lecture2_d3
TRANSCRIPT
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MA 106 - Linear Algebra
Neela Nataraj
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai [email protected]
January 10, 2013
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Outline of the lecture
Gauss Elimination Method
Row-Echelon Form, Row Reduced Echelon Form
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Example 1 (Determined System)
Solve
x1 + x2 + 2x3 = 23x1 x2 + x3 = 6
x
1+ 3x
2+ 4x
3= 4.
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Example 1 (Determined System)
Solve
x1 + x2 + 2x3 = 23x1 x2 + x3 = 6
x
1+ 3x
2+ 4x
3= 4.
Solution : The linear system of equations can be expressed as
1 1 23
1 1
1 3 4x1x2x3 =
264
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2... 2
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
O
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2... 2
0 2
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
E h l F i R O i
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2... 2
0 2 7... 12
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
E h l F i R O i
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2... 2
0 2 7... 12
0 2
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
E h l F i R O ti
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2... 2
0 2 7... 12
0 2 2... 2
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
E h l F i R O ti
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2... 2
0 2 7... 12
0 2 2... 2
R3
R3+(
1)R2
1The circled element is the pivot element at each stage and the
elements in the boxes need to be eliminated.Neela Nataraj Lecture 2
E h l F si R O ti s
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2..
. 20 2 7
... 12
0 2 2... 2
R3
R3+(
1)R2 1 1 2
... 2
0 2 7... 12
1The 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
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2R2+(3)R1R3R3+(1)R1
1 1 2..
. 20 2 7
... 12
0 2 2... 2
R3
R3+(
1)R2 1 1 2
... 2
0 2 7... 12
0 0 5 ... 10
1The 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
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Echelon Form using Row Operations
The augmented matrix is given by
A = 11
1 2
..
. 23 1 1 ... 6-1 3 4
... 4
R2
R2+(3)R1R3R3+(1)R1
1 1 2..
. 20 2 7
... 12
0 2 2... 2
R3
R3+(
1)R2 1 1 2
... 2
0 2 7... 12
0 0 5 ... 10
(Row-Echelon Form)1The 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
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Back Substitution
The given system has the same solution as the system
x1 + x2 + 2x3 = 22x2 + 7x3 = 12
5x3 = 10.
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Back Substitution
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Back Substitution
The given system has the same solution as the system
x1 + x2 + 2x3 = 22x2 + 7x3 = 12
5x3 = 10.
and this system can be solved easily as
5x3 = 10 = x3 = 2;
Neela Nataraj Lecture 2
Back Substitution
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Back Substitution
The given system has the same solution as the system
x1 + x2 + 2x3 = 22x2 + 7x3 = 12
5x3 = 10.
and this system can be solved easily as
5x3 = 10 = x3 = 2;2x2 + 7x3 = 12 = 2x2 + 14 = 12 = x2 = 1;
Neela Nataraj Lecture 2
Back Substitution
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Back Substitution
The given system has the same solution as the system
x1 + x2 + 2x3 = 22x2 + 7x3 = 12
5x3 = 10.
and this system can be solved easily as
5x3 = 10 = x3 = 2;2x2 + 7x3 = 12 = 2x2 + 14 = 12 = x2 = 1;
x1 + x2 + 2x3 = 2 =
x1
1 + 4 = 2 =
x1 = 1.
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Back Substitution
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Back Substitution
The given system has the same solution as the system
x1 + x2 + 2x3 = 22x2 + 7x3 = 12
5x3 = 10.
and this system can be solved easily as
5x3 = 10 = x3 = 2;2x2 + 7x3 = 12 = 2x2 + 14 = 12 = x2 = 1;
x1 + x2 + 2x3 = 2 =
x1
1 + 4 = 2 =
x1 = 1.
Hence, x1 = 1, x2 = 1, x3 = 2.
Neela Nataraj Lecture 2
Back Substitution
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Back Substitution
The given system has the same solution as the system
x1 + x2 + 2x3 = 22x2 + 7x3 = 12
5x3 = 10.
and this system can be solved easily as
5x3 = 10 = x3 = 2;2x2 + 7x3 = 12 = 2x2 + 14 = 12 = x2 = 1;
x1 + x2 + 2x3 = 2 =
x1
1 + 4 = 2 =
x1 = 1.
Hence, x1 = 1, x2 = 1, x3 = 2.THIS IS AN EXAMPLE OF A CONSISTENT SYSTEMWITH A UNIQUE SOLUTION (DETERMINED SYSTEM).
