Mathematics · 2019
JEE Main · 9 January 2019, Shift 2 · Q64
If A= [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], then A is:
If
$$A=\left[\begin{array}{ccc}
e^t & e^{-t} \cos t & e^{-t} \sin t \\
e^t & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^t & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right],
$$
then A is :
Official answer
(3)
invertible for all $\displaystyle \mathrm{t} \in \mathbf{R}$.
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.