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MatricesAbu Bakar

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MatrixTwo Dimensional Array (2 dimensional ordered data)

Apricots Damsons MangoesAslam 2 3 4Babar 3 0 5Chishti 1 4 2

43 0

3,

2

25

1 4A

4

3 0 51 4 2

23B

A B


Order of a Matrix

• Order=Size

1R2R

3R

4R

5R

6R

1C 2C 3C 4C 5C 6C 7C 8C

Rows

Columns

R C

R CSquare Matrix


General Form of a Matrix11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

1 2 3

1 2 3

. . . . . .

. . . . . .

. . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .. . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .. . . . . .

j n

j n

j n

i i i ij in

m m m mj mn

a a a a a

a a a a a

a a a a a

a a a a a

a a a a a

A


m nijaA

( , ) th element of the matrix ija i j A

element lying in the

intersection of & of the matrix i j

ij

R C A

a


Algebra of Matrices

,2 3

23 21

53 46

12A B

13 2 5

62

35 7

2 52 3 74 5

51

A B

42 3

3 2 112 5 3

6 11

53

3 2 1A B


342

13

2C

( 3) ( 3) ( 3)3 3

( 3) ( 3) (3 4 4

3)2 22

3

2

6

1 13

9

3

91

3

2

6

C


44

33

3 1 24 2

3 2

3

21 2

,A B

( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )2 4

( )( )3 3 3 1 3 22 3 2

2 3 24 4 4 2 4 33 4 3 2 3 3

21 4 1 1 23 1 22

AB

33 11 1022 7 7

AB


AB

,A B AB

, pn pnm m

, ( , )i jR C ji


Operations on Matrices, ,

m n m nijij b FB kA a

m n ij im n i mi j nj jbA a aB b

m n ij im n i mi j nj jbA a aB b

m n m nij ijAk k ka a


, n nj pi ijm

A B ba

1pm

mij kj

n

n ikk

ij np

AB a ba b



cost/pc in $1000

0.5 0.0.3 0.41.2 1.

6

6,A

total cost in $10 million

5.3.3 3.2 3.

13.2 12.8 13.6 15.

1 5.2 5.4 64 3.

69

.3C AB

Production figures in 10,000 units

836 2

64

93B

PC1086 PC1186

Raw Components

Labor

Miscellaneous

PC1086

Q-1 Q-2 Q-3 Q-4

PC1186

Q-1 Q-2 Q-3 Q-4

Raw Components

Labor

Miscellaneous

,A B AB

, pn pnm m www.TheStuffPoint.Com

Determinant of a Square Matrix• The Number associated with the square

matrix

1 2 12 1 2

0 21,1

2B B

1 2 12 1 12

120 2

1 23 4

2

1 23 4

2,A A


Minor of ija

11 12 13

21 22 23

31 32 33

a a aa a aa a a

ijM

Minor of ij jiM a

12 1321

32 321

3

Minor of a

aa

Maa

11 1332

21 332

2

Minor of a

aa

Maa

11 12 13

21 22 23

31 32 33

a a aa a aa a a

A

11 12 13

21 22 23

31 32 33

a a aa a aa a a

21 2213

31 213

3

Minor of a

aa

Maa


( 1)ij jij

iMA

ija Cofactor of ijA

if is even,

if is odd.

ij

ij

ij

i j

i j

M

MA

1 if is even,1 if is d

(o

1)d

i j i ji j


11 12 13

21 22 23

31 32 33

a a aa a aa a a

A

11 11 12 12 13 13 1

13 13 23 23 33 33 3

expanding by ,expanding by ,

.

.

.

