homogenous system

University of Regina MATH 122 - Linear Algebra I




MATH 122 - Linear Algebra I
§2.2. Associated homogeneous system

Martin Frankland

September 26, 2022

Example 1 (#2.2.5(c)). Express every solution of the system as a sum of a specific solution
plus a solution of the associated homogeneous system:


 x1 + x2 − x3 − 5x5 = 2

 x2 + x3 − 4x5 = −1


 x2 + x3 + x4 − x5 = −1
2x1 − 4x3 + x4 + x5

= 6.


Solution. Let us write the augmented matrix of the system A⃗x = ⃗b and apply Gaussian
elimination:
   
1 1 −1 0 −5 2 1 1 −1 0 −5 2
h i 0 1 1 0 −4 −1 R −2R 0 1 1 0 −4 −1 −R2
 R3∼
A ⃗b = 

 4∼ 1 
0 1 1 1 −1 −1 0 1 1 1 −1 −1 R4 +2R2

2 0 −4 1 1 6 0 −2 −2 1 11 2
     
1 1 −1 0 −5 2 1 1 −1 0 −5 2 1 0 −2 0 −1 3
0 1 1 0 −4 −1 R4 −R3 0
  1 1 0 −4 −1 R1 −R2 0
  1 1 0 −4 −1

0 0 ∼  ∼  .
0 1 3 0 0 0 0 1 3 0 0 0 0 1 3 0
0 0 0 1 3 0 0 0 0 0 0 0 0 0 0 0 0 0
We use the free variables x3 and x5 as parameters, setting:
x3 = s, x5 = t.
The three equations yield
x1 = 3 + 2x3 + x5
= 3 + 2s + t
x2 = −1 − x3 + 4x5
= −1 − s + 4t
x4 = −3x5
= −3t

© 2022 Martin Frankland All Rights Reserved 1

, University of Regina MATH 122 - Linear Algebra I



so that the general solution is
         
x1 3 + 2s + t 3 2 1
x2  −1 − s + 4t −1 −1 4
         
x3  =  s
⃗x =    =  0  + s 1  + t 0 
      
x4   −3t  0 0 −3
x5 t 0 0 1

for any s, t ∈ R. Here  
3
−1
 
0
⃗xp =  
0
0
is a particular solution and   
2 1
−1 4
   
 1  + t 0 
⃗xh = s    
0 −3
0 1
is the general solution of the associated homogeneous system A⃗x = ⃗0.




© 2022 Martin Frankland All Rights Reserved 2