Linear Algebra: Gauss Elimination Method
Uploaded by
Aman ChaudharyLinear Algebra: Gauss Elimination Method
Uploaded by
Aman ChaudharyMA 106 - Linear Algebra
Neela Nataraj
Department of Mathematics,
Indian Institute of Technology Bombay,
Powai, Mumbai 76
neela@math.iitb.ac.in
January 10, 2013
Neela Nataraj Lecture 2
Outline of the lecture
Gauss Elimination Method
Row-Echelon Form, Row Reduced Echelon Form
Neela Nataraj Lecture 2
Example 1 (Determined System)
Solve
x
1
+ x
2
+ 2x
3
= 2
3x
1
x
2
+ x
3
= 6
x
1
+ 3x
2
+ 4x
3
= 4.
Solution : The linear system of equations can be expressed as
_
_
1 1 2
3 1 1
1 3 4
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
6
4
_
_
Neela Nataraj Lecture 2
Example 1 (Determined System)
Solve
x
1
+ x
2
+ 2x
3
= 2
3x
1
x
2
+ x
3
= 6
x
1
+ 3x
2
+ 4x
3
= 4.
Solution : The linear system of equations can be expressed as
_
_
1 1 2
3 1 1
1 3 4
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
6
4
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3
1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1
3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2
7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2
2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1
1
1 2
.
.
. 2
3 1 1
.
.
. 6
-1 3 4
.
.
. 4
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 2 2
.
.
. 2
_
_
R
3
R
3
+(1)R
2
_
1 1 2
.
.
. 2
0 2 7
.
.
. 12
0 0 5
.
.
. 10
_
_
(Row-Echelon Form)
1
The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ x
2
+ 2x
3
= 2
2x
2
+ 7x
3
= 12
5x
3
= 10.
and this system can be solved easily as
5x
3
= 10 = x
3
= 2;
2x
2
+ 7x
3
= 12 = 2x
2
+ 14 = 12 =x
2
= 1;
x
1
+ x
2
+ 2x
3
= 2 = x
1
1 + 4 = 2 =x
1
= 1.
Hence, x
1
= 1, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT SYSTEM
WITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ x
2
+ 2x
3
= 2
2x
2
+ 7x
3
= 12
5x
3
= 10.
and this system can be solved easily as
5x
3
= 10 = x
3
= 2;
2x
2
+ 7x
3
= 12 = 2x
2
+ 14 = 12 =x
2
= 1;
x
1
+ x
2
+ 2x
3
= 2 = x
1
1 + 4 = 2 =x
1
= 1.
Hence, x
1
= 1, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT SYSTEM
WITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ x
2
+ 2x
3
= 2
2x
2
+ 7x
3
= 12
5x
3
= 10.
and this system can be solved easily as
5x
3
= 10 = x
3
= 2;
2x
2
+ 7x
3
= 12 = 2x
2
+ 14 = 12 =x
2
= 1;
x
1
+ x
2
+ 2x
3
= 2 = x
1
1 + 4 = 2 =x
1
= 1.
Hence, x
1
= 1, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT SYSTEM
WITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ x
2
+ 2x
3
= 2
2x
2
+ 7x
3
= 12
5x
3
= 10.
and this system can be solved easily as
5x
3
= 10 = x
3
= 2;
2x
2
+ 7x
3
= 12 = 2x
2
+ 14 = 12 =x
2
= 1;
x
1
+ x
2
+ 2x
3
= 2 = x
1
1 + 4 = 2 =x
1
= 1.
Hence, x
1
= 1, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT SYSTEM
WITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ x
2
+ 2x
3
= 2
2x
2
+ 7x
3
= 12
5x
3
= 10.
and this system can be solved easily as
5x
3
= 10 = x
3
= 2;
2x
2
+ 7x
3
= 12 = 2x
2
+ 14 = 12 =x
2
= 1;
x
1
+ x
2
+ 2x
3
= 2 = x
1
1 + 4 = 2 =x
1
= 1.
Hence, x
1
= 1, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT SYSTEM
WITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ x
2
+ 2x
3
= 2
2x
2
+ 7x
3
= 12
5x
3
= 10.
and this system can be solved easily as
5x
3
= 10 = x
3
= 2;
2x
2
+ 7x
3
= 12 = 2x
2
+ 14 = 12 =x
2
= 1;
x
1
+ x
2
+ 2x
3
= 2 = x
1
1 + 4 = 2 =x
1
= 1.