.

a A a A a A Ra A a A a A C

A

11 12 13

21 22 23

31 32 33

a a aa a aa a a

A

Determinant of An www.TheStuffPoint.Com

11 12

21 22

aa

Aa

a

11 1211 22 21 12

21 22

a aa a a

a aA a

dA

a bc

a bad cb

c dA

3 25 4

A

Determinant of A2

(3)( 4) ( 5)( 2) 12 10 22A


11 12 13

21 22 23

31 32 33

a a aa a aa a a

A

11 12 1

21 22 23

31 32 33

3

a aa a a

a aA a

a

aaaa a

22 23

31

321

3

a aaa

aa a

aa2

12 13

21

3

2 23

3

1

1 2

1

33

11 13

221 232

31 32 3

12

3

a aa a

aa

a

aa 21 2

1

2

31 32

11 12

23

33

3a aaa

a a aa

a

expanding by R1

Determinant of A3

aaaa a

21 22

23

311

3

a

aaa a

21 23

32

311

3

A


3 2 44 1 12 5 2

1 1 4 1 4 14 1 1

5 2 2 2 2 52 5 2

2 5 8

( ) ( )

( ) ( )

3 2 43 2 4

3 2 2 20 23 10 22

9 20 8899

( )( ) ( ) ( )

43 2 4

A

3 2 44 1 12 5 2

A

4 12

2

2

3 41

5

1 1 4 13

4 15 2 2

2 45

)2

( ( )2

99

3 2( ) ( ) ( )2042 5 8 2 2

3 103( ) ( )2 4( 2 )2

9 20 88

expanding by R1

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3 2 44 1 12 5 2

A

3 24 1

41

22 5

4 1( ) ( )4 1 3 2 3 2

2 42

2 5 5 1

( ) ( ) (4 120 2 15 4 32 )8

( ) (22 11)1 2 )14 1(

88 11 22

99

expanding by C3

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3 2 44 1 12 5 2

A

21

3 44 1

52 2

4 1 3 4 3 42 2

( ) (2 14

52 2 1

)

( ) (8 2 6 8 3 1(1 62 ) )5

10 14( ) ( )2 1 5 3)1(

20 14 65

99

expanding by C2

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A B C

37 39 44

?

Q-1:


. . . . . (S)

.

2 3 2 373 2 2 395 1 1 44

xyz

Let x, y and z be the cost/scoop of Pineapple, Strawberry and Vanilla flavors, respectively. Then according to the given conditions

In matrix form, we can write it as:.x y z 5 1 1 44

,x y z 2 3 2 37,3 2 2 39x y z


.

2 3 2 373 2 2 395 1 1 44

xyz

, , .x

A x y bz

2 3 23 2 25 1 1

373944

Ax b


2 3 2 373 2 2 , , 39 .5 1 1 44

xA x y b

z

1

3 22 2 ,1

373944 1

A

2

373944

2 23 2 ,5 1

A

3

2 33 25 1

373944

A

2 3 23 2 25 1 1

A

ww

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,7A

2 3 2 373 2 2 , , 39 .5 1 1 44

xA x y b

z

= 2(2x1-1x2) -3(3x1-5x2) +2(3x1-5x2)

,A 1 49 = 37(2x1-1x2) -3(39x1-44x2) +2(39x1-44x2)

A2 3 23 2 25 1 1

A1

373

3 22 29

144 1

,A 2 35 A2

2 23 2

3739

5 144 = 2(39x1-44x2) -37(3x1-5x2) +2(3x1-5x39)

,3 28A A3

2 33

3739245 1 4

= 2(2x44-1x39) -3(3x44-5x39) +37(3x44-5x2)


( , , ) ( , , ). 7 5 4x y z

,7A, ,1 2 349 35 28A A A

According to Cramer’s Rule

2 3 2 373 2 2 , , 39 .5 1 1 44

xA x y b

z

.zAA

3 28 47

,yAA

2 35 57

,xAA

1 49 77


A yarn merchant sells 3 brands: B1, B2 and B3 of yarn, each of which is a blend of Pakistani, Egyptian and American cotton in ratios:1:2:1, 2:1:1 and 2:0:2.If cost/kg of B1, B1 and B3 is Rs. 40, 50 and 60 respectively,

Q-2:

find the cost/kg of cotton of each country.


