1) If A= [122212221] then A−1= A) 4l-A B) A-4l C) 15(A−4l) D) 15(4l−A) Answer: Option CExplanation:We have , A= [122212221] |A|= 1(1-4)-2(2-4)+2(4-2)=-3+4+4=5 Adj A= [−3222−3222−3] A−1=1|A|AdjA=15[−3222−3222−3] A−1=15[122212221]-15[400040004] A−1=15[A−4l]