Neela Nataraj Lecture 2
Example 2 (Over Determined System)
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Example 2 (Over Determined System)
Solve :
x1 + 2x2 = 2
3x1 + 6x2 x3 = 8x1 + 2x2 + x3 = 0
2x1 + 5x2 2x3 = 9.
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Example 2 (Over Determined System)
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p (O Sy )
Solve :
x1 + 2x2 = 2
3x1 + 6x2 x3 = 8x1 + 2x2 + x3 = 0
2x1 + 5x2 2x3 = 9.Solution : The linear system of equations can be expressed as
1 2 03 6
1
1 2 12 5 2
x1x2x3 =
2809
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Example 2 (Over Determined System)
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p ( y )
Solve :
x1 + 2x2 = 2
3x1 + 6x2 x3 = 8x1 + 2x2 + x3 = 0
2x1 + 5x2 2x3 = 9.Solution : The linear system of equations can be expressed as
1 2 03 6
1
1 2 12 5 2
x1x2x3 =
2809
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Echelon Form using Row Operations
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g p
The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2 ... 9
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Echelon Form using Row Operations
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g p
The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2 ... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
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Echelon Form using Row Operations
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g
The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2 ... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 2
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2 ... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 2
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 20 1 2 ... 5
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 20 1 2 ... 5
no x2 in second equation;
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 20 1 2 ... 5
no x2 in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x2
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 20 1 2 ... 5
no x2 in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x2
R2 R4
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 20 1 2 ... 5
no x2 in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x2
R2 R4PARTIAL PIVOTING
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
1 2 0 ... 2
3 6 1 ... 81 2 1
... 0
2 5 2... 9
R2R2+(3)R1R3R3+(1)R1
R4R4+(2)R1
1 2 0
... 2
0 0 1 ... 20 0 1
.
.. 20 1 2 ... 5
no x2 in second equation;
interchange with 4th eqn.
as 3rd also doesnt have x2
R2 R4PARTIAL PIVOTING
Neela Nataraj Lecture 2
Contd...
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R2R4
1 2 0 ... 2
0 1 2 ... 5
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Contd...
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R2R4
1 2 0 ... 2
0 1 2 ... 50 0 1
... 2
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Contd...
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R2R4
1 2 0 ... 2
0 1 2 ... 50 0 1
... 20 0 1
.
.. 2
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Contd...
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R2R4
1 2 0 ... 2
0 1 2 ... 50 0 1
... 20 0 1
.
.. 2
1 2 0
... 2
0 1 2 ... 5
0 0 1
.
.. 2
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Contd...
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R2R4
1 2 0 ... 2
0 1 2 ... 50 0 1
... 20 0 1
.
.. 2
1 2 0
... 2
0 1 2 ... 5
0 0 1
.
.. 20 0 -1
... 2
Neela Nataraj Lecture 2
Contd...
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R2R4
1 2 0 ... 2
0 1 2 ... 50 0 1
... 20 0 1
.
.. 2
1 2 0
... 2
0 1 2 ... 5
0 0 1
.
.. 20 0 -1
... 2
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Contd..
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R4
R4+(1)R3
1 2 0... 2
0 1 2..
. 50 0 1
... 20 0 0
... 0
Neela Nataraj Lecture 2
Contd..
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R4
R4+(1)R3
1 2 0... 2
0 1 2..
. 50 0 1
... 20 0 0
... 0
ROW-ECHELON FORM
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Back Substitution
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The given system has the same solution as the system
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Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
x3 = 2;
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
x3 = 2;x2 2x3 = 5 = x2 = 5 + 2 2 = x2 = 1;
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
x3 = 2;x2 2x3 = 5 = x2 = 5 + 2 2 = x2 = 1;x1 + 2x2 = 2 =
x1 = 2
2 =
x1 = 0.
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
x3 = 2;x2 2x3 = 5 = x2 = 5 + 2 2 = x2 = 1;x1 + 2x2 = 2 =
x1 = 2
2 =
x1 = 0.
Hence, x1 = 0, x2 = 1, x3 = 2.
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
x3 = 2;x2 2x3 = 5 = x2 = 5 + 2 2 = x2 = 1;x1 + 2x2 = 2 =
x1 = 2
2 =
x1 = 0.
Hence, x1 = 0, x2 = 1, x3 = 2.THIS IS AN EXAMPLE OF A CONSISTENTOVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Back Substitution
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The given system has the same solution as the system
x1 + 2x2 = 2x2 2x3 = 5
x3 = 2.
and this system can be solved easily as
x3 = 2;x2 2x3 = 5 = x2 = 5 + 2 2 = x2 = 1;x1 + 2x2 = 2 =
x1 = 2
2 =
x1 = 0.