1B 2B B3

PEA

1:2:1, 2:1:1, 2:0:2

E P P P PA AEA

40 50 60

Let x, y and z be the cost/kg of Pakistani, Egyptian and American Cotton respectively. Then according to the given conditions

. . . . . (S )

,x y z 1 2 1 404 4 4

,x y z 2 1 1 504 4 4

.x z 2 2 604 4


. . . . . (S,.

),x y z

x y zx z

1 2 1 1602 1 1 2001 1 120

. . . . . (

,

,

.

S )

x y z

x y z

x z

1 2 1 404 4 42 1 1 504 4 42 2 604 4

.xyz

1 2 1 1602 1 1 2001 0 1 120

In matrix form, we can write it as:

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, , .x

A x y bz

1602

1 2 12 1 1 00

11 00 1 2

Ax b

.xyz

1 2 1 1602 1 1 2001 0 1 120www.TheStuffPoint.C

om

xA x y b

z

1 2 1 1602 1 1 , , 200 .1 0 1 120

A1

16020012

2 11 1 ,

0 0 1

A2

1602001

1 12 1 ,1 20 1

A3

1 22 11 0

160200120

A1 2 12 1 11 0 1

ww

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,A 2

,A 1 120

A1 2 12 1 11 0 1

A1

16020

2 11 10

1120 0

,A 2 40 A2

1602

1 12 1001 1120

,A 3 120 A3

1 22 1

160200

01 0 12

xA x y b

z

1 2 1 1602 1 1 , , 200 .1 0 1 120

= 1(1x1-0x1) -2(2x1-1x1) +1(2x1-1x1)

= 160(1x1-0x1) -2(200x1-120x1) +1(200x1-120x1)

= 1(200x1-120x1) -160(2x1-1x1) +1(2x1-1x200)

= 1(1x120-0x200) -2(2x120-1x200) +160(2x120-1x1)


x y z ( , , ) ( , , 2060 60).

,A 2, ,A A A 1 3 3120 40 120

According to Cramer’s Rule

.zAA

3 120 602

,yAA

2 40 202

,xAA

1 120 602

xA x y b

z

1 2 1 1602 1 1 , , 200 .1 0 1 120


Solutions of System of Linear Equations

(System of m linear equations in n unknowns)

+ + +...++ + +...++ + +...+

. . . . . . (S )

.

.+ + +...+

11 12 13 1

21 22 23 2

31 32 33

1 2 3

3

1 2

1 2

2

2 33

3

1 3

1

n

n

n

m m

n

n

n

mn nm

a a a aa a a aa

x x x xx x x xx x x x

x x x

a a a

a a a a x


In matrix form, we can write it as:

+ + +...++ + +...++ + +...+

. . . . . . (S )

.

.+ + +...+

11 12 13 1

21 22 23 2

31 32 33

1 2 3

3

1 2

1 2

2

2 33

3

1 3

1

n

n

n

m m

n

n

n

mn nm

a a a aa a a aa

x x x xx x x xx x x x

x x x

a a a

a a a a x

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

1 2 3

1 2 3

. . . . . .

. . . . . .

. . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .. . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .. . . . . .

j n

j n

j n

i i i ij in

m m m mj mn

a a a a a

a a a a a

a a a a a

a a a a a

a a a a a

1 1

2 2

3 3

. . .

. .

. .

n n

x bx bx b

x b


11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

1 2 3

1 2 3

. . . . . .

. . . . . .

. . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .. . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .. . . . . .

j n

j n

j n

i i i ij in

m m m mj mn

a a a a a

a a a a a

a a a a a

a a a a a

a a a a a

A

,

Ax b(System of m linear equations in n unknowns)

1

2

3

. .

.

.

n

bbb

b

b

1

2

3

. ,

.

.

n

xxx

x

x


• Gauss Jordan Method

Ax b ???x

1x bA

Echelon Form

Reduced Echelon Form

Row Operations

• Elimination Method,• Cramer’s Rule,• Matrix Inverse Method,

• Gauss Elimination Method,


How can we purchase two things exactly in $15 such that the price of the first thing is twice of that of the other?

