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MATH ZC 161Engineering Mathematics-I

Lecture-1By

Dr. Deepmala Agarwal

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Learning Objectives

27/7/2011 MATH ZC161 EngineeringMathematics I

2

To understand :Operations on matricesSpecial types of matrices

Use of matrices to study system of linear equationsRow echelon and row reduced echelon formsGaussian and Gauss-J ordan elimination as application.Rank using row reduced echelon form.

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3

2nd row

2nd column


An m x n matrix has m rows and nColumns, as shown here.

nm)(a

:wayeAlternativ

ij

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4

Row Vector

vectorrowcalledis

)(

matrix1A

21 naaaA

n

=


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5

Column Vector

vector.columncalledis

2

1

=

ma

a

a

A

matrix1A n


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6

Square Matrix

.33orderofmatrixsquareais

312

163

745

:exampleFor

=B

A matrix in which number of rowsis same as number of columns


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7

Matrix Addition and Subtraction


size)Samehave(All)(

,)(

thensize),(sameandIf

nmijij

nmijij

nmijnmij

baBA

baBA

)(bB)(aA

=

+=+

==

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8

Addition

A = a11 a12

a21 a22

B = b11 b12

b21 b22

C = a11 +b11 a12 +b12a21 +b21 a 22 +b22

If

and

then


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9

Matrix Addition Example

A + B = 3 4

5 6

+

1 2

3 4

=

4 6

8 10

= C


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scalar.acalledis

.912

15633435323

34523

:exampleFor

.)(

thennumber,complexorrealaisIf

k

kakA

)(aAk

nmij

nmij

=

=

=

=

:tionMultiplicaScalar

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12

Matrix Multiplication

Matrices A and B have these dimensions:

[r x c] and [s x d]


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13


Matrices A and B can be multiplied if:

[r x c] and [s x d]

c = s


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14


The resulting matrix will have the dimensions:

[r x c] and [s x d]

r x d


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23

Zero Matrix

0 0...... 0

0 0...... 0

0 0...... 00

. . .

. . .0 0...... 0

=


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24

Triangular Matrix

1 2 3 4 1 0 0 0

0 5 6 8 1 6 0 0

0 0 11 5 7 9 3 0

0 0 0 9 2 3 4 6

Upper TriangularMatrix

Lower TriangularMatrix


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26

Symmetric Matrix

A matrix An n is said to be symmetric if

AT

= A

Ex: A=

1 2 7

2 5 6

7 6 4


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Systems of Linear Equations

A system ofn linear equations in n variables,

can be expressed as a matrix equationAx =b:

Ifb =0, then system is homogeneous; otherwise it is

nonhomogeneous.

=

nnnnnn

n

n

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

,2,1,

,22,21,2

,12,11,1

,,22,11,

2,222,211,2

1,122,111,1

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

=+++

=+++

=+++

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Matrix in Row Echelon form :

1.1st non-zero entry of any non-zero row is 1.2.In consecutive non-zero rows, 1st entry 1 in the lowerrow appears to the right of 1st entry 1 of the upper row.3. Rows containing all 0s are at the bottom of the matrix.

Matrix in Reduced Row Echelon form :

This is matrix which is in row echelon form with further

property that if a column contains a 1st

entry 1 of anyrow then all other entries of that column are 0.

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Elimination methods :

We have two ways :

1. Gaussian elimination : Use elementary row operations toconvert the augmented matrix to a row echelon form.Solve the resulting system corresponding to row echelonform using back substitution.These are solutions are the solutions of the original system.

2. Gauss-Jordan Elimination : Use elementary row operations toconvert the augmented matrix to the reduced row echelon form.Advantage :No back substitution required to solve this system.

27/7/2011 34MATH ZC161 Engineering Math. I

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Remark : In Gaussian elimination, the sequence ofrow operations required may not be unique, also

the result, a row echelon form, may not be unique.

In Gauss-J ordan elimination, the sequence of row

operations may not be unique, bur reduced rowechelon form is unique.


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.113217532

3111

issystemtheofmatrixaugmentedThe.1132

7532

3exist.notdoessolutionshow the

orsystemthesolvetoneliminatioJordan-Gauss

orneliminatioGaussianeitherUse:

321

321

321

=+

=++

=

xxx

xxx

xxx

366)Ex.5(p


lli i iG i

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form.echelonrowaisThis.

1100

8410

3111

272700

8410

3111

13750

8410

3111

13750

84103111

8410

137503111

11321

13750

3111

11321

7532

3111

form.echelonrow

amatrix toaugmentedisconvert thtooperations

rowelementaryuseWe:

27

5

2

3

232

2313

12

R

RRR

RRR

RR

neliminatioGaussian


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also.systemoriginalsolnisThis.0giveseqn

1stinvariables2theseofvaluesthengSubstituti

.4getweeqn,2ndinthisngSubstituti.

givesequationlaston,substitutibackbysolveTo

.1

84

3

isformechelon

rowthistoingcorrespondsystemresultingThe

1

23

3

32

321

=

=

=

==

x

xx

x

xx

xxx


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form.echelonrow

reducedgettoneliminatioGaussianinobtained

formechelonreduced-rowtheonoperationsrow

elementaryuseWe:neliminatioJordan-Gauss

soln.(same)get thedirectlyweandrequirednotis

onsubstitutibackthatseeweneliminatioGaussianinas

systemtheFormingform.echelonrowreducedisThis

.

1100

4010

0001

1100

8410

0001

1100

8410

5501

1100

8410

3111

3231

21

45

R

++

+

RRRR

R


J dGli i tiG iU366)9(E

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.

4111-

311-1

81-1-1

issystemtheofmatrixaugmentedThe

4

3

8

exist.notdoessolutionthe

showorequationslinearofsystemthesolvetoneliminatio

Jordan-GaussorneliminatioGaussianUse:

321

321

321

=++

=+

=

xxx

xxx

xxx

366)9(pEx


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form.echelonrowisThis10005/2-100

81-1-1

12000

5/2-10081-1-1

12000

5-20081-1-1

4111-

5-200

81-1-1

4111-

311-1

81-1-1

12

2

3

2

23

12

+

R

RRR

RR

Gaussian elimination : We use elementary row operations toconvert this augmented matrix to row echelon form.


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.102/5

8

isformechelonrow

thistoingcorrespondequationslinearofsystemThe

3

321

==

=

x

xxx

The system has no solution, as the last equation can never besatisfied.

Thus the original system is also inconsistent. (has no solution)

Try Gauss-J ordan elimination yourself.27/7/2011 42MATH ZC161 Engineering Math. I

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Example

.3,2

7,

2

5

solutiontrivialnonahas

020342

unknownsin threeequationstwoofsystemhomgeneousThe

321

321

321

===

=+

=+

xxx

xxx

xxx


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Rank using echelon form

The definition of the rank of a matrix is more involved andunsuitable for our purpose. For all practical reasons we canuse its characterization involving Reduced row echelon form.

Steps : To find rank of an m x n matrix AReduce A to the reduced row echelon formusing a sequence ofelementary row operations

Count the number ofnon-zero rows in it to get the rank of A


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Exercise.Try the following problems based on lectures 1On p 349: 38, 39, 40. On p 359: 8, 10 ,17 ,19. On p 364: 3, 4,6, 7, 8.

27/7/2011 52MATH ZC161 Engineering Math I

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