Math 415 - Final ExamFriday, December 12, 2014Circle your section:Philipp Hieronymi 2pm 3pmArmin Straub 9am 11amName:NetID:UIN:To be completed by the grader:0 1 2 3 4 5 MCP/1 /? /? /? /? /? /? /??Good luck!Instructions• No notes, personal aids or calculators are permitted.• This exam consists of ? pages. Take a moment to make sure you have all pages.• You have 180 minutes.• Answer all questions in the space provided. If you require more space to write youranswer, you may continue on the back of the page (make it clear if you do).• Explain your work! Little or no points will be given for a correct answer with noexplanation of how you got it.• In particular, you have to write down all row operations for full credit.1Important Note• The collection of problems below is not representative of the final exam!• The first three problems cover the material since the third midterm exam, andproblems on the final exam on these topics will be of similar nature.• Problems 4 and 5 are a good start to review the material we covered earlier; however,on the exam itself you should expect questions of the kind that we had on theprevious midterms.• In other words, to prepare for the final, you need to also prepare our pastmidterm exams and practice exams.• In particular, a basic understanding of Fourier series or the ability to work withspaces of polynomials are expected.Problem 1. Find a solution to the initial value problem (that is, differential equation plusinitial condition)ddtu =1 1 01 0 10 1 1u, u(0) =210.Simplify your solution as far as possible.Solution. The solution is u(t) = eAtu(0), where A =1 1 01 0 10 1 1. In order to compute eAt, wehave to find eigenvalues and corresponding eigenvectors of A. We have:det1 − λ 1 01 −λ 10 1 1 − λ= (1−λ)(λ(λ−1)−1)−(1−λ) = (1−λ)(λ2−λ−2) = (1−λ)(−1−λ)(2−λ)Hence, the eigenvalues of A are 2, 1, and −1. For λ = 2:−1 1 01 −2 10 1 −1R2→R2+R1,R3→R3+R2,R1→R1+R2−−−−−−−−−−−−−−−−−−−−−−→−1 0 10 −1 10 0 0Hence, the corresponding eigenspace is span111.For λ = 1:0 1 01 −1 10 1 0R3→R3−R1,R2→R2+R1−−−−−−−−−−−−−−→0 1 01 0 10 0 0Hence, the corresponding eigenspace is span10−1.For λ = −1:2 1 01 1 10 1 2R2→R2−1/2R1,R2→R2−1/2R3,R1→R1−R3,R1→1/2R1−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→1 0 −10 0 00 1 22Hence, the corresponding eigenspace is span1−21.Hence, A = P DP−1where P =1 1 11 0 −21 −1 1and D =2 0 00 1 00 0 −1. (columns of P arelinearly independent eigenvectors of A, and entries on the main diagonal of D are the corre-sponding eigenvalues)Note that the columns of P are pairwise orthogonal so we can get P−1by dividing rows of PTby the square of the length of each row, i.e.:P−1=131313120 −1216−1316(Don’t worry if you did not see this, and used Gauss–Jordan instead.) Since A = P DP−1, wehave eAt= P eDtP−1. Hence,u(t) = eAtu(0) = P eDtP−1u(0) =1 1 11 0 −21 −1 1e2t0 00 et00 0 e−t131313120 −1216−1316210=1 1 11 0 −21 −1 1e2t0 00 et00 0 e−t110=1 1 11 0 −21 −1 1e2tet0=e2t+ ete2te2t− etProblem 2. The processors of a supercomputer are inspected weekly in order to determinetheir condition. The condition of a processor can either be perfect, good, reasonable or bad.A perfect processor is still perfect after one week with probability 0.7, with probability 0.2the state is good, and with probability 0.1 it is reasonable. A processor in good conditionsis still good after one week with probability 0.6, reasonable with probability 0.2, and badwith probability 0.2. A processor in reasonable condition is still reasonable after one weekwith probability 0.5 and bad with probability 0.5. A bad processor must be repaired. Thereparation takes one week, after which the processor is again in perfect condition.In the steady state, what is percentage of processors in perfect condition?Solution. We consider four states: perfect, good, reasonable, badThe transition matrix is:0.7 0 0 10.2 0.6 0 00.1 0.2 0.5 00 0.2 0.5 0The steady state is the eigenvector corresponding to the eigenvalue 1 (with the extra conditionthat summation of the entries of the vector should be 1; since the states are percentages). We3have:0.7 − 1 0 0 10.2 0.6 − 1 0 00.1 0.2 0.5 − 1 00 0.2 0.5 0 − 1=−0.3 0 0 10.2 −0.4 0 00.1 0.2 −0.5 00 0.2 0.5 −1R2→R2+2/3R1,R3→R3+1/3R1−−−−−−−−−−−−−−−−−−→−0.3 0 0 10 −0.4 0230 0.2 −0.5130 0.2 0.5 −1R3→R3+1/2R2,R4→R4+1/2R2,R4→R4+R3−−−−−−−−−−−−−−−−−−−−−−−−−→−0.3 0 0 10 −0.4 0230 0 −0.5230 0 0 0Hence, the eigenspace corresponding to eigenvalue 1 is span10353431. Therefore, the steadystate is1022522422322. In particular, in the steady state (almost) 45% of processors are in perfectcondition.Problem 3. Determine the PageRank vector for the following system of webpages, and rankthe webpages accordingly.AB CDEFSolution. The transition matrix is:0 0 012120120 0120 00120 0 0 0012130 0 0120130 0 10 01301204The steady state is the eigenvector corresponding to the eigenvalue 1 (with the extra conditionthat summation of the entries of the vector should be 1). We have:−1 0 01212012−1 0120 0012−1 0 0 001213−1 0 0120130 −1 10 013012−1R2→R2+1/2R1,R5→R5+1/2R1−−−−−−−−−−−−−−−−−−→−1 0 0121200 −1 034140012−1 0 0 001213−1 0 00 01314−3410 013012−1R3→R3+1/2R2,R4→R4+1/2R2−−−−−−−−−−−−−−−−−−→−1 0 0121200 −1 0341400 0 −1381800 013−581800 01314−3410
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