then A is:
[JEE Main Online Paper (Held On 09-Jan-2019 Evening]
A) invertible only if \[t=n\]
B) invertible for all \[t\,\,\in \,\,R\]
C) invertible only if \[t=\frac{\pi }{2}\]
D) not invertible for any \[t\text{ }\in \text{ }R\]
Correct Answer: B
Solution :
A=\[\left[ \begin{align} & {{e}^{t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{e}^{-t}}\cos \,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{e}^{-t}}\,\sin \,t\,\,\,\,\,\,\, \\ & {{e}^{t}}\,\,\,\,-{{e}^{-t}}\cos \,t\,\,-{{e}^{-t}}\,\sin \,t\,\,\,\,\,\,-{{e}^{-t}}\sin \,t\,\,+{{e}^{-t}}\,\cos \,t \\ & {{e}^{t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{e}^{-t}}\,\sin \,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,2{{e}^{-t}}\cos \,t\,\, \\ \end{align} \right]\,\] \[\left| A \right|={{e}^{t}}.{{e}^{-t}}.{{e}^{-t}}\,\left| \begin{align} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \,t \\ & 1\,\,\,\,\,\,\,-\cos \,t-\sin \,t\,\,\,\,\,\,-\sin \,t+\cos \,t \\ & 1\,\,\,\,\,\,\,\,\,\,\,\,2\,\sin \,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2\cos \,\,t\,\, \\ \end{align} \right|\] Apply operations \[{{R}_{2}}<{{R}_{2}}-{{R}_{1}},\text{ }{{R}_{3}}<{{R}_{3}}-{{R}_{1}},\text{ }{{R}_{1}}<{{R}_{1}}\] \[\left| A \right|\,\,=\,\,{{e}^{-t}}\,\left| \begin{align} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \,t \\ & 0\,\,\,\,\,\,\,-\sin \,t\,-\,\,2\,\cos \,t\,\,\,\,\,\,\,\,\,-2\sin \,t\,+\,\,\cos \,t \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,2\,\sin \,t-\cos \,t\,\,\,\,\,\,\,-2\cos \,\,t-\sin \,t\,\, \\ \end{align} \right|\]Open the determinant by \[{{R}_{1}}\] \[\left| A \right|=5{{e}^{-t}}\] Invertible for all \[t\text{ }\in \text{ }R\]
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