# By the same argument 4 4 y x will be the negative of

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By the same argument,44yx, will be the negative of22yxetc.To summarize, starting at00yx, thekkyxsequence repeats in blocks of 4, and in each such block the second pair of vectors is the negative ofthe first:20,410,20,410,20,410,20,410,20Another slightly more elegant way to make the above argument is to start with the observation thatIA==10012.It follows immediately from this that=++kkkkyxyx22for allk.And again thatgives us the wholekkyxsequence if we start with the first two terms.page6
MATH 111 Final Exam Apr 2010page7of 11 pages6.[8 marks]LetAbe the matrix:A=112020001.To save you some calculation, let me give you the firstfew powers ofA:=1000100010A,=1120200011A,=1340400012A,=1760800013A,=115801600014A(a)Based on these (and others if needed), conjecture a formula forAnand check it withA5.Solution.We conjecture the formula:An=1122020001nnnTo check this, calculate:===1311003200011158016000111202000145AAAAnd this fits the pattern forn= 5.(b) Use mathematical induction to prove that your formula forAnis correct.Solution.We let P(n) be the statementAn=1122020001nnn.P(0) is the assertion==100010001112)0(2020001000Aand this is the case sinceA0=I.Now assume that P(n) holds.To establish P(n+1) we calculate:An+1=AAn=1122020001112020001nnn(by the induction hypothesis)+=++=++112)1(2020001112222022000111nnnnnnnand this is the result we wanted.page7
MATH 111 Final Exam Apr 2010page8of 11 pages7.[12 marks]For the game diagramed at the right, the counter starts on 1 andat each move the counter moves according to the probabilities shown on thearrows.Once the counter gets to L or W the game is over.

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Term
Fall
Professor
TAYLOR
Tags
Probability theory, Eigenvalue eigenvector and eigenspace, Det
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