Gauss' Method for Solving EquationsDate: 12/11/98 at 04:48:07 From: Tania Jacob Subject: Gauss' method for solving equations How and why does Gauss' method work? (The method where you get all the elements of the lower left corner of a matrix to equal zero...) Tania Date: 12/11/98 at 08:30:37 From: Doctor Jerry Subject: Re: Gauss' method for solving equations Hi Tania, Gauss' method is based on three properties of the systems of equations to which it is applied. Each of these properties has a corresponding property of the associated matrices. By the solution set of a system I mean the set of all possible solutions. The solution set might be empty. P1. The solution set of a system doesn't change if two equations are interchanged. P2. The solution set of a system doesn't change if an equation is multiplied by a nonzero constant. P3. The solution set of a system doesn't change if an equation is replaced by the sum of that equation and the product of another equation by a constant. In matrix terms (I'm now talking about the augmented matrix), these correspond to the three row operations. If these row operations are done, the solution set doesn't change. R1. Any two rows may be interchanged. R2. Any row may be multiplied by a nonzero constant. R3. To any row may be added a constant times another row. Gauss' method is nothing more than a systematic use of these row operations to transform a given augmented matrix to one that is triangular. Suppose we have the system: x + 2y = 1 -x + y + 3z = 2 2x + z = 3 Let A be: 1 2 0 1 -1 1 3 2 2 0 1 3 First use row 1 to make the elements in the first column 0, except for the first. Add to 2nd row 1 times the first. Then add to third row -2 times the first: 1 2 0 1 0 3 3 3 0 -4 1 1 Next, multiply 2nd row by 1/3, to make the (2,2) element 1: 1 2 0 1 0 1 1 1 0 -4 1 1 Now multiply 2nd row by 4 and add to 3rd row. This gives: 1 2 0 1 0 1 1 1 0 0 5 5 The system can be easily solved at this point. 5z = 5 implies z = 1. From second equation: y = 1 - z = 1 - 1 = 0 From first equation: x = 1 - 2y = 1 - 2(0) = 1 - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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