Let
A=[etetete−t(sint−2cost)e−t(2sint+cost)e−tcoste−t(−2sint−cost)e−t(sint−2cost)e−tsint]
Given matrix A is invertible if
∣A∣=0
⇒∣etetete−t(sint−2cost)e−t(2sint+cost)e−tcoste−t(−2sint−cost)e−t(sint−2cost)e−tsint∣=0
⇒et⋅e−t⋅e−t∣111sint−2cost2sint+costcost−2sint−costsint−2costsint∣=0
Applying R1→R1−R2 then R2→R2−R3, ee get
e−t∣001−sint−3cost2sintcost−3sint+cost−2costsint∣=0
By expanding we have,
e−t×(2sintcost+6cos2t+6sin2t−2sintcost)=0
⇒e−t×6=0 for ∀t∈R