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Gauss' Method for Solving Equations


Date: 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/   
    
Associated Topics:
High School Linear Algebra
High School Logic

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