Hence, x1 = 0, x2 = 1, x3 = 2.THIS IS AN EXAMPLE OF A CONSISTENTOVER-DETERMINED SYSTEM.
Neela Nataraj Lecture 2
Example 3 (Determined, Inconsistent System)
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Solve
x2 + 3x3 = 1x1 + 2x2 x3 = 8x1 + x2 + 2x3 = 1
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Example 3 (Determined, Inconsistent System)
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Solve
x2 + 3x3 = 1x1 + 2x2 x3 = 8x1 + x2 + 2x3 = 1
Solution : The linear system of equations can be expressed as
0 1 31 2 11 1 2
x1x2
x3
=
181
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
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Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
Neela Nataraj Lecture 2
Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
interchange with 2nd equation
Neela Nataraj Lecture 2
Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
interchange with 2nd equation
R2 R1 : PARTIAL PIVOTING
Neela Nataraj Lecture 2
Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
interchange with 2nd equation
R2 R1 : PARTIAL PIVOTING
R1R2
1 2 1 ... 8
Neela Nataraj Lecture 2
Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
interchange with 2nd equation
R2 R1 : PARTIAL PIVOTING
R1R2
1 2 1 ... 80 1 3 ... 1
Neela Nataraj Lecture 2
Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
interchange with 2nd equation
R2 R1 : PARTIAL PIVOTING
R1R2
1 2 1 ... 80 1 3 ... 1
1 1 2
..
. 1
Neela Nataraj Lecture 2
Echelon Form using Row Operations
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The augmented matrix is given by
A =
0 1 3 ... 11 2 1 ... 8
1 1 2
..
. 1
no x1 in the first equation;
interchange with 2nd equation
R2 R1 : PARTIAL PIVOTING
R1R2
1 2 1 ... 80 1 3 ... 1
1 1 2
..
. 1
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Contd...
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R3R3+(1)R1
1 2 1 ... 80 -1 3
... 1
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R3R3+(1)R1
1 2 1 ... 80 -1 3
... 10 -1 3
... 7
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Contd...
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R3R3+(1)R1
1 2 1 ... 80 -1 3
... 10 -1 3
... 7
R3R3+(1)R2
1 2 1 ... 80 1 3 ... 10 0 0
... 8
Neela Nataraj Lecture 2
Contd...
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R3R3+(1)R1
1 2 1 ... 80 -1 3
... 10 -1 3
... 7
R3R3+(1)R2
1 2 1 ... 80 1 3 ... 10 0 0
... 8
The third equation implies INCONSISTENCY. Hence,NO SOLUTION .
Neela Nataraj Lecture 2
Example 4 : Consistent, determined system (Infinitelymany solutions)
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Solve :
x1 + 2x2 = 2
3x1 + 6x2 x3 = 8x
1+ 2x
2+ x
3= 0
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Example 4 : Consistent, determined system (Infinitelymany solutions)
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Solve :
x1 + 2x2 = 2
3x1 + 6x2 x3 = 8x
1+ 2x
2+ x
3= 0
Solution : The linear system of equations can be expressed as
1 2 03 6
1
1 2 1 x1x2x3 =
280
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Echelon Form using Row OperationsThe augmented matrix is given by
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Neela Nataraj Lecture 2
Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
Neela Nataraj Lecture 2
Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 8
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
Neela Nataraj Lecture 2
Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1 ... 2
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1 ... 20 0 1
... 2
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1 ... 20 0 1
... 2 Candidate for second pivot is 0
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1 ... 20 0 1
... 2 Candidate for second pivot is 0Go to next column
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1 ... 20 0 1
... 2 Candidate for second pivot is 0Go to next column
1 2 0
... 2
0 0 -1... 2
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1... 2
0 0 1... 2
Candidate for second pivot is 0Go to next column
1 2 0
... 2
0 0 -1... 2
0 0 1... 2
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Echelon Form using Row OperationsThe augmented matrix is given by
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A = 1 2 0
... 2
3 6 1 ... 81 2 1
... 0
R2R2+(3)R1
R3R3+(1)R11 2 0
... 2
0 0 1... 2
0 0 1... 2
Candidate for second pivot is 0Go to next column
1 2 0
... 2
0 0 -1... 2
0 0 1... 2
Neela Nataraj Lecture 2
Contd..
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R3R3+(1)R2
Neela Nataraj Lecture 2
Contd..
