Page 67 - 35Linear Algebra

2.5 Solution Sets for Systems of Linear Equations 67

for any µ 1 , µ 2 ∈ R. Choosing µ 1 = µ 2 = 0, we obtain

P
Mx = v .

P
This is why x is an example of a particular solution.
P
Setting µ 1 = 1, µ 2 = 0, and subtracting Mx = v we obtain
H
Mx = 0 .
1
Likewise, setting µ 1 = 0, µ 2 = 1, we obtain

H
Mx = 0 .
2
Here x H and x H are examples of what are called homogeneous solutions to
1 2
the system. They do not solve the original equation Mx = v, but instead its
associated homogeneous equation My = 0.
We have just learnt a fundamental lesson of linear algebra: the solution
set to Ax = b, where A is a linear operator, consists of a particular solution
plus homogeneous solutions.

{Solutions} = {Particular solution + Homogeneous solutions}

Example 33 Consider the matrix equation of example 32. It has solution set

      
 1 −1 1 
 
1
 
   1   −1 
S =   + µ 1   + µ 2   : µ 1 , µ 2 ∈ R .
 0  1   0 
 

 
 
0 0 1
 
1
1
P
 
Then Mx = v says that   is a solution to the original matrix equation, which is
0
 
0
certainly true, but this is not the only solution.
 
−1
1 
H

Mx = 0 says that   is a solution to the homogeneous equation.
1
 1 
0
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