x′[23−1−2]x⟹x1′(t)=2x1(t)−x2(t)x2′=3x1(t)−2x2(t)this gives us two solutionsa) show that the given vector functions are indeed solutions[etet]?=[3−22−1][etet]⟹ this works because it equals b) show thatx(t)=c1x1+c2x2 is also a solution for any consatnt c1,c2[c1etc1et]+[c2et3c2et]=