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

detailed solution

Correct option is B

|A|=ete−tcoste−tsintet−etcost−e−tsint−e−tsint+e−tcostet2e−tsint−2e−tcost=ete−te−t1costsint1−cost−sint−sint+cost12sint−2cost(taking common from each column) Applying R2→R2−R1 and R3→R3−R1, we got ∵et−t=e0=1=e−t1costsint0−2cost−sint−2sint+cost02sint−cost−2cost−sint=e−t2cost+sint2+2sint−cost2( expanding along column 1)=e−t5cos2⁡t+5sin2⁡t=5e−t ∵cos2⁡t+sin2⁡t=1⇒|A|=5e−t≠0  for all t∈R∴A is invertible for all t∈R[∵If|A|≠0, then A is invertible ]