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MA 106 - Linear Algebra

Neela Nataraj

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai 76neela@math.iitb.ac.in

January 10, 2013

Neela Nataraj Lecture 2

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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.

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.

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.

Neela Nataraj Lecture 2

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

Neela Nataraj Lecture 2

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

Neela Nataraj Lecture 2

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

Neela Nataraj Lecture 2

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

Neela Nataraj Lecture 2

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

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

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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

Neela Nataraj Lecture 2

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

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

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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

Neela Nataraj Lecture 2

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

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

Solution : The linear system of equations can be expressed as

1 2 03 6

1

1 2 1 x1x2x3 =

280

Neela Nataraj Lecture 2

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

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

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

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

0 0 1 ... 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 ... 81 2 1

... 0

R2R2+(3)R1

R3R3+(1)R11 2 0

... 2

0 0 1 ... 20 0 1

... 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 ... 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

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

0 0 1 ... 20 0 1

... 2 Candidate for second pivot is 0Go to next column

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

0 0 1 ... 20 0 1

... 2 Candidate for second pivot is 0Go to next column

1 2 0

... 2

0 0 -1... 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 ... 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

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).

Neela Nataraj Lecture 2

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

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

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

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

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

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,

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

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,

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 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

Neela Nataraj Lecture 2

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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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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

http://find/

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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

http://find/

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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).

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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).

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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/

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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/