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R3R3+(1)R2
1 2 0
... 2
Neela Nataraj Lecture 2
Contd..
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R3R3+(1)R2
1 2 0
... 2
0 0 1... 2
Neela Nataraj Lecture 2
Contd..
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R3R3+(1)R2
1 2 0
... 2
0 0 1... 2
0 0 0... 0
Neela Nataraj Lecture 2
Contd..
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R3R3+(1)R2
1 2 0... 2
0 0 1... 2
0 0 0... 0
ROW-ECHELON FORM
Neela Nataraj Lecture 2
Back SubstitutionThat is,
2
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x3 = 2;
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Back SubstitutionThat is,
2
-
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x3 = 2;x1 + 2x2 = 2 = x1 = 2 2x2.
Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).
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Back SubstitutionThat is,
2
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x3 = 2;x1 + 2x2 = 2 = x1 = 2 2x2.
Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t,
Neela Nataraj Lecture 2
Back SubstitutionThat is,
x 2;
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x3 = 2;x1 + 2x2 = 2 = x1 = 2 2x2.
Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t, then x1 = 2
2t.
Neela Nataraj Lecture 2
Back SubstitutionThat is,
x3 = 2;
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x3 = 2;x1 + 2x2 = 2 = x1 = 2 2x2.
Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t, then x1 = 2
2t.
The solution vector is
Neela Nataraj Lecture 2
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Back SubstitutionThat is,
x3 = 2;
-
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x3 = 2;
x1 + 2x2 = 2 = x1 = 2 2x2.Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t, then x1 = 2
2t.
The solution vector is
2 2t
t
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Back SubstitutionThat is,
x3 = 2;
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x3 = 2;
x1 + 2x2 = 2 = x1 = 2 2x2.Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t, then x1 = 2
2t.
The solution vector is
2 2t
t
2
Neela Nataraj Lecture 2
Back SubstitutionThat is,
x3 = 2;
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x3 2;
x1 + 2x2 = 2 = x1 = 2 2x2.Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t, then x1 = 2
2t.
The solution vector is
2 2t
t
2
=
20
2
+ t
210
Neela Nataraj Lect re 2
Back SubstitutionThat is,
x3 = 2;
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x3 2;
x1 + 2x2 = 2 = x1 = 2 2x2.Here, x1 and x3 are basic variables and x2 is the free variable (nopivot).If we choose x2 = t, then x1 = 2
2t.
The solution vector is
2 2t
t
2
=
20
2
+ t
210
THIS IS AN EXAMPLE OF A CONSISTENTDETERMINED SYSTEM WITH INFINITELY MANYSOLUTIONS.
N l N t j L t 2
Example 5 : Consistent, determined system (Infinitelymany solutions)
Exercise : Solve
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2x1 + 3x2 2x3 + 4x4 = 26x1 + 9x2 + 7x3 8x4 = 3
4x1 + 6x2 x3 + 20x4 = 13
N l N t j L t 2
Example 5 : Consistent, determined system (Infinitelymany solutions)
Exercise : Solve
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2x1 + 3x2 2x3 + 4x4 = 26x1 + 9x2 + 7x3 8x4 = 3
4x1 + 6x2 x3 + 20x4 = 13
The solution vector is
Neela Nataraj Lecture 2
Example 5 : Consistent, determined system (Infinitelymany solutions)
Exercise : Solve
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2x1 + 3x2 2x3 + 4x4 = 26x1 + 9x2 + 7x3 8x4 = 3
4x1 + 6x2 x3 + 20x4 = 13
The solution vector is
4 32
t 6st
3 4ss
=
40
30+ t
3/21
00+ s
60
41
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
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2 3 2 46 9 7 84 6 1 20
x1
x2x3x4
= 2313
The augmented matrix is given by
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
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2 3 2 46 9 7 84 6 1 20
x1
x2x3x4
= 2313
The augmented matrix is given by
A =
2 3 2 4 ... 2
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
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2 3 2 46 9 7 84 6 1 20
x1x2x3x4
= 2313
The augmented matrix is given by
A =
2 3 2 4 ... 2-6 9 7 8 ... 3
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
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2 3 2 46 9 7 84 6 1 20
x1x2x3x4
= 2313
The augmented matrix is given by
A =
2 3 2 4 ... 2-6 9 7 8 ... 34 6 1 20
.
.. 13
Neela Nataraj Lecture 2
Solution :
The linear system of equations can be expressed as
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2 3 2 46 9 7 84 6 1 20
x1x2x3x4
= 2313
The augmented matrix is given by
A =
2 3 2 4 ... 2-6 9 7 8 ... 34 6 1 20
.
