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Griven, vectors{:[v_(1)=[[-1],[2],[2]]],[v_(2)=[[-2],[-3],[-3]]],[v_(3)=[[-3],[2],[1]]]:}Let, A=[[-(1)/(2),-2,-3],[2,-3,2],[2,-3,1]]-Now, we have to calculate rowreduced form for finding basis elements.{:[^(R_(3)^('))=R_(3)+2R_(1)^(R_(2)^(')=R_(2)+2R_(1))[[-1,-2,-3],[0,-7,-4],[0,-7,-5]]],[longrightarrow^(R_(2)^(')=-(1)/(7)R_(2))[[-1,-2,-3],[0,1,(4)/(7)],[0,-7,-5]]],[longrightarrow^(R_(3)^(')=R_(3)+7R_(2))[[-1,-2,-3],[0,1,4],[0,0,-1]]],[" Therefore, basis set "={[-1],[-1],[0]]","[[-2],[1],[0]]","[[-3],[(4)/(1)],[-1]]]:}therefore,dimension of domain space =3Also given, omega_(1)=[[3],[0],[5],[-5],[-1]]quadomega_(2)=[[5],[-3],[-2],[1],[2]]quadomega_(3)=[[-2],[4],[-3],[0],[-5]](ANS)w_(1)=[[3],[0],[5],[-5],[-1]]quadomega_(2)=[[-5],[-3],[-2],[1],[2]]quadw_(3)=[[-2],[4],[-3],[0],[-5]][1(5)/(3)-(2)/(3):}{:[-(31)/(3),(1)/(3)],[(28)/(3),-(10)/(3)],[(11)/(3),-(17)/(3)]:}[[1,(5)/(3),-(2)/(3)],[0,1,-(4)/(3)],[0,0,-(121)/(9)],[ ... See the full answer