# solutions1.1 - Math 307: Problems for section 1.1 1. Use...

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Math 307: Problems for section 1.11.Use Gaussian elimination to find the solution(s) toAx=b where(a)A=1234123456785678b=1111,(b)A=1111111111000011b=3101.The process of Gaussian elimination is not unique — there is more than one way to do it.We create the augmented matrixC= [A|b] and perform Gaussian elimination onC(a)12341123415678156781(b)11113111111100000111mapsto→12341040825678156781mapsto→11113002221100000111mapsto→1234104082567810120162mapsto→11113000001100000111mapsto→12341040820481240120162mapsto→11113110000000000111mapsto→1234104082008420120162mapsto→11113021130000000111mapsto→12341040820084200084mapsto→11113021130011100000The solution to the system will be the same regardless of how the Gaussian elimination is done:(a)x1x2x3x4=01/201/2(b)x1x2x3x4=1110+s00111
2.Use MATLAB/Octave to find the solution(s) toAx=b where(a)A=1111111111000011b=1111,(b)A=103242165011311,b=475.(a)> A=[1 1 1 1; 1 1 -1 -1 ; 1 -1 0 0; 0 0 1 -1];> b=[1 1 1 1]’;> C=[A b];> rref(C)ans =1 0 0 0 10 1 0 0 00 0 1 0 0.50 0 0 1 -0.5and so the answer isx1x2x3x4=100.50.5We could have instead done> A=[1 1 1 1; 1 1 -1 -1 ; 1 -1 0 0; 0 0 1 -1];> b=[1 1 1 1]’;> A\bans =100.5-0.5and found the same answer — be careful though (see next part).(b)> A=[1 0 3 2 -4; 2 1 6 5 0; -1 1 -3 -1 1];> b = [4 7 -5]’;> C=[A b];> rref(C)ans =1 0 3 2 0 40 1 0 1 0 -10 0 0 0 1 0The solution isx1x2x3x4x5=41000+s21010+t301002
If we had instead used> A\bthen MATLAB/Octave would have given only one of the many solutions.3.Compute the time it takes to solveAx=b usingA\bon your computer for five or morerandom problems of increasing size.Make a plot of time vs.size.Repeat using themethodA^(-1)*band plot on the same graph. (If your calculation is taking too long, itcan be interrupted by typing<ctrl> c.)

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