Engr 231 Lab 3 solution

Engr 231 Lab 3 solution - -25.0000-55.0000 -110.0000...

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Lab 3 - In Class Assignment, Version A Table of Contents Problem Statement . ..................................................................................................... 1 Put the system in and reduce it . ...................................................................................... 1 Form the parametric Solution . ....................................................................................... 1 Plotting the solution . .................................................................................................... 2 Problem Statement Given a system of eq's, put it in an augmented matrix, reduce it, find the parametric solution, verify it, and plot it in 3D. Put the system in and reduce it Using what we learned from lab 2 A = [-26.5000 -53.0000 53.5000 -130.5000 -5.0000 -10.0000 10.0000
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Unformatted text preview: -25.0000-55.0000 -110.0000 111.0000 -271.0000] rref(A) A =-26.5000-53.0000 53.5000 -130.5000-5.0000-10.0000 10.0000-25.0000-55.0000 -110.0000 111.0000 -271.0000 ans = 1 2 13 1 4 Form the parametric Solution C is the coefficient matrix, p is the particular solution, and v is the homogeneous solution. Cp = b and Cv = 0. C = A(:,1:3) p = [13 0 4]' v = [-2 1 0]' 1 C =-26.5000-53.0000 53.5000-5.0000-10.0000 10.0000-55.0000 -110.0000 111.0000 p = 13 4 v =-2 1 Verifying the solutions Cv=C*v Cp=C*p - A(:,4) Cv = Cp = Plotting the solution For c =0 and c = 1: sol = [p v+p] plot3(sol(1,:),sol(2,:),sol(3,:)) grid on sol = 13 11 1 4 4 Lab 3 - In Class Assignment, Version A 2 Published with MATLAB 7.11 Lab 3 - In Class Assignment, Version A 3...
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This note was uploaded on 11/11/2011 for the course ENGR 231 taught by Professor Olehtretiak during the Fall '10 term at Drexel.

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Engr 231 Lab 3 solution - -25.0000-55.0000 -110.0000...

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