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if $A = [\begin{matrix} e^{t} & e^{- t} \cos t & e^{- t} \sin t \\ e^{t} & - e^{- t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & 2 e^{- t} \sin t & - 2 e^{- t} \cos t \end{matrix}]$ Then A is

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Invertible only when t=π

invertible for every t∈R

Not invertible for any t∈R

invertible only when t=π2

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detailed solution

Correct option is B

We have|A|=ete-te-t1costsint1-cost-sint-sint+cost12sint-2cost apply R2→R2-R1,R3→R3-R1⇒|A|=e-t1costsint0-(2cost+sint)cost-2sint02sint-cost-(2cost+sint) =e-t(2cost+sint)2+(cost-2sint)2=e-t·5≠0∴A≠0 ∀t∈R ⇒A is invertible ∀t∈R

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if A=ete−tcoste−tsintet−e−tcost−e−tsint−e−tsint+e−tcostet2e−tsint−2e−tcost Then A is

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if $A = [\begin{matrix} e^{t} & e^{- t} \cos t & e^{- t} \sin t \\ e^{t} & - e^{- t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & 2 e^{- t} \sin t & - 2 e^{- t} \cos t \end{matrix}]$ Then A is