1. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
Lecture 3
Now we are in the stage to tackle the question.
How to solve a linear system AX = B?
First: Row operations
The key to solve a system of linear equations is to
transform the original augmented matrix to some
matrix with some properties via a few elementary
row operations. As a matter of fact, we can solve
any system of linear equations by transforming the
associate augmented matrix to a matrix in some
form.
The form is referred to as the reduced row echelon
form.
1.1 Gauss-elimination method (REF)
Step 1: Form augmented matrix bA :
Step 2: Transform bA : to row echelon form matrix
DC : using row operations
Step 3: Solve the system corresponding to DC : ,
using back substitution
2. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
Example: Solve the following system of equations
3-z-y2-x3
1yx
5z3y-2
x
using Gauss elimination.
Sol.
3:123
1:011
5:312
11 R
2
1 R
3:123
1:011
2
5:
2
3
2
11
R33R
RR
13
212
R
R
2/21:2/112/10
2
3:
2
3
2
30
2
5:
2
3
2
11
33
22
R2
R
3
2
R
R
21:1110
1:110
2
5:
2
3
2
11
323 RR R
22:1200
1:110
2
5:
2
3
2
11
33 R
12
1 R
6
11:100
1:110
2
5:
2
3
2
11
So, z = 11/6 , y = 5/6, x = 1/6
1.2 Gauss-Jordan Reduction Method (RREF)
Step 1: Form augmented matrix bA :
Step 2: Transform bA : to reduced row echelon
form matrix FH : using row operations
3. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
Step 3: for each nonzero row in FH : , solve the
corresponding equations
Example: Solve the following linear system of
equations
3z-x3
8zy2x
9z32y
x
using Gauss-Jordan reduction
method
Sol.
3:103
8:112
9:321
313
212
R3R-
R2R
R
R
24:1060
10:550
9:321
22 R
5
1 R
24:1060
2:110
9:321
121
323
R2R
R6R
R
R
12:400
2:110
5:101
33 R
4
1 R
3:100
2:110
5:101
131
232
RR
RR
R
R
3:100
1:010
2:001
So, x = 2, y = -1, z = 3
5. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
Examples
I. Exactly one solution:
Solve for the following system:
33
82
932
31
321
321
xx
xxx
xxx
[Solution:]The Gauss-Jordan reduction is as follows:
Step 1: The augmented matrix is
3
8
9
1
1
3
0
1
2
3
2
1
.
Step 2:The matrix in reduced row echelon form is
3
1
2
1
0
0
0
1
0
0
0
1
Step 3: The solution is 3,1,2 321 xxx
II.Infinite number of solutions:
Solve for the following system:
153
0242
21
321
xx
xxx
[Solution:]The Gauss-Jordan reduction is as follows:
6. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
Step 1: The augmented matrix is
1
0
0
2
5
4
3
2
Step 2: The matrix in reduced row echelon form
is
1
2
3
5
1
0
0
1
Step 3: The linear system corresponding to the
matrix in reduced row echelon form is
13
25
32
31
xx
xx
The solutions are 1 3 2 32 5 , 1 3x x x x
3x is free variable or parameter and let 3 ,x t t R
therefore
Rttxtxtx ,,31,52 321
III. No solution:
Solve for the following system:
62
1753
5422
431
4321
4321
xxx
xxxx
xxxx
[Solution:]The Gauss-Jordan reduction is as follows:
Step 1:The augmented matrix is
6
11
5
2
7
4
1
5
3
0
3
2
1
1
1
Step 2: The matrix in reduced row echelon form
is
7. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
1
0
0
0
3
2
0
2
1
0
1
0
0
0
1
Step 3: The linear system corresponding to the
matrix in reduced row echelon form is
10
032
02
432
431
xxx
xxx
Since ,10 there is no solution
Example: Solve for the following linear system:
5723
-1132
-3952
352
4321
4321
4321
4321
xxxx
xxxx
xxxx
xxxx
[Solution:] The Gauss-Jordan reduction is as
follows:
Step 1: The augmented matrix is
5
11
3
3
7
3
9
5
2
1
1
2
3
1
5
1
1
2
2
1
Step 2:After elementary row operations, the
matrix in reduced row echelon form is
0
3
2
5
0
2
3
2
0
1
0
0
0
0
1
0
0
0
0
1
.
Step 3:The linear system corresponding to the
matrix in reduced row echelon form is
32
23
52
43
42
41
xx
xx
xx
8. Science Department in Engineering MTH 2311/ 2132 Dr .Gharib S. Mahmoud
The solutions are 1 4 2 4 3 45 2 , 2 3 , 3 2x x x x x x
Rttxtxtxtx ,,23,32,25 4321
Example: Find conditions on a such that the
following system has no solution, one solution, or
infinitely many solutions. 1
( 2) 1
2 2 ( 2) 1
x ay z
x a y z
x y a z
1 1 1 1 1 1 1 1 1
1 2 1 1 0 2 2 0 0 0 2 2 0 0
2 2 2 1 0 2 2 1 0 0 1
1 1 1 1
Case1: 1 0 0 1 1
0 0 0 0
1 1 1
Case2: 1 0 1 0 0
0 0 1
1 0 1 1
(a) 0 0 1 0 0
0 0 0 1
a a a
a a a
a a a a
a
a
a
a
a
1 1 1
(b) 0 0 1 0 0
10 0 1
a
a
a
1 : has infinitely many solutions.
0 : has no solutions.
1 and 0 : has exactly one solution.
a
a
a a
)sol
