Page 45 - 35Linear Algebra

2.1 Gaussian Elimination 45

Advanced Elimination

It is important that you are able to convert RREF back into a system
of equations. The first thing you might notice is that if any of the numbers
r
b k+1 , . . . , b in 2.1.3 are non-zero then the system of equations is inconsistent
and has no solutions. Our next task is to extract all possible solutions from
an RREF augmented matrix.

2.1.4 Solution Sets and RREF

RREF is a maximally simplified version of the original system of equations
in the following sense:
• As many coefficients of the variables as possible are 0.

• As many coefficients of the variables as possible are 1.
It is easier to read off solutions from the maximally simplified equations than
from the original equations, even when there are infinitely many solutions.

Example 19 (Standard approach from a system of equations to the solution set)
    
x + y + 5w = 1  1 1 0 5 1 1 0 0 3 −5
y + 2w = 6 ⇔  0 1 0 2 6  ∼  0 1 0 2 6 
z + 4w = 8  0 0 1 4 8 0 0 1 4 8

 
 x
  = −5 − 3w 
x + 3w = −5  


   y = 6 − 2w 
⇔ y + 2w = 6 ⇔
 z =
  8 − 4w 
z + 4w = 8  


w = w
 
     
x −5 −3
     
 .
 y  6 −2
⇔   =   + w 
 z  8 −4
w 0 1
There is one solution for each value of w, so the solution set is

   
 −5 −3 
 
 
    
 6 −2
 : α ∈ R .
  + α 
 8 −4 
 
 
0 1
 
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