

, the determinant never vanishes, so

A

is invertible for all

t \in R

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Matrices

Solution

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