.. 13
Neela Nataraj Lecture 2
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Echelon Form using Row Operations
R2R2+(3)R1
-
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( )
R3R3+(2)R1
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Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4.. 2
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R3R3+(2)R12 3
2 4 . 2
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Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R12 3
2 4 . 2
0 0 1 4 ... 3
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R12 3
2 4 . 2
0 0 1 4 ... 3
0 0 3 12... 9
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
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Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Go to next column and choose the pivot
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Go to next column and choose the pivot
R3R3+(3)R2
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Go to next column and choose the pivot
R3R3+(3)R2
2 3 2 4 ... 2
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Go to next column and choose the pivot
R3R3+(3)R2
2 3 2 4 ... 2
0 0 1 4
..
. 3
Neela Nataraj Lecture 2
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Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
-
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Go to next column and choose the pivot
R3R3+(3)R2
2 3 2 4 ... 2
0 0 1 4
..
. 30 0 0 0
... 0
x4 is also a free variable
Neela Nataraj Lecture 2
Echelon Form using Row Operations
R2R2+(3)R1 2 3 2 4... 2
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R3R3+(2)R1 0 0 1 4 ... 30 0 3 12
... 9
Candidate for second pivot is 0,
x2 is a free variable
Go to next column and choose the pivot
R3R3+(3)R2
2 3 2 4 ... 2
0 0 1 4
..
. 30 0 0 0
... 0
x4 is also a free variable
Row-Echelon Form
Neela Nataraj Lecture 2
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Back Substitution
Here, x2 & x4 are free variables
-
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Neela Nataraj Lecture 2
Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
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Neela Nataraj Lecture 2
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Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
(F i bl d h b i i f i h
-
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(Free variables do not occur at the beginning of any equation whenthe system has been reduced to echelon form).If we choose x2 = t &
Neela Nataraj Lecture 2
Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
(F i bl d h b i i f i h
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(Free variables do not occur at the beginning of any equation whenthe system has been reduced to echelon form).If we choose x2 = t & x4 = s,
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Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
(F i bl d t t th b i i f ti h
-
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(Free variables do not occur at the beginning of any equation whenthe system has been reduced to echelon form).If we choose x2 = t & x4 = s, then x3 = 3 4s and
2x1 = 2
3x2 + 2x3
4x4
Neela Nataraj Lecture 2
Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
(Free ariables do not occ r at the beginning of an eq ation hen
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(Free variables do not occur at the beginning of any equation whenthe system has been reduced to echelon form).If we choose x2 = t & x4 = s, then x3 = 3 4s and
2x1 = 2
3x2 + 2x3
4x4
= 2 3t+ 2(3 4s) 4s
Neela Nataraj Lecture 2
Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
(Free variables do not occur at the beginning of any equation when
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(Free variables do not occur at the beginning of any equation whenthe system has been reduced to echelon form).If we choose x2 = t & x4 = s, then x3 = 3 4s and
2x1 = 2
3x2 + 2x3
4x4
= 2 3t+ 2(3 4s) 4s= 8 3t 12s
Neela Nataraj Lecture 2
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Back Substitution
Here, x2 & x4 are free variables and x1 & x3 are the basic variables.
(Free variables do not occur at the beginning of any equation when
-
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(Free variables do not occur at the beginning of any equation whenthe system has been reduced to echelon form).If we choose x2 = t & x4 = s, then x3 = 3 4s and
2x1 = 2
3x2 + 2x3
4x4
= 2 3t+ 2(3 4s) 4s= 8 3t 12s = x1 = 4 3
2t 6s.
Neela Nataraj Lecture 2
Solution Vector
Hence, the solution vector is
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Neela Nataraj Lecture 2
Solution Vector
Hence, the solution vector is
3
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4 32
t 6st
3 4ss
=
4030
+ t
3/2100
+ s
60
41
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Solution Vector
Hence, the solution vector is
3
-
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4 32
t 6st
3 4ss
=
4030
+ t
3/2100
+ s
60
41
THIS IS AN EXAMPLE OF A CONSISTENTDETERMINED SYSTEM WITH INFINITELY MANYSOLUTIONS.