Hence, x
1
= 1, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT SYSTEM
WITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Example 2 (Over Determined System)
Solve :
x
1
+ 2x
2
= 2
3x
1
+ 6x
2
x
3
= 8
x
1
+ 2x
2
+ x
3
= 0
2x
1
+ 5x
2
2x
3
= 9.
Solution : The linear system of equations can be expressed as
_
_
1 2 0
3 6 1
1 2 1
2 5 2
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
8
0
9
_
_
Neela Nataraj Lecture 2
Example 2 (Over Determined System)
Solve :
x
1
+ 2x
2
= 2
3x
1
+ 6x
2
x
3
= 8
x
1
+ 2x
2
+ x
3
= 0
2x
1
+ 5x
2
2x
3
= 9.
Solution : The linear system of equations can be expressed as
_
_
1 2 0
3 6 1
1 2 1
2 5 2
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
8
0
9
_
_
Neela Nataraj Lecture 2
Example 2 (Over Determined System)
Solve :
x
1
+ 2x
2
= 2
3x
1
+ 6x
2
x
3
= 8
x
1
+ 2x
2
+ x
3
= 0
2x
1
+ 5x
2
2x
3
= 9.
Solution : The linear system of equations can be expressed as
_
_
1 2 0
3 6 1
1 2 1
2 5 2
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
8
0
9
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
2 5 2
.
.
. 9
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
R
4
R
4
+(2)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
0 1 2
.
.
. 5
_
_
_
_
no x
2
in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x
2
R
2
R
4
PARTIAL PIVOTING
Neela Nataraj Lecture 2
Contd...
R
2
R
4
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 -1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd...
R
2
R
4
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 -1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd...
R
2
R
4
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 -1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd...
R
2
R
4
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 -1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd...
R
2
R
4
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 -1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd...
R
2
R
4
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 -1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd..
R
4
R
4
+(1)R
3
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Contd..
R
4
R
4
+(1)R
3
_
1 2 0
.
.
. 2
0 1 2
.
.
. 5
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
The given system has the same solution as the system
x
1
+ 2x
2
= 2
x
2
2x
3
= 5
x
3
= 2.
and this system can be solved easily as
x
3
= 2;
x
2
2x
3
= 5 = x
2
= 5 + 2 2 =x
2
= 1;
x
1
+ 2x
2
= 2 = x
1
= 2 2 =x
1
= 0.
Hence, x
1
= 0, x
2
= 1, x
3
= 2.
THIS IS AN EXAMPLE OF A CONSISTENT
OVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Example 3 (Determined, Inconsistent System)
Solve
x
2
+ 3x
3
= 1
x
1
+ 2x
2
x
3
= 8
x
1
+ x
2
+ 2x
3
= 1
Solution : The linear system of equations can be expressed as
_
_
0 1 3
1 2 1
1 1 2
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
1
8
1
_
_
Neela Nataraj Lecture 2
Example 3 (Determined, Inconsistent System)
Solve
x
2
+ 3x
3
= 1
x
1
+ 2x
2
x
3
= 8
x
1
+ x
2
+ 2x
3
= 1
Solution : The linear system of equations can be expressed as
_
_
0 1 3
1 2 1
1 1 2
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
1
8
1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
0 1 3
.
.
. 1
1 2 1
.
.
. 8
1 1 2
.
.
. 1
_
_
_
_
no x
1
in the rst equation;
interchange with 2nd equation
R
2
R
1
: PARTIAL PIVOTING
R
1
R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
1 1 2
.
.
. 1
_
_
Neela Nataraj Lecture 2
Contd...
R
3
R
3
+(1)R
1
_
1 2 1
.
.
. 8
0 -1 3
.
.
. 1
0 -1 3
.
.
. 7
_
_
R
3
R
3
+(1)R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
0 0 0
.
.
. 8
_
_
The third equation implies INCONSISTENCY. Hence,
NO SOLUTION .
Neela Nataraj Lecture 2
Contd...
R
3
R
3
+(1)R
1
_
1 2 1
.
.
. 8
0 -1 3
.
.
. 1
0 -1 3
.
.
. 7
_
_
R
3
R
3
+(1)R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
0 0 0
.
.
. 8
_
_
The third equation implies INCONSISTENCY. Hence,
NO SOLUTION .
Neela Nataraj Lecture 2
Contd...
R
3
R
3
+(1)R
1
_
1 2 1
.