Elementary Row Operations

Let x be the price of the first and y be the price (both in $) of the second thing, then according to the given conditions:


. . . . .,

. (S )

215x

yy

x

.......( )

.......( )

. . . . . (S ),

.1

21

15

2 0

x y

x y

.......( )

.......( )

. . . . . (S.

), 1

22

2

15

0

x

x

y

y

12R

ijR

1 2 01 1 15

1 1 151 2 0

Elimination Method:


.......( )

.......( )

. . . . .,

. (S )1

210

2 2 30

2x y

x y

( )) ??( ?12 2

1 2 01 1 15

.......( )

.......( )

. . . . . (S ),

.1

21

15

2 0

x y

x y

1 2 02 2 30

( ) ?2 1 12R

,ikR0k

y


( )) (21 2

3 30x

10x

2 12R R

.......( )

.......( )

. . . . .,

. (S )1

210

2 2 30

2x y

x y

1 2 01 1 15

( ?) )( 2 12

,j iR kR

3 0 301 1 15

i j


ijR

,j iR kR i j

,ikR 0k

i jR R We can interchange any two (distinct) rows

We can multiply a row by a non zero scalar

Multiple of a row can be added to some other row

Produces Zeros in a column


Echelon Form/Reduced Echelon Form

Row Leader:The first non zero element of a row is said to be the leader of the row

An Approach to Solution of System of Linear Equations

Leading Zeros of a Row:The consecutive zeros lying before the row leader of a row are called the leading zeros of the row

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.Com

Echelon Form

• Each row contains more leading zeros than the preceding row• Zero row(s) is also acceptable at the end

Steps:1 row le( ader)

0



1 row le( ader)

0

0

• Row rows are also acceptable at the end

Echelon Form1 row le( ader)

0


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Converting a Matrix to Echelon/Reduced Echelon

Form

Row OperationsA A AA

is Row Equivalent to


Examples Echelon From

00 0 00 0

11

110 0

00 00 0

11

110

00 00 0 0 0 0

11

1

00 0 00 0 0 0 0

11

1

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Examples Reduced Echelon From

00 0 00 0

11

11

0 0 00 0

00 0

00 00 0

11

110

0 0 00 0

0

00 00 0 0 0 0

1010

1

0

00 0 00 0 0 0 0

01 0 0

11

Reduce Echelon Form may contain Identity Matrix

4I

3I


Q-1:

y b 10 5 2 330 6 1 3

14

0 21 1R

.

2 3 2 373 2 2 395 1 1 44

xyz

2 3 2 373 2 2 395 1 1 44

bA

by 1 23 2 2 395 1 1 4

1

4

1 0 2R R

, by 2 1 3 10 5 2 330 6 1 34

3 51 1 0 2

R R R R


y b 10 5 2 330 6 1 3

14

0 21 1R

b y 2 3

1 1 0 2

0 6 1 30 1 1 1

4R R

the y n b 3 2 20 1 1 6 10 0 7 28

111 0 2

R R R

by 310 1 17

0

1 0 211

10 4R


by 310 1 17

0

1 0 211

10 4R

1

11

1 0 210 1

40 0

which is Echelon Form of Ab

(M4): Gauss Elimination Method

E . . . . . (S ).4z

,2x y ,1y z

( ) (, , , .),7 5 4x y z Backward substitution


00 0

1 0 21 1

4

11

1bA

00 0

11

1

1 0 21 1

4bA

b y 1 20 1 10 0 4

011

1

1 3R R

,by 1 3 1 200 0 4

110 0 7

01

5 R R R R

11

1

0 00 0 5

40 0

7

(M5): Gauss Jordan Method

which is Reduced Echelon Form of Ab

( ) (, , , .),7 5 4x y z

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Date post:	13-Apr-2017
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