Sec. 1.5 Solutions Sets of Linear Systems.doc

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1.5 Solution Sets of Linear Systems
A solution of a linear system is a vector or a
collection of vectors.
We begin by looking at a special type of
equation to see what solution sets are like.
Definition: A homogeneous system of linear
equations is a system of the form Ax = 0.
A is and m x n matrix, and 0 is the zero vector
in Rm.
Every homogeneous system has at least one
solution: x = 0.
We call this the trivial solution.
Example:
x1  3 x2  0
2 x1  6 x2  0
The corresponding matrix equation is:
1 3  x1  0
2 6  x   0

 2   
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The trivial solution is
 x1  0
x  
 x2  0
►Non-zero solutions are called non-trivial
solutions.
“Do non-trivial solutions for this system exist?”
Put the associated augmented matrix into
reduced row echelon form:
 1 3 0  1 3 0
 2 6 0 ~  0 0 0

 

Yes, there are infinitely many solutions.
Why?
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Fact: The homogeneous equation Ax = 0 has
a non-trivial solution if and only if the equation
has at least one free variable.
To see this:
Recall: Theorem 1.2 part 2:
If a linear system is consistent, then the
solution contains either
a) a unique solution (when there are no free
variables)
b) infinitely many solutions (when there is at
least one free variable).
 If the homogeneous equation Ax = 0 has a
non-trivial solution, it has more than one
solution, hence infinitely many solutions,
hence at least one free variable.
 If the equation has at least one free
variable,  infinitely many solutions, hence
more than just the trivial one.
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Example: Determine whether the
homogeneous system has non-trivial
solutions and describe the solution set.
2 x1  4 x2  6 x3  0
4 x1  8 x2  10 x3  0
“Is there a free variable?”
Write the matrix associated to Ax = 0.
2 4 - 6
4 6 - 10

0  1 2 0 0
~

0  0 0 1 0
x1  2 x2
x2 is free
x3  0
We can write this as a vector:
 x1   2 x2 
x   x2    x2 
 x3   0 
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Factor out x2 to get
  2
x  x2  1  or x  x2 v
 0 
Note 1: a non-trivial solution can have zero
entries.
Note 2: since x2 can be any number, the
solution set is Span{v}.
►Geometrically, this is a line through 0 and v.
Example: A non-homogeneous system:
2 x1  4 x2  6 x3  0
4 x1  8 x2  10 x3  4
The only difference is b in Ax = b.
2 4 - 6
4 6 - 10

0  1 2 0 6
~

4  0 0 1 2
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Solution:
x1  6  2 x2
x3  2
x2 is free
We can write this as a vector:
 x1  6  2 x2 
x   x2    x2 
 x3   2 
We can separate this vector into two vectors
and factor out x2 to get
6 
  2
x  0  x2  1  or x  p  x2 v
.
2
 0 
Note: This solution set is the line parallel to v
through p. It is a translation of x2 v by p.
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►You can think of solution sets of matrix
equations as lines (if you have one free
variable) or planes (if you have two).
►We call x  x2 v and x  p  x2 v
parametric vector equations. Sometimes we
write x  p  tv to emphasize that t can be any
real number.
Theorem 6: Suppose the equation Ax = b is
consistent for some given b, and let p be a
solution. Then the solution set is the set of all
vectors of the form
w = p + vh ,
where vh is any solution of the homogeneous
equation Ax = 0.
Example: Describe the solution set of
2 x1  4 x2  4 x3  0
and compare it to the solution set of
2 x1  4 x2  4 x3  6 .
“Are there free variables?”
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Answer: These are systems of one equation
each in three variables.
For the homogeneous system:
[2 –4 –4 0]~[1 –2 –2 0]
The vector form of the solution is
 x1  2 x2  2 x3 

x   x2    x2

 x3   x3

We can separate this vector into two vectors
 2
 2
x  x2 1   x3 0 or x  x2u  x3 v
0
1 
u
v
The non-homogenous situation is:
[2 –4 –4 6]~[1 –2 –2 3]
The vector form of the solution is
 x1  3  2 x2  2 x3 

x   x2   
x2

 x3  

x3
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Separate this vector into three vectors
 3
 2
 2
x  0  x2 1  x3 0
0
0
1
or x  p  x2u  x3 v
Geometrically, the solution set of the
homogeneous equation is a plane in R3
through 0, u, and v: Span{u,v}.
The solution of the non-homogeneous
equation is a plane parallel to Span{u,v} and
shifted three units in the x1 direction.
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