.
. 8
0 -1 3
.
.
. 1
0 -1 3
.
.
. 7
_
_
R
3
R
3
+(1)R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
0 0 0
.
.
. 8
_
_
The third equation implies INCONSISTENCY. Hence,
NO SOLUTION .
Neela Nataraj Lecture 2
Contd...
R
3
R
3
+(1)R
1
_
1 2 1
.
.
. 8
0 -1 3
.
.
. 1
0 -1 3
.
.
. 7
_
_
R
3
R
3
+(1)R
2
_
1 2 1
.
.
. 8
0 1 3
.
.
. 1
0 0 0
.
.
. 8
_
_
The third equation implies INCONSISTENCY. Hence,
NO SOLUTION .
Neela Nataraj Lecture 2
Example 4 : Consistent, determined system (Innitely
many solutions)
Solve :
x
1
+ 2x
2
= 2
3x
1
+ 6x
2
x
3
= 8
x
1
+ 2x
2
+ x
3
= 0
Solution : The linear system of equations can be expressed as
_
_
1 2 0
3 6 1
1 2 1
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
8
0
_
_
Neela Nataraj Lecture 2
Example 4 : Consistent, determined system (Innitely
many solutions)
Solve :
x
1
+ 2x
2
= 2
3x
1
+ 6x
2
x
3
= 8
x
1
+ 2x
2
+ x
3
= 0
Solution : The linear system of equations can be expressed as
_
_
1 2 0
3 6 1
1 2 1
_
_
_
_
x
1
x
2
x
3
_
_
=
_
_
2
8
0
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
The augmented matrix is given by
A =
_
_
1 2 0
.
.
. 2
3 6 1
.
.
. 8
1 2 1
.
.
. 0
_
_
R
2
R
2
+(3)R
1
R
3
R
3
+(1)R
1
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 1
.
.
. 2
_
_
_
Candidate for second pivot is 0
Go to next column
_
1 2 0
.
.
. 2
0 0 -1
.
.
. 2
0 0 1
.
.
. 2
_
_
Neela Nataraj Lecture 2
Contd..
R
3
R
3
+(1)R
2
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Contd..
R
3
R
3
+(1)R
2
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Contd..
R
3
R
3
+(1)R
2
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Contd..
R
3
R
3
+(1)R
2
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Contd..
R
3
R
3
+(1)R
2
_
1 2 0
.
.
. 2
0 0 1
.
.
. 2
0 0 0
.
.
. 0
_
_
_
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t,
then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Back Substitution
That is,
x
3
= 2;
x
1
+ 2x
2
= 2 = x
1
= 2 2x
2
.
Here, x
1
and x
3
are basic variables and x
2
is the free variable (no
pivot).
If we choose x
2
= t, then x
1
= 2 2t.
The solution vector is
_
_
2 2t
t
2
_
_
=
_
_
2
0
2
_
_
+ t
_
_
2
1
0
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Example 5 : Consistent, determined system (Innitely
many solutions)
Exercise : Solve
2x
1
+ 3x
2
2x
3
+ 4x
4
= 2
6x
1
+9x
2
+ 7x
3
8x
4
= 3
4x
1
+ 6x
2
x
3
+ 20x
4
= 13
The solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
Neela Nataraj Lecture 2
Example 5 : Consistent, determined system (Innitely
many solutions)
Exercise : Solve
2x
1
+ 3x
2
2x
3
+ 4x
4
= 2
6x
1
+9x
2
+ 7x
3
8x
4
= 3
4x
1
+ 6x
2
x
3
+ 20x
4
= 13
The solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
Neela Nataraj Lecture 2
Example 5 : Consistent, determined system (Innitely
many solutions)
Exercise : Solve
2x
1
+ 3x
2
2x
3
+ 4x
4
= 2
6x
1
+9x
2
+ 7x
3
8x
4
= 3
4x
1
+ 6x
2
x
3
+ 20x
4
= 13
The solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
_
_
2 3 2 4
6 9 7 8
4 6 1 20
_
_
_
_
x
1
x
2
x
3
x
4
_
_
=
_
_
2
3
13
_
_
The augmented matrix is given by
A =
_
_
2 3 2 4
.
.
. 2
-6 9 7 8
.
.
. 3
4 6 1 20
.
.
. 13
_
_
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
_
_
2 3 2 4
6 9 7 8
4 6 1 20
_
_
_
_
x
1
x
2
x
3
x
4
_
_
=
_
_
2
3
13
_
_
The augmented matrix is given by
A =
_
_
2 3 2 4
.