Neela Nataraj Lecture 2
Illustration
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EXERCISE : TUTORIAL SHEET 2 - QN. 1Neela Nataraj Lecture 2
Stage 1: Forward Elimination Phase
The steps in Gaussian elimination can be summarized as follows:
1. Search the first column of [A|b] from the top to the bottom for the firstnon-zero entry, and then if necessary, the second column (the case whereall the coefficients corresponding to the first variable are zero) and then
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y, y, (all the coefficients corresponding to the first variable are zero), and thenthe third column, and so on. The entry thus found is called the currentpivot.
2. Interchange, if necessary, the row containing the current pivot with thefirst row.
3. Keeping the row containing the pivot (that is, the first row) untouched,subtract appropriate multiples of the first row from all the other rows toobtain all zeroes below the current pivot in its column.4. Repeat the preceding steps on the submatrix consisting of all thoseelements 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 m n coefficient matrix A of the linear system Ax = b is thusreduced to an (m n) row echelon matrix U and the augmentedmatrix [A|b] is reduced to[U
|c] = 0 . . . p1 . . . c1
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|
0 . . . p1 . . . c10 . . . 0 . . . p2 . . . c20 . . . 0 0 0 . . . p3 . . . c3...
......
... 0 0 . . ....
......
...
0 . . . 0 0 0 0 0 0 pr . . . cr0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...
......
......
......
......
......
0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm
.
Neela Nataraj Lecture 2
REF
The m n coefficient matrix A of the linear system Ax = b is thusreduced to an (m n) row echelon matrix U and the augmentedmatrix [A|b] is reduced to[U
|c] = 0 . . . p1 . . . c1
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|
p1 10 . . . 0 . . . p2 . . . c20 . . . 0 0 0 . . . p3 . . . c3...
......
... 0 0 . . ....
......
...
0 . . . 0 0 0 0 0 0 pr . . . cr0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...
......
......
......
......
......
0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm
.
The entries denoted by and the cis are real numbers; they may or maynot be zero.
Neela Nataraj Lecture 2
REF
The m n coefficient matrix A of the linear system Ax = b is thusreduced to an (m n) row echelon matrix U and the augmentedmatrix [A|b] is reduced to[U
|c] = 0 . . . p1 . . . c1
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|
p1 10 . . . 0 . . . p2 . . . c20 . . . 0 0 0 . . . p3 . . . c3...
......
... 0 0 . . ....
......
...
0 . . . 0 0 0 0 0 0 pr . . . cr0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...
......
......
......
......
......
0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm
.
The entries denoted by and the cis are real numbers; they may or maynot be zero. The pis denote the pivots; they are non-zero.
Neela Nataraj Lecture 2
REF
The m n coefficient matrix A of the linear system Ax = b is thusreduced to an (m n) row echelon matrix U and the augmentedmatrix [A|b] is reduced to[U
|c] = 0 . . . p1 . . . c1
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|
p0 . . . 0 . . . p2 . . . c20 . . . 0 0 0 . . . p3 . . . c3...
......
... 0 0 . . ....
......
...
0 . . . 0 0 0 0 0 0 pr . . . cr0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...
......
......
......
......
......
0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm
.
The entries denoted by and the cis are real numbers; they may or maynot be zero. The pis denote the pivots; they are non-zero. Note thatthere is exactly one pivot in each of the first r rows of U and that any
column of U has at most one pivot.
Neela Nataraj Lecture 2
REF
The m n coefficient matrix A of the linear system Ax = b is thusreduced to an (m n) row echelon matrix U and the augmentedmatrix [A|b] is reduced to[U
|c] = 0 . . . p1 . . . c1
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|
0 . . . 0 . . . p2 . . . c20 . . . 0 0 0 . . . p3 . . . c3...
......
... 0 0 . . ....
......
...
0 . . . 0 0 0 0 0 0 pr . . . cr0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...
......
......
......
......
......
0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm
.
The entries denoted by and the cis are real numbers; they may or maynot be zero. The pis denote the pivots; they are non-zero. Note thatthere is exactly one pivot in each of the first 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 ofequations)
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Neela Nataraj Lecture 2
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Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number ofequations) and cr+k = 0 for some k 1, then the (r + k)th rowcorresponds to the self-contradictory equation 0 = cr+k
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Neela Nataraj Lecture 2
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-
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Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number ofequations) and cr+k = 0 for some k 1, then the (r + k)th rowcorresponds to the self-contradictory equation 0 = cr+k and so thesystem has no solutions (inconsistent system) .
-
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y o so t o s ( y )
If (i) r = m or (ii) r < m and cr+k = 0 for all k 1, then thereexists a solution of the system (consistent system) .
