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hw05_soln - Back-filling b =-3(3 11 = 2 a = 12 – 2 –...

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SOLUTION EGR115 2008 Spring Due: End of class Wed, April 2 / Thu, April 3 HW05: Class Exercise a + b + 3c = 12 2a + 3b + 6c = 26 4a + b + 3c = 15 (1) Solve this set of linear equations using algebraic substitution (not Gaussian Elimination – see the next problem). Show your work here: Rearranging (1): (4) a = 12 – b – 3c Rearranging (3): (5) b = 15 – 4a – 3c Substitute (4) into (5): b = 15 – 4(12 – b – 3c) - 3c b = 15 – 48 + 4b + 12c - 3c b = -33 + 4b + 12c - 3c -3b = 9c - 33 (6) b = -3c + 11 Rearranging (2): (7) 6c = 26 – 2a - 3b Substitute (4) and (6) into (7): (8) 6c = 26 – 2(12-b-3c) - 3(-3c+11) 6c = 26 – 24 + 2b + 6c + 9c – 33 6c = 2 + 2b + 15c – 33 6c = 2b + 15c – 31 Substitute (6) again: 6c = 2(-3c+11) + 15c – 31 6c = -6c + 22 + 15c – 31 6c = 9c – 9 2c = 3c – 3 -c = -3 c = 3
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Unformatted text preview: Back-filling: b = -3(3) + 11 = 2 a = 12 – 2 – 3(3) = 1 Therefore: a=1, b=2, c=3 is the solution to this set of linear equations A = [1 1 3; 2 3 6; 4 1 3] A = 1 1 3 2 3 6 4 1 3 >> B = [12;26;15] B = 12 26 15 >> A\B ans = 1.0000 2.0000 3.0000 (2) Solve this set of linear equations using Gaussian Elimination (use Wikipedia. ..). Show your work here. a + b + 3c = 12 (1) 2a + 3b + 6c = 26 (2) 4a + b + 3c = 15 (3) Subtract 2* (1) from (2) , and 4* (1) from (3) : a + b + 3c = 12 (1) b = 2 (4)-3b - 9c = -33 (5) Add 3* (4) to (5) : a + b + 3c = 12 b = 2- 9c = -27 Thus: c = 3 b = 2 a = 12 - 2 + 3(3) = 1 (3) Clear your command window before you start. Use MATLAB's left divide operator to solve this set of linear equations. Attach a printout of your MATLAB work....
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