.
. 2
-6 9 7 8
.
.
. 3
4 6 1 20
.
.
. 13
_
_
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
_
_
2 3 2 4
6 9 7 8
4 6 1 20
_
_
_
_
x
1
x
2
x
3
x
4
_
_
=
_
_
2
3
13
_
_
The augmented matrix is given by
A =
_
_
2 3 2 4
.
.
. 2
-6 9 7 8
.
.
. 3
4 6 1 20
.
.
. 13
_
_
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
_
_
2 3 2 4
6 9 7 8
4 6 1 20
_
_
_
_
x
1
x
2
x
3
x
4
_
_
=
_
_
2
3
13
_
_
The augmented matrix is given by
A =
_
_
2 3 2 4
.
.
. 2
-6 9 7 8
.
.
. 3
4 6 1 20
.
.
. 13
_
_
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
_
_
2 3 2 4
6 9 7 8
4 6 1 20
_
_
_
_
x
1
x
2
x
3
x
4
_
_
=
_
_
2
3
13
_
_
The augmented matrix is given by
A =
_
_
2 3 2 4
.
.
. 2
-6 9 7 8
.
.
. 3
4 6 1 20
.
.
. 13
_
_
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R
2
R
2
+(3)R
1
R
3
R
3
+(2)R
1
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 3 12
.
.
. 9
_
_
_
Candidate for second pivot is 0,
x
2
is a free variable
Go to next column and choose the pivot
R
3
R
3
+(3)R
2
_
2 3 2 4
.
.
. 2
0 0 1 4
.
.
. 3
0 0 0 0
.
.
. 0
_
_
_
_
x
4
is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables
and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t &
x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s,
then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s
and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s
=x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Back Substitution
Here, x
2
& x
4
are free variables and x
1
& x
3
are the basic variables.
(Free variables do not occur at the beginning of any equation when
the system has been reduced to echelon form).
If we choose x
2
= t & x
4
= s, then x
3
= 3 4s and
2x
1
= 2 3x
2
+ 2x
3
4x
4
= 2 3t + 2(3 4s) 4s
= 8 3t 12s =x
1
= 4
3
2
t 6s.
Neela Nataraj Lecture 2
Solution Vector
Hence, the solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Solution Vector
Hence, the solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Solution Vector
Hence, the solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Solution Vector
Hence, the solution vector is
_
_
4
3
2
t 6s
t
3 4s
s
_
_
=
_
_
4
0
3
0
_
_
+ t
_
_
3/2
1
0
0
_
_
+ s
_
_
6
0
4
1
_
_
THIS IS AN EXAMPLE OF A CONSISTENT
DETERMINED SYSTEM WITH INFINITELY MANY
SOLUTIONS.
Neela Nataraj Lecture 2
Illustration
EXERCISE : TUTORIAL SHEET 2 - QN. 1
Neela Nataraj Lecture 2
Stage 1: Forward Elimination Phase
The steps in Gaussian elimination can be summarized as follows:
1. Search the rst column of [A|b] from the top to the bottom for the rst
non-zero entry, and then if necessary, the second column (the case where
all the coecients corresponding to the rst variable are zero), and then
the third column, and so on. The entry thus found is called the current
pivot.
2. Interchange, if necessary, the row containing the current pivot with the
rst row.
3. Keeping the row containing the pivot (that is, the rst row) untouched,
subtract appropriate multiples of the rst row from all the other rows to
obtain all zeroes below the current pivot in its column.
4. Repeat the preceding steps on the submatrix consisting of all those
elements which are below and to the right of the current pivot.
5. Stop when no further pivot can be found.
Neela Nataraj Lecture 2
REF
The mn coecient matrix A of the linear system Ax = b is thus
reduced to an (m n) row echelon matrix U and the augmented
matrix [A|b] is reduced to
[U|c] =
_
_
0 . . . p
1
. . . c
1
0 . . . 0 . . . p
2
. . . c
2
0 . . . 0 0 0 . . . p
3
. . . c
3
.
.
.
.
.
.
.
.
.
.
.
. 0 0 . . .
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 p
r
. . . c
r
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
r +1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
m
_
_
.
The entries denoted by and the c
i
s are real numbers; they may or may
not be zero. The p
i
s denote the pivots; they are non-zero. Note that
there is exactly one pivot in each of the rst r rows of U and that any
column of U has at most one pivot. Hence r m and r n.
