Gaussian EliminationDate: 01/11/2000 at 06:47:18 From: Michael Harper Subject: Gaussian Elimination I am attempting to do Gaussian elimination and I am completely stuck. Date: 01/11/2000 at 13:55:03 From: Doctor Rob Subject: Re: Gaussian Elimination Thanks for writing to Ask Dr. Math, Michael. You are given a system of n or more simultaneous linear equations involving n unknowns. Pick one of the unknowns, called the pivot variable. Find an equation in which it appears, called the pivot equation. Use that equation to eliminate the pivot variable from all other equations in the following way. If an equation doesn't contain the pivot variable, leave it alone. If an equation does contain the pivot variable, subtract the right multiple of the pivot equation from that equation to make the pivot variable's coefficient zero. When you have eliminated the pivot variable from all the rest of the equations, set aside the pivot equation, and you will have a system of one fewer simultaneous linear equations involving n-1 unknowns. Repeat this process, driving down the number of unknowns appearing until it reaches 0. Any remaining equations must look like c = c for some constant c, and can be discarded (if not, the original system had no solution). Now look at the equations that you set aside and used as pivot equations. The last equation will allow you to solve for the pivot variable. Substitute that into all the remaining equations, set it aside, and repeat. When you are done with this process, you'll have each pivot variable solved for, and you are done. Example: -4*w + 3*x - 4*y - z = -37 -2*w - 5*y + 3*z = -20 -w - x - 3*y - 4*z = -27 -3*w + 2*x + 4*y - z = 7 Let w be the first pivot variable, and the first equation the first pivot equation. w appears in the second equation, so we subtract (-2)/(-4) = 1/2 times the first equation from the second, getting -2*w - 5*y + 3*z = -20, -2*w + (3/2)*x - 2*y - (1/2)*z = -37/2 -------------------------------------- -(3/2)*x - 3*y + (7/2)*z = -3/2 w appears in the third equation so we subtract (-1)/(-4) times the first equation from the third. w appears in the fourth equation, so we subtract (-3)/(-4) times the first equation from the fourth. This leaves -4*w + 3*x - 4*y - z = -37 -(3/2)*x - 3*y + (7/2)*z = -3/2 -(7/4)*x - 2*y - (15/4)*z = -71/4 -(1/4)*x + 7*y - (1/4)*z = 139/4 Now the last three equations are a system of three equations in three unknowns, and we treat that similarly. Let x be the next pivot variable, and the second equation the pivot equation. Since x appears in the third equation, we subtract (-7/4)/(-3/2) = 7/6 times the second equation from the third. Since x appears in the fourth equation we subtract (-1/4)/(-3/2) = 1/6 times the second equation from the fourth. This leaves -4*w + 3*x - 4*y - z = -37 -(3/2)*x - 3*y + (7/2)*z = -3/2 (3/2)*y - (47/6)*z = -16 (15/2)*y - 16*z = 35 Now the last two equations are a system of two equations in two unknowns, and we treat that similarly. Let y be the next pivot variable, and the third equation the next pivot equation. Since y appears in the fourth equation, we subtract (15/2)/(3/2) = 5 times the third equation from the fourth. This leaves -4*w + 3*x - 4*y - z = -37 -(3/2)*x - 3*y + (7/2)*z = -3/2 (3/2)*y - (47/6)*z = -16 (115/3)*z = 115 This completes the first phase. Now we start with the last equation, and solve for z. This gives z = 3, and we substitute that into the preceding equations: -4*w + 3*x - 4*y = -34 -(3/2)*x - 3*y = -12 (3/2)*y = 15/2 The last remaining equation tells us that y = 5, and we substitute that into the preceding equations: -4*w + 3*x = -14 -(3/2)*x = 3 The last remaining equation tells us that x = -2, and we substitute that into the preceding equation: -4*w = -8 That gives us the value w = 2, and so the solution is (w,x,y,z) = (2,-2,5,3) Understood? - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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