If A=etetcos⁡te−tsin⁡tet−etcos⁡t−e−tsin⁡t−e−tsin⁡t+e−tcos⁡tet2e−tsin⁡t−2e−tcos⁡t then A is

|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 ]