Neela Nataraj Lecture 2
REF
The mn coecient matrix A of the linear system Ax = b is thus
reduced to an (m n) row echelon matrix U and the augmented
matrix [A|b] is reduced to
[U|c] =
_
_
0 . . . p
1
. . . c
1
0 . . . 0 . . . p
2
. . . c
2
0 . . . 0 0 0 . . . p
3
. . . c
3
.
.
.
.
.
.
.
.
.
.
.
. 0 0 . . .
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 p
r
. . . c
r
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
r +1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
m
_
_
.
The entries denoted by and the c
i
s are real numbers; they may or may
not be zero.
The p
i
s denote the pivots; they are non-zero. Note that
there is exactly one pivot in each of the rst r rows of U and that any
column of U has at most one pivot. Hence r m and r n.
Neela Nataraj Lecture 2
REF
The mn coecient matrix A of the linear system Ax = b is thus
reduced to an (m n) row echelon matrix U and the augmented
matrix [A|b] is reduced to
[U|c] =
_
_
0 . . . p
1
. . . c
1
0 . . . 0 . . . p
2
. . . c
2
0 . . . 0 0 0 . . . p
3
. . . c
3
.
.
.
.
.
.
.
.
.
.
.
. 0 0 . . .
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 p
r
. . . c
r
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
r +1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
m
_
_
.
The entries denoted by and the c
i
s are real numbers; they may or may
not be zero. The p
i
s denote the pivots; they are non-zero.
Note that
there is exactly one pivot in each of the rst r rows of U and that any
column of U has at most one pivot. Hence r m and r n.
Neela Nataraj Lecture 2
REF
The mn coecient matrix A of the linear system Ax = b is thus
reduced to an (m n) row echelon matrix U and the augmented
matrix [A|b] is reduced to
[U|c] =
_
_
0 . . . p
1
. . . c
1
0 . . . 0 . . . p
2
. . . c
2
0 . . . 0 0 0 . . . p
3
. . . c
3
.
.
.
.
.
.
.
.
.
.
.
. 0 0 . . .
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 p
r
. . . c
r
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
r +1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
m
_
_
.
The entries denoted by and the c
i
s are real numbers; they may or may
not be zero. The p
i
s denote the pivots; they are non-zero. Note that
there is exactly one pivot in each of the rst r rows of U and that any
column of U has at most one pivot.
Hence r m and r n.
Neela Nataraj Lecture 2
REF
The mn coecient matrix A of the linear system Ax = b is thus
reduced to an (m n) row echelon matrix U and the augmented
matrix [A|b] is reduced to
[U|c] =
_
_
0 . . . p
1
. . . c
1
0 . . . 0 . . . p
2
. . . c
2
0 . . . 0 0 0 . . . p
3
. . . c
3
.
.
.
.
.
.
.
.
.
.
.
. 0 0 . . .
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 p
r
. . . c
r
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
r +1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 . . . 0 0 0 0 0 0 0 0 . . . 0 c
m
_
_
.
The entries denoted by and the c
i
s are real numbers; they may or may
not be zero. The p
i
s denote the pivots; they are non-zero. Note that
there is exactly one pivot in each of the rst r rows of U and that any
column of U has at most one pivot. Hence r m and r n.
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations)
and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1,
then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable;
otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number of
equations) and c
r +k
= 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = c
r +k
and so the
system has no solutions (inconsistent system) .
If (i) r = m or (ii) r < m and c
r +k
= 0 for all k 1, then there
exists a solution of the system (consistent system) .
(Basic & Free Variables)
If the j th column of U contains a pivot, then x
j
is called a basic
variable; otherwise x
j
is called a free variable.
In fact, there are n r free variables , where n is the number of
columns (unknowns) of A (and hence of U).
Neela Nataraj Lecture 2
Stage 2: (Back Substitution Phase)
In the case of a consistent system, if x
j
is a free variable, then it
can be set equal to a parameter s
j
which can assume arbitrary
values.
If x
j
is a basic variable, then we solve for x
j
in terms of
x
j +1
, . . . , x
m
, starting from the last basic variable and working our
way up row by row.
Neela Nataraj Lecture 2
Stage 2: (Back Substitution Phase)
In the case of a consistent system, if x
j
is a free variable, then it
can be set equal to a parameter s
j
which can assume arbitrary
values.
