The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

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...)


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 

P1. The solution set of a system doesn't change if two equations are 

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 

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   
Associated Topics:
High School Linear Algebra
High School Logic

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.