(Basic & Free Variables)
If the jth column of U contains a pivot, then xj is called a basicvariable;
Neela Nataraj Lecture 2
Consistent & Inconsistent Systems
If r < m (the number of non-zero rows is less than the number ofequations) and cr+k = 0 for some k 1, then the (r + k)th row
corresponds to the self-contradictory equation 0 = cr+k and so thesystem has no solutions (inconsistent system) .
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y ( y )
If (i) r = m or (ii) r < m and cr+k = 0 for all k 1, then thereexists a solution of the system (consistent system) .
(Basic & Free Variables)
If the jth column of U contains a pivot, then xj is called a basicvariable; otherwise xj is called a free variable.
Neela Nataraj Lecture 2
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Stage 2: (Back Substitution Phase)
In the case of a consistent system, if xj is a free variable, then it
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can be set equal to a parameter sj which can assume arbitraryvalues.
Neela Nataraj Lecture 2
Stage 2: (Back Substitution Phase)
In the case of a consistent system, if xj is a free variable, then it
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can be set equal to a parameter sj which can assume arbitraryvalues.If xj is a basic variable, then we solve for xj in terms of
xj+1, . . . , xm, starting from the last basic variable and working ourway up row by row.
Neela Nataraj Lecture 2
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REF - A revisit
The forward elimination phase of the Gauss elimination methodleads to the row echelon form of a matrix which can be definedas follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the following
-
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matrix) if it has a staircase like pattern characterized by the followingproperties:
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination methodleads to the row echelon form of a matrix which can be definedas follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the following
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) p y gproperties:(a) The all-zero rows (if any) are at the bottom.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination methodleads to the row echelon form of a matrix which can be definedas follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the following
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) p y gproperties:(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 theleading entry of the preceding row.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination methodleads to the row echelon form of a matrix which can be definedas follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the following
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) p y gproperties:(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 theleading entry of the preceding row.(c) All entries in a column below a leading entry is zero.
Neela Nataraj Lecture 2
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REF - A revisit
The forward elimination phase of the Gauss elimination methodleads to the row echelon form of a matrix which can be definedas follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the following
-
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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 theleading 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 satisfies the following additional conditions,then it is in row reduced echelon form:(d) The leading entry in each nonzero row is 1.
Neela Nataraj Lecture 2
REF - A revisit
The forward elimination phase of the Gauss elimination methodleads to the row echelon form of a matrix which can be definedas follows:
(Row-Echelon Form (REF))
A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the following
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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 theleading 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 satisfies 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
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Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
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Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
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0 0 1 20
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8 is in RREF.
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0 0 1 20
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
0 0 1 20
is in RREF.
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0 0 1 20
Question : How to obtain RREF?
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
0 0 1 20
is in RREF.
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0 0 1 20
Question : How to obtain RREF?
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
0 0 1 20
is in RREF.
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0 0 1 20
Question : How to obtain RREF?Remarks :
1 Any row reduced echelon matrix is also a row-echelon matrix.
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
0 0 1 20
is in RREF.
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0 0 1 20
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 formthat one obtains from a matrix is unique.
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
0 0 1 20
is in RREF.
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0 0 1 20
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 formthat one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used asneeded to create zeros with the help of row operations.
Neela Nataraj Lecture 2
Examples
1
2 3 2 40 9 7 80 0 0 200 0 0 0
is in REF.
21 0 0 40 1 0 8
0 0 1 20
is in RREF.
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0 0 1 20
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 formthat one obtains from a matrix is unique.
3
A pivot is a nonzero number in a pivot position that is used asneeded to create zeros with the help of row operations.4 Different sequences of row operations might involve a
different 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.
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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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1.
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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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations
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echelon form by a sequence of elementary row operations.
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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations T
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echelon form by a sequence of elementary row operations. T4 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.
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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
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echelon form by a sequence of elementary row operations. T4 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. F5 If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, thenthe system is inconsistent.
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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
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y q y p4 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. F5 If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, thenthe system is inconsistent. T
6 If the reduced row echelon form of the augmented matrix of aconsistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables.
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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
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y q y p4 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. F5 If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, thenthe system is inconsistent. T
6 If the reduced row echelon form of the augmented matrix of aconsistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables. T
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. F2 If a matrix is in reduced row echelon form, then the leading
entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row
echelon form by a sequence of elementary row operations. T
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y q y p4 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. F5 If the only nonzero entry in some row of an augmented matrix
of a system of linear equations lies in the last column, thenthe system is inconsistent. T
6 If the reduced row echelon form of the augmented matrix of aconsistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables. T (number of free variables = n r).
