Linear Algebra Exercises-n-Answers.pdf

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156 Linear Algebra, by Hefferon n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.16638 0 0 0 0.40837 0 0 0 0.33412 0 0 0 0.09112 0 0 0 0 0.09151 0.09151 0.09151 0 0.29948 0 0 0 0.36754 0 0 0 0.20047 0 0 0 0.04101 0.04101 0.04101 0 0 0.16471 0.16471 0 0 0.33691 0 0 0 0.27565 0 0 0 0.09021 0.09021 0 0 0 0.18530 0 0 0 0.30322 0 0 0 0.12404 0 0 0 0 0 0 0 0 0-0 1-0 0-1 2-0 1-1 0-2 3-0 2-1 1-2 0-3 4-0 3-1 2-2 1-3 0-4 4-1 3-2 2-3 1-4 4-2 3-3 2-4 4-3 3-4 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.55000 0.45000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.30250 0.49500 0.20250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 and these are the p = .60 vectors. 0-0 1-0 0-1 2-0 1-1 0-2 3-0 2-1 1-2 0-3 4-0 3-1 2-2 1-3 0-4 4-1 3-2 2-3 1-4 4-2 3-3 2-4 4-3 3-4 0 0 0 0 0 0 0 0 0 0 0.09151 0 0 0 0.04101 0.16471 0 0 0.09021 0.18530 0 0.12404 0.16677 0.13645 n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 7 1 0 0 0 0 0 0 0 0 0.60000 0 0 0 0 0 0 0 0.40000 0 0 0 0 0 0 0 0 0.36000 0 0 0 0 0 0 0 0.48000 0 0 0 0 0 0 0 0.16000 0 0 0 0 0 0 0 0 0.21600 0 0 0 0 0 0 0 0.43200 0 0 0 0 0 0 0 0.28800 0 0 0 0 0 0 0 0.06400 0 0 0 0 0 0 0 0 0.12960 0.12960 0.12960 0.12960 0 0 0 0 0.34560 0 0 0 0 0 0 0 0.34560 0 0 0 0 0 0 0 0.15360 0 0 0 0 0 0 0 0.02560 0.02560 0.02560 0.02560 0 0 0 0 0 0.20736 0.20736 0.20736 0 0 0 0 0 0.34560 0 0 0 0 0 0 0 0.23040 0 0 0 0 0 0 0 0.06144 0.06144 0.06144 0 0 0 0 0 0 0.20736 0.20736 0 0 0 0 0 0 0.27648 0 0 0 0 0 0 0 0.09216 0.09216 0 0 0 0 0 0 0 0.16589 0 0 0 0 0 0 0 0.11059 (a) The script from the computer code section can be easily adapted. # Octave script file to compute chance of World Series outcomes. function w = markov(p,v) q = 1-p; A=[0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-0 p,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-0 q,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-1_ 0,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-0 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-1 0,0,q,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-2__ 0,0,0,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-0
Answers to Exercises 157 0,0,0,q,p,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-1 0,0,0,0,q,p, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-2_ 0,0,0,0,0,q, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-3 0,0,0,0,0,0, p,0,0,0,1,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 4-0 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-1__ 0,0,0,0,0,0, 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-2 0,0,0,0,0,0, 0,0,q,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-3 0,0,0,0,0,0, 0,0,0,q,0,0, 0,0,1,0,0,0, 0,0,0,0,0,0; # 0-4_ 0,0,0,0,0,0, 0,0,0,0,0,p, 0,0,0,1,0,0, 0,0,0,0,0,0; # 4-1 0,0,0,0,0,0, 0,0,0,0,0,q, p,0,0,0,0,0, 0,0,0,0,0,0; # 3-2 0,0,0,0,0,0, 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0; # 2-3__ 0,0,0,0,0,0, 0,0,0,0,0,0, 0,q,0,0,0,0, 1,0,0,0,0,0; # 1-4 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,p,0, 0,1,0,0,0,0; # 4-2 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,q,p, 0,0,0,0,0,0; # 3-3_ 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,q, 0,0,0,1,0,0; # 2-4 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,p,0,1,0; # 4-3 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,q,0,0,1]; # 3-4 v7 = (A**7) * v; w = v7(11)+v7(16)+v7(20)+v7(23) endfunction Using this script, we get that when the American League has a p = 0.55 probability of winning each game then their probability of winning the first-to-win-four series is 0.60829. When their probability of winning any one game is p = 0.6 then their probability of winning the series is 0.71021. (b) This Octave session octave:1> v0=[1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0]; octave:2> x=(.01:.01:.99)’; octave:3> y=(.01:.01:.99)’; octave:4> for i=.01:.01:.99 > y(100*i)=markov(i,v0); > endfor octave:5> z=[x, y]; octave:6> gplot z yields this graph. By eye we judge that if p > 0.7 then the team is close to assurred of the series. 1 line 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 7 (a) They must satisfy this condition because the total probability of a state transition (including back to the same state) is 100%. (b) See the answer to the third item. (c) We will do the 2×2 case; bigger-sized cases are just notational problems. This product has these two column sums ( ) ( ) a1,1 a 1,2 b1,1 b 1,2 = a 2,1 a 2,2 b 2,1 b 2,2 ( ) a1,1 b 1,1 + a 1,2 b 2,1 a 1,1 b 1,2 + a 1,2 b 2,2 a 2,1 b 1,1 + a 2,2 b 2,1 a 2,1 b 1,2 + a 2,2 b 2,2 (a 1,1 b 1,1 + a 1,2 b 2,1 ) + (a 2,1 b 1,1 + a 2,2 b 2,1 ) = (a 1,1 + a 2,1 ) · b 1,1 + (a 1,2 + a 2,2 ) · b 2,1 = 1 · b 1,1 + 1 · b 2,1 = 1 and (a 1,1 b 1,2 + a 1,2 b 2,2 ) + (a 2,1 b 1,2 + a 2,2 b 2,2 ) = (a 1,1 + a 2,1 ) · b 1,2 + (a 1,2 + a 2,2 ) · b 2,2 = 1 · b 1,2 + 1 · b 2,2 = 1 as required.
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