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

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