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
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Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
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Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF,
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Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
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Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
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0 0 0 1 3
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF??
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0 0 0 1 3
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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0 0 0 1 3
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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0 0 0 1 3
3 The matrix
Neela Nataraj Lecture 2
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More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
-
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0 0 0 1 3
3 The matrix
2 1 2 1 50 1 1 3 30 2 0 0 50 0 0 3 2
is NOT
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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0 0 0 1 3
3 The matrix
2 1 2 1 50 1 1 3 30 2 0 0 50 0 0 3 2
is NOT in REF.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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0 0 0 1 3
3 The matrix
2 1 2 1 50 1 1 3 30 2 0 0 50 0 0 3 2
is NOT in REF.4
1 0 0 1/2 0 10 0 1
1/3 0 2
0 0 0 0 1 3 Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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0 0 0 1 3
3 The matrix
2 1 2 1 50 1 1 3 30 2 0 0 50 0 0 3 2
is NOT in REF.4
1 0 0 1/2 0 10 0 1
1/3 0 2
0 0 0 0 1 3 is in RREF
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.
2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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0 0 0 1 3
3 The matrix
2 1 2 1 50 1 1 3 30 2 0 0 50 0 0 3 2
is NOT in REF.4
1 0 0 1/2 0 10 0 1
1/3 0 2
0 0 0 0 1 3 is in RREF and hence in REF.
Neela Nataraj Lecture 2
More Examples - REF, RREF
Determine whether the following matrices in REF, RREF :
1
1 2 3 4 50 0 1 1 20 0 0 1 5
is an example of a matrix in REF, not
RREF.2
1 0 0 0 10 1 0 0 20 0 0 1 3
is in RREF. Is it in REF?? YES.
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3 The matrix
2 1 2 1 50 1 1 3 30 2 0 0 50 0 0 3 2
is NOT in REF.4
1 0 0 1/2 0 10 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 identitymatrix.
Neela Nataraj Lecture 2
Solution given REF
Example
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 2
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 2
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots)
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Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
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variables).
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
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a ab es)
Note that x3 is a free variable
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
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)
Note that x3 is a free variable and x1, x2, x4 are basic variables.
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
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)
Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
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)
Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3 = 1 5s
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3 = 1 5sx1 + x2 + 2x3 + 3x4 = 2
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/http://goback/ -
7/29/2019 lecture2_D3
211/219
Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3 = 1 5sx1 + x2 + 2x3 + 3x4 = 2 = x1 = 2 (1 5s) 2s 3 2
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3 = 1 5sx1 + x2 + 2x3 + 3x4 = 2 = x1 = 2 (1 5s) 2s 3 2
= x1 = 5 + 3s.
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3 = 1 5sx1 + x2 + 2x3 + 3x4 = 2 = x1 = 2 (1 5s) 2s 3 2
= x1 = 5 + 3s.We get infinitely many solutions
Neela Nataraj Lecture 2
Solution given REF
Example
Consider the REF form of the augmented matrix given by
1 1 2 3 20 1 5 0 1
0 0 0 1 20 0 0 0 0
.
In this case r = 3 (number of pivots) and n = 4 (number ofvariables).
N h i f i bl d b i i bl
http://find/ -
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Note that x3 is a free variable and x1, x2, x4 are basic variables.
Back substitution yields x4 = 2,
x2 + 5x3 = 1 = x2 = 1 5x3 = 1 5sx1 + x2 + 2x3 + 3x4 = 2 = x1 = 2 (1 5s) 2s 3 2
= x1 = 5 + 3s.We get infinitely many solutions (since x3 is a free parameter).
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we definerank(A) = number of non-zero rows in REF of A
http://find/http://goback/ -
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Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we definerank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
http://find/ -
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1. IfA = 1 0 50 1 30 0 0
,
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we definerank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
http://find/ -
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1. IfA = 1 0 50 1 30 0 0
, then rank (A) =2,
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we definerank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1 0 5
http://goforward/http://find/http://goback/ -
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1. IfA = 1 0 50 1 30 0 0
, then rank (A) =2, nullity (A)=1.
Neela Nataraj Lecture 2
RANK, NULLITY - The Two Important Invariants
For a matrix A, we definerank(A) = number of non-zero rows in REF of A
nullity(A) = number of free variables in the solution of AX = 0.
Example
1 0 5
http://find/http://goback/ -
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1. IfA = 1 0 50 1 30 0 0
, then rank (A) =2, nullity (A)=1.
Neela Nataraj Lecture 2
http://find/