1.5 Solutions Sets of Linear Systems
Math 2331 – Linear Algebra
1.5 Solution Sets of Linear Systems
Alex Bearden
Department of Mathematics, University of Houston
[email protected]
math.uh.edu/∼cabearde/math2331
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
1 / 18
1.5 Solutions Sets of Linear Systems
Goals of this Section
The main goals of this section are to:
1
show how to write the solutions of linear systems in an explicit
form called “parametric vector form”;
2
give geometric interpretations for solutions to linear systems.
Along the way, we will use the notions of “homogeneous” and
“nonhomogeneous” linear systems to help describe what we learn.
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
2 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Linear Systems
Definition
A system of linear equations is said to be homogeneous if it can
be written in the form Ax = 0, where A is an m × n matrix and 0
is the zero vector in Rm .
A system is called nonhomogeneous if it is not homogeneous.
A homogeneous system Ax = 0 always has at least one
solution—namely x = 0. (Why?) This zero solution is sometimes
called the trivial solution. A nonzero solution is called a
nontrivial solution.
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
3 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Systems: Example 1
Example (1)
Consider the following homogeneous linear system:
x1 + 10x2 = 0
2x1 + 20x2 = 0
The corresponding matrix equation Ax = 0 is:
1 10
x1
0
=
x2
2 20
0
Check that the trivial solution x1 = 0, x2 = 0 is indeed a solution.
Can you find any nontrivial solutions?
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
4 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Linear Systems (cont.)
Since a consistent linear system either has exactly one solution or
infinitely many solutions, a homogeneous system either only has
the trivial solution or has the trivial solution plus infinitely many
nontrivial solutions.
A linear system has infinitely many solutions if and only if it has at
least one free variable. Thus we have the following useful fact:
Fact
A homogeneous linear system has a nontrivial solution if and only
if the system has at least one free variable.
(Recall that a free variable is a variable corresponding to a column
with no pivot position in the augmented matrix.)
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
5 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Systems: Example 1, Part 2
Example (1)
Use row reduction to figure out if the homogeneous linear system
x1 + 10x2 = 0
2x1 + 20x2 = 0
has any nontrivial solutions.
Solution:
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
6 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Systems: Example 2
Example (2)
Determine if the following homogeneous system has nontrivial
solutions and then describe the solution set.
3x1 + 5x2 − 4x3 = 0
−3x1 − 2x2 + 4x3 = 0
6x1 − x2 − 8x3 = 0
Solution: Let A be the coefficient matrix of the
system.
To solve
the problem, first reduce the augmented matrix A 0 to an
echelon form to figure out if the system has any free variables:
To describe the solution set, continue the row reduction of A 0
down to reduced echelon form and write down the corresponding
linear system:
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
7 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Systems: Example 2 (cont.)
Example (2 cont.)
Solve for the basic variables
to obtain the solution

x1
x2

x3
As a vector, the general solution of Ax = 0 we found in the
previous example can be rewritten as
 
   
 
x1
x = x2  =   = x3   = x3 v, where v =  
x3
In other words, the solutions of Ax = 0 are exactly the scalar
multiples of v. (The trivial solution is obtained by letting x3 =
What does this solution set look like geometrically?
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
.)
8 / 18
1.5 Solutions Sets of Linear Systems
Parametric Vector Form
In the previous example, we figured out that the solution of
Ax = 0 was all vectors of the form
x = cv
(c in R),
where v was a certain fixed vector. When written this way, we say
that the solution is in parametric vector form, meaning that it is
written as a linear combination of vectors where some of the
weights (also called the “parameters” here) range over all scalars.
In the example above, only one vector v showed up in the
parametric vector form of the solution, but sometimes a solution
will have more than one vector. In the next example, the solution
in parametric vector form will have two vectors, so it will look like:
x = su + tv
Alex Bearden, University of Houston
(s, t in R)
Math 2331, Linear Algebra
Spring 2017
9 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Systems: Example 3
Example (3)
A single linear equation can be treated as a very simple linear
system. Describe all solution of the homogeneous “system”
10x1 − 3x2 − 2x3 = 0
in parametric vector form.
Solution:
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
10 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous Systems: Example 4
Example (4)
Determine if the following homogeneous system has nontrivial
solutions and then describe the solution set in parametric vector
form.
2x1 + 4x2 − 6x3 = 0
4x1 + 8x2 − 10x3 = 0
Solution:
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
11 / 18
1.5 Solutions Sets of Linear Systems
Homogeneous System: Example 4 (cont.)
The general solution to the homogeneous equation Ax = 0 in the
previous example was found to be
 
−2
x = x2  1  = x2 v (x2 in R).
0
This can be graphically represented as a line through 0 in R3 :
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
12 / 18
1.5 Solutions Sets of Linear Systems
Nonhomogeneous System: Example 5
When a nonhomogeneous linear system has more than one
solution, the general solution can be written in parametric vector
form as one vector plus a linear combination of vectors with
weights ranging over the scalars.
Example (5)
Describe the solution set of the following in parametric vector form.
2x1 + 4x2 − 6x3 = 0
4x1 + 8x2 − 10x3 = 4
Solution: Row reduce the augmented matrix to reduced echelon
form:
Convert back to equations, then write the general solution in
parametric vector form:
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
13 / 18
1.5 Solutions Sets of Linear Systems
Nonhomogeneous System: Example 5 (cont.)
The general solution to the nonhomogeneous equation Ax = b in
the previous example was found to be
 


6
−2
x =  0  + x2  1  = p+x2 v
2
0
This can be graphically represented as a line in R3 parallel to the
line forming the solution set of the homogeneous equation Ax = 0.
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
14 / 18
1.5 Solutions Sets of Linear Systems
Recap of Previous Two Examples
Here is a recap of the solutions to the previous two examples:
Example (4. Solution of Ax = 0)


−2
The solution in parametric vector form was x = x2  1  = x2 v,
0
which was represented as a line passing through 0 and v.
Example (5. Solution of Ax = b)
The 
solution
 in parametric

 vector form was
6
−2
x =  0  + x2  1  = p+x2 v, which was represented as a line
2
0
passing through p parallel to v.
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
15 / 18
1.5 Solutions Sets of Linear Systems
Nonhomogeneous System: Theorem
Parallel solution sets of
Ax = b and Ax = 0
Theorem
Suppose the equation Ax = b is consistent for some given b, and
let p be a solution. Then the solution set of Ax = b is the set of
all vectors of the form w = p + vh , where vh is any solution of the
homogeneous equation Ax = 0.
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
16 / 18
1.5 Solutions Sets of Linear Systems
Nonhomogeneous System: Example 6
Example (6)
Describe the solution set of 2x1 − 4x2 − 4x3 = 0 and compare it to
the solution set of 2x1 − 4x2 − 4x3 = 6.
Solution:
Row reduce the augmented matrix for 2x1 − 4x2 − 4x3 = 0:
Write the solution in parametric vector form:
Row reduce the augmented matrix for 2x1 − 4x2 − 4x3 = 6:
Write the solution in parametric vector form:
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
17 / 18
1.5 Solutions Sets of Linear Systems
Nonhomogeneous System: Example 6 (cont.)
The solution set of 2x1 − 4x2 − 4x3 = 0 is a
.
The solution set of 2x1 − 4x2 − 4x3 = 6 is the plane
the solution
set of 2x1 − 4x2 − 4x3 = 0 and passing through the
point
.
Here’s a graph of the two solution sets:
Parallel solution sets of the two systems above.
Alex Bearden, University of Houston
Math 2331, Linear Algebra
Spring 2017
18 / 18