Page 19 - Matrix Analysis & Applied Linear Algebra

1.2 Gaussian Elimination and Matrices 11
Example 1.2.1

Problem: Solve the following system using Gaussian elimination with back sub-
stitution:

v − w = 3,
−2u +4v − w = 1,
−2u +5v − 4w = − 2.

Solution: The associated augmented matrix is

 
01 −1 3
 −24 −1 1  .
−25 −4 −2

Since the first pivotal position contains 0, interchange rows one and two before
eliminating below the first pivot:
   
1 −1 3 4 −1 1
0
-2
Interchange R 1 and R 2
 −24 −1 1  −−−−−−−−→  01 −1 3 
−25 −4 −2 −25 −4 −2 R 3 − R 1
   
−2 4 −1 1 −24 −1 1
1
−→  0 −1 3  −→  01 −1 3  .
0 1 −3 −3 R 3 − R 2 00 −2 −6
Back substitution yields
−6
w = =3,
−2
v =3 + w =3 + 3=6,
1 1
u = (1 − 4v + w)= (1 − 24+3) = 10.
−2 −2

Exercises for section 1.2

1.2.1. Use Gaussian elimination with back substitution to solve the following
system:
x 1 + x 2 + x 3 =1,
x 1 +2x 2 +2x 3 =1,
x 1 +2x 2 +3x 3 =1.