1.I.3 - General Solutions
MTH 214 - Linear Algebra
Dr. Lew
0 / 11
Solution sets of a linear system fit a general pattern. To see what this
pattern is, consider the following example.
x+y+ z− w= 1
y − z + w = −1
3x
+ 6z − 6w = 6
−y+ z− w= 1
Solving this, we will find the solution set is parameterized as
 

 
 


2
−2
2












−1

1
−1
  + z   + w   : z, w ∈ R
0
1
0




 
 
 




 0

0
1
1 / 11
This parameterization can be broken down into two parts: a particular
solution and an unrestricted combination.
 
 
 
2
2
−2
 
 
 
 
−1
1
 
  + w −1
+
z
0
0
1
 
 
 
1
0
0
particular solution
unrestricted combination
A particular solution is a specific solution to the linear system. The
unrestricted combination refers to the fact that you vary the
parameter(s) to any value with no restriction or conditions.
It is a fundamental result for linear systems that the solution set will
always take on this form.
2 / 11
Solution Set of a Linear System
The solution set of a linear system has the form
n
o
~
~
~
p~ + c1 β1 + c2 β2 + · · · + ck βk : c1 , c2 , . . . , ck ∈ R
where p~ is a particular solution to the linear system and the number k
is equal to the number of free variables of the linear system after Gauss
reduction.
To understand the structure more clearly, let’s focus on the underlying
nature of the unrestricted combination.
Consider the following linear system:
3 / 11
x+y+ z− w=0
y− z+ w=0
3x
+ 6z − 6w = 0
−y+ z− w=0
Note that this linear system has the same coefficients as the linear system
from the first slide, but all of the constant terms are equal to zero.
Homogeneous System
A linear system in which all constant terms are equal to zero is called a
homogenous linear system. A linear system that is not homogenous is
called nonhomogeneous.
The zero vector is always a solution to a homogenous system. This
means that a homogeneous linear system is always consistent.
4 / 11
x+y+ z− w=0
y− z+ w=0
3x
+ 6z − 6w = 0
−y+ z− w=0
Setting up the augmented matrix to the homogenous system and
applying Gauss reduction will give us the following echelon form:




1 1
1 −1 0
1 1 1 −1 0




0 1 −1 1 0
 0 1 −1 1 0

 −→ · · · −→ 

3 0


6 −6 0
0 0

0 0 0

0 −1 1 −1 0
0 0 0
0 0
The two free variables are z and w. The first row of the echelon matrix
gives x + y + z − w = 0 and the second row gives y − z + w = 0.
5 / 11
x + y + z − w = 0 and
y−z+w =0
Solve these for x and y in terms of z and w. We get:
x = −2z + 2w
y =z−w
z is free
w is free
So the solution set to this homogeneous system is
  

 
 
x
−2z + 2w
−2
2
  

 
 
y   z − w 
 
 
 = 
 = z  1  + w −1
z  
1
0

z
  

 
 
w
w
0
1
6 / 11
 
 
 
2
−2
x
 
 
 
 
 
y 
  = z  1  + w −1
0
1
z 
 
 
 
1
0
w
Where have we seen this before? This appears to be the unrestricted
combination portion of the solution set to the nonhomogeneous linear system that
we solved earlier that has the same coefficient matrix. This is always true....
General = Particular + Homogeneous
Any solution to a linear system has the form p~ + ~h, where p~ is a particular solution to the linear system and ~h is a solution to the associated
homogeneous system.
The solution set to the linear system is called its general solution.
7 / 11
Example
Determine the general solution to the linear system
x
+ z= 4
x − y + 2z = 5
4x − y + 5z = 17
Answer. Set up the augmented matrix and using Gauss reduction to
get it to echelon form:




1 0 1 4
1 0 1
4




1 −1 2 5  −→ · · · −→  0 1 −1 −1
4 −1 5 17
0 0
0
0
So x and y are leading variables, and z is a free variable.
8 / 11


1 0 1
4


0 1 −1 −1
0 0 0
0
The first row gives the equation x + z = 4 and the second row gives
y − z = −1. Solving for x and y in terms of z, we see that any solution
to the linear system has the form
  
  
 
x
4−z
4
−1
  
  
 
y  = −1 + z  = −1 + z  1  = p~ + z~h
z
z
0
1
It is easy to check that p~ is a particular solution to the original system
and ~h is a solution to the homogeneous system.
9 / 11
If a nonhomogeneous linear system is to be consistent, then it MUST
have a particular solution. If it doesn’t, then it will not have any solutions
whatsoever.
If it does, then whether or not a consistent nonhomogeneous linear
system has infinitely many solutions is determined by whether or not the
corresponding homogeneous system has infinitely many solutions.
These observations are summarize in the chart below:
10 / 11
From the chart notice that a nonhomogeneous system has exactly one
solution when there is a particular solution and the homogeneous system
has only the zero vector as a solution.
This means that there are no free variables when the coefficient matrix is
Gauss reduced to an echelon form. But this occurs when every variable is
a leading variable.
In this case we refer to the matrix of coefficients coming from the linear
system nonsingular.
We will explore this idea and the more general notion of matrices in
detail in a later section.
11 / 11