If x
j
is a basic variable, then we solve for x
j
in terms of
x
j +1
, . . . , x
m
, starting from the last basic variable and working our
way up row by row.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination method
leads to the row echelon form of a matrix which can be dened
as follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelon
matrix) if it has a staircase-like pattern characterized by the following
properties:
(a) The all-zero rows (if any) are at the bottom.
(b) If we call the left most non-zero entry of a non-zero row its leading
entry, then the leading entry of each non-zero row is to the right of the
leading entry of the preceding row.
(c) All entries in a column below a leading entry is zero.
( Row Reduced Echelon Form (RREF))
If a matrix in echelon form satises the following additional conditions,
then it is in row reduced echelon form:
(d) The leading entry in each nonzero row is 1.
(e) Each leading 1 is the only nonzero entry in its column.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
Examples
1
_
_
2 3 2 4
0 9 7 8
0 0 0 20
0 0 0 0
_
_
is in REF.
2
_
_
1 0 0 4
0 1 0 8
0 0 1 20
_
_
is in RREF.
Question : How to obtain RREF?
Remarks :
1
Any row reduced echelon matrix is also a row-echelon matrix.
2
Any nonzero matrix may be row reduced into more than one
matrix in echelon form. But the row reduced echelon form
that one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used as
needed to create zeros with the help of row operations.
4
Dierent sequences of row operations might involve a
dierent set of pivots.
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1.
F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T (number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1.
T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T (number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations.
T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T (number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent.
F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T (number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent.
T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T (number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables.
T (number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T
(number of free variables = n r ).
Neela Nataraj Lecture 2
EXAMPLES:
Determine whether the following statements are true or false.
1
If a matrix is in row echelon form, then the leading entry of
each nonzero row must be 1. F
2
If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T
3
Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
4
If the reduced row echelon form of the augmented matrix of a
system of linear equations contains a zero row, then the
system is consistent. F
5
If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, then
the system is inconsistent. T
6
If the reduced row echelon form of the augmented matrix of a
consistent system of m linear equations in n variables contains
r nonzero rows, then its general solution contains r basic
variables. T (number of free variables = n r ).
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF,
not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF??
YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT
in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF
and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
_
_
1 2 3 4 5
0 0 1 1 2
0 0 0 1 5
_
_
is an example of a matrix in REF, not
RREF.
2
_
_
1 0 0 0 1
0 1 0 0 2
0 0 0 1 3
_
_
is in RREF. Is it in REF?? YES.
3
The matrix
_
_
2 1 2 1 5
0 1 1 3 3
0 2 0 0 5
0 0 0 3 2
_
_
is NOT in REF.
4
_
_
1 0 0 1/2 0 1
0 0 1 1/3 0 2
0 0 0 0 1 3
_
_
is in RREF and hence in REF.
Note that the left side of a matrix in RREF need not be identity
matrix.
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots)
and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable
and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields
x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1
= x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2
= x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions
(since x
3
is a free parameter).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
_
_
1 1 2 3 2
0 1 5 0 1
0 0 0 1 2
0 0 0 0 0
_
_
.
In this case r = 3 (number of pivots) and n = 4 (number of
variables).
Note that x
3
is a free variable and x
1
, x
2
, x
4
are basic variables.
Back substitution yields x
4
= 2,
x
2
+ 5x
3
= 1 = x
2
= 1 5x
3
= 1 5s
x
1
+ x
2
+ 2x
3
+ 3x
4
= 2 = x
1
= 2 (1 5s) 2s 3 2
= x
1
= 5 + 3s.
We get innitely many solutions (since x
3
is a free parameter).
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we dene
rank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1. If A =
_
_
1 0 5
0 1 3
0 0 0
_
_
, then rank (A) =2, nullity (A)=1.
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we dene
rank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1. If A =
_
_
1 0 5
0 1 3
0 0 0
_
_
,
then rank (A) =2, nullity (A)=1.
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we dene
rank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1. If A =
_
_
1 0 5
0 1 3
0 0 0
_
_
, then rank (A) =2,
nullity (A)=1.
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we dene
rank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1. If A =
_
_
1 0 5
0 1 3
0 0 0
_
_
, then rank (A) =2, nullity (A)=1.
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we dene
rank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1. If A =
_
_
1 0 5
0 1 3
0 0 0
_
_
, then rank (A) =2, nullity (A)=1.
Neela Nataraj Lecture 2
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