Preparation problems for the discussion sections on December 4th and 9th1. Suppose there is an epidemic in which, every month, half of those who are well become sick,and a quarter of those who are sick become dead. Set up a 3 × 3 transition matrix and find thelong term equilibrium (steady state).Solution: The transition graph is as follows:Well Sick Dead0.50.50.750.251Hence, the transition matrix isA =.5 0 0.5 .75 00 .25 1.Letut=ut,Wut,Sut,D ,where ut,Wis the percentage of people that are well, ut,Sis the percentage of people that aresick and ut,Dthe percentage of people that have died, at time t. Note thatut+1= Aut.The long term equlibrium is a state u∞such thatAu∞= u∞.For that we need to find an eigenvector of A to the eigenvalue 1:−.5 0 0 0.5 −.25 0 00 .25 0 0→−.5 0 0 00 −.25 0 00 .25 0 0→−.5 0 0 00 −.25 0 00 0 0 0→1 0 0 00 1 0 00 0 0 0Hence, the eigenspace of λ = 1 is spanned by001. Since this vector is already scaled so thatits entries add up to 1, we concludeu∞=001.This means all people will die from the epidemic.12. Find the PageRank vector for the following system of webpages, and rank the six webpagesaccordingly:AB CDEFSolution: The PageRank matrix (that is, the transition matrix for the random surfer) isT =0 0 0 1 1 0120 0 0 0 00120 0 0 0012130 0 0120130 0 10 0130 0 0.The PageRank vector is an eigenvector of T to the eigenvalue 1. We row reduce T − I as follows(do it!):−1 0 0 1 1 0 0.5 −1 0 0 0 0 00 .5 −1 0 0 0 00 .513−1 0 0 0.5 0130 −1 1 00 0130 0 −1 0→1 0 0 0 0 −12 00 1 0 0 0 −6 00 0 1 0 0 −3 00 0 0 1 0 −4 00 0 0 0 1 −8 00 0 0 0 0 0 0.Hence an eigenvector of T to the eigenvalue 1 is given by1263481,and the corresponding normalized PageRank vector is1341263481.For the purpose of ranking this normalizing is immaterial (and it is OK if you don’t normalizePageRank vectors on the exam). The ranking is A, E, B, D, C, F.23. Solve the differential equationdudt=1 −1−1 1uwith initial condition u(0) =31.Solution: Set A =1 −1−1 1. We first diagonalize A. We start by calculating the eigenvaluesof A:det(A − λI) =(1 − λ) −1−1 (1 − λ)= (1 − λ)2− 1 = 1 − 2λ + λ2+ 1 = λ(λ − 2).Hence, the eigenvalues of A are 0, 2. We now calculate the corresponding eigenvectors. Forλ = 0,1 −1 0−1 1 0→1 −1 00 0 0.Hence,11is an eigenvector of A corresponding to the eigenvalue 0. For λ = 2,−1 −1 0−1 −1 0→1 1 00 0 0.Hence,−11is an eigenvector of A to the eigenvalue 2. SetD =0 00 2, P =1 −11 1.Then A = P DP−1. Note thatP−1=1212−1212.The solution to the above differential equation with initial condition isu(t) = eAtu(0) = P eDtP−1u(0)=1 −11 1e0t00 e2t1212−121231=1 −11 11 00 e2t2−1=1 −11 12−e2t=2 + e2t2 −
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