(ii).If X₁ (t) = e¹tU₁₂,X₂(t) = e^t (U₂ + tU)... X₁ (t) = e¹t (U₁ + tU₁ k-1+...+u2tk-1/ (k-1)!)
Are solutions of X' = AX, then X1....Xk are linearly independent,i.e.
C₁X₂ + C₂X₂ + + CX = 0 for some arbitrary constants C, s. [4 marks]