61IfAB=B A, thencw+dy=ay+cz; that is,c(w−z)=y(a−d).There are two cases to consider. Ifa=d, then defineB=0111.The 2,1 entry ofABisd=a, while the 2,1 entry ofB Aisa+c. Asc=0,we haveAB=B A. Ifa=d, defineB=1110.Now the 2,1 entry ofABisc+d, while the 2,1 entry ofB Aisc+a. Sincea=d, these entries are different, and soAB=B Ain this case as well.Therefore,Ais not in the center of GL(2,R). A similar argument holds ifb=0, (This result is generalized ton×nmatrices in Corollary 4.86. Theproof using linear transformations is much shorter and much simpler.)2.85Letζ=e2πi/nbe a primitiventh root of unity, and defineA=ζ00ζ−1andB=0 11 0.(i)Prove thatAhas ordernand thatBhas order 2.Solution.It is clear thatAn=I=B2. If 1≤k<n, thenAk=I, for the 1,1 entry ofAkisζk=1.(ii)Prove thatB AB=A−1.Solution.One multiplies the matrices.(iii)Prove that the matrices of the form
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