Unit-VII: Linear Algebra-I - BookSpar · Unit-VII: Linear Algebra-I ... row matrix from the Table 1...

Unit-VII: Linear Algebra-I

� Purpose of lession :

� To show what are the matrices , why they are useful, how they are classified as various types

and how they are solved.

� Introduction: Matrices is a powerful tool of modern Mathematics and its study is becoming

important day by day due to its wide applications in almost every branch of science and

especially in Physics and Engineering. Matrices are used by Sociologists in the study of

dominance within a group, by Demographers in the study of births and deaths, etc, by

Economist in the study of inter-industry economists, input-output tables and for various

practical business purposes, by statisticians in the study of ‘design of experiments and for

various practical business purposes, by statisticians in the study of ‘design of experiments and

multivariate analysis by Engineers in the study of ‘network analysis’ which is used in electrical

and communication engineering.

Application:

The following table shows the number of transistors and resistors purchased by a manufacturer

from suppliers A and B for the first week of January

A B

Transistor

Resistors

400

600

800

500

a) Write the data in the table as 2 x 2 matrix S1

b) b)Use scalar multiplication to find a matrix S2 whose entries are all 10 % large than the

corresponding entries of S1

c) Suppose that S2 is the supply matrix for the second week of January. Find S1 + S2 and explain

what its entries represent.

Solution: a) The supply table can be written as the following 2 x 2 matrix

400 800

1 600 500S

=

www.bookspar.com | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS | FORUMS

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b) If the entries of S2 are 10 % large than the entries of S1, then

S2 = S1 + 0.10 S1 = 1.1 S1

S2=1.1

c) The entries of S1 + S2 give the total number of transistors and resistors purchased from

suppliers A and B for the first two weeks of January.

Example : The following table 1 Shows the number of economy, mid-size and large cars rented by

individuals and corporations at a rental agency in a single day Table 1

Economy mid-size large

Individuals 3 2 6

Corporations 5 2 4

Table 2

BBoonnuuss ppooiinnttss FFrreeee mmiilleess

EEccoonnoommyy

MMiidd--ssiizzee

LLaarrggee

2200

3300

4400

5500

110000

115500

Table 2 shows the number of bonus points and free miles given in a promotional program for each of

the three car types. Find

a) Total bonus points for individuals

b) Total free miles for individuals

c) Total bonus points for corporations

d) Total free miles for corporations

440 880400 800

660 550600 500

=

1 2

440 880 840 1680400 800

660 550 1260 1050600 500s s

+ = + =

[ ]3 2 6



The 1 x 3

row matrix from the Table 1 –

by Individuals

• The 3 x 1 column matrix

From table 2 represents the bonus points given for each economy, mid

rented.

The product of these two matrices is a 1 x 1 matrix whose entry is the sum of the products of the

corresponding entries

The product of above two matrices, gives the total number of bonus points given to individuals on the

rental of the 11 cars.

Traffic control in the future:

The above picture shows the intersections of four one way streets. 400 cars per hour want to enter

intersection P from the north on First Avenue while 300 cars per hour want to

intersection Q on Elm Street. The letters w, x, y, and z represent the number of cars per hour passing

the four points between these four intersections, as shown in picture.

A) Find values for w, x, y and z that would realize this desired

[3 2 6 30 360

Eco Mid Lar

3 2 6

5 2 4

Bonus Pts Free Miles

X

– Represents the number of economy, medium size and large rented

The 3 x 1 column matrix

From table 2 represents the bonus points given for each economy, mid-size and large car that is




intersection P from the north on First Avenue while 300 cars per hour want to


the four points between these four intersections, as shown in picture.

A) Find values for w, x, y and z that would realize this desired traffic flow.

20

30

40

] [ ]20

3 2 6 30 360

40

=

Bonus Pts Free Miles

Eco

Mid

Lar

20 50360 1250

30 100320 1050

40 150

X

=

Represents the number of economy, medium size and large rented

size and large car that is




intersection P from the north on First Avenue while 300 cars per hour want to head east from


360 1250

320 1050



B) If construction on Oak Street limits z to 300 cars per hour, then how many cars per hour would have

to pass w, x, and y.

Solution: The solution to the problem is based on the fact that the number of cars entering an

intersection per hour must equal the number leaving that intersection per hour, if the traffic is to keep

flowing

Since 900 cars ( 400 + 500) enter intersection P, 900 must leave, x + w = 900. Writing a similar equation

for each intersection yields the system of equation

W + x = 900

W + y = 1000

X + z = 850

y + z = 950

w + x = 900 w = 50 + z

x + z = 850 x = 850 - z

y + z = 950 y = 950 – z

The system is a dependent system does not have a unique solution.(z is non negative integer)

b)If z is limited to 300 because of construction, then the solution is ( 350,550,650,300) To keep traffic

flowing when z = 300, the system must route 350 cars past w, 550 past x and 650

9501100

8501010

10000101

9000011

−≈

00000

1000110

8501010

9000011



hour must equal the number leaving that intersection per hour, if the traffic is to keep


for each intersection yields the system of equations.

R4=R4 – R2 R23

nt system does not have a unique solution.(z is non negative integer)


flowing when z = 300, the system must route 350 cars past w, 550 past x and 650 past y.

≈

00000

9501100

8501010

9000011

−

−≈

1000110

8501010

1000110

9000011

−≈

000

010

110

011



hour must equal the number leaving that intersection per hour, if the traffic is to keep


R23

nt system does not have a unique solution.(z is non negative integer)


past y.

000

85010

10001

90000



Definition Matrices and Types of Matrices:

Twenty six female students and twenty-four male students responded to a survey on income in a college

algebra class. Among the female students , 5 classified themselves as low-income, 10 as middle-income,

and 11 as high-income. Among the male students, 9 were low-income, 2 were middle-income, and 13

were high-income. Each student is classified in two ways, according to gender and income. This

information can be written in a matrix:

L M H

Female

Male

Square matrix: In this matrix we can see the class of makeup according to gender and income. A matrix

provides a convenient way to organize a two way classification of data.

A matrix is a rectangular array of real numbers. The rows of a matrix run horizontally, and the

columns run vertically.

A matrix with m rows and n columns has size m x n (read “ m by n”). The number of rows is

always given first. For example, the matrix used to classify the students is a 2 x 3 matrix

When m = n i. e the number of rows is the same as the number of columns the array is called a square

matrix. The others are called rectangular matrix.

Row-matrix: In a matrix if there is only one row it is called a row matrix

Column matrix: In a matrix if there is only one column it is called a column matrix

Null matrix : In a matrix, if all the elements of matrix are zero, it is called null matrix and is denoted by o

Equal Matrix: Two matrices are said to be equal if

i) They are of the same type i.e they have same number of rows and columns

ii) The elements in the corresponding positions of the two matrices are equal.

Location of an element: To locate any particular element of a matrix the elements are usually denoted

by a letter followed by two suffixes which respectively specify the row and the column in which it

appears. Thus the element occurring in pth

row and qth

column will be written as apq.

Note: The matrices are generally denoted by capital letters.

Diagonal Elements of a Matrix: An element aij of a square matrix is said to be a diagonal element if i = j.

1329

11105



Scalar Multiple of a matrix: If k is a number and A is a matrix, then kA is defined as the matrix each

element of which is k times the corresponding element of A.

Triangular Matrix: if every element above (or below ) the diagonal is zero, the matrix is called a

triangular matrix

Example:

Transposed Matrix: The matrix of order n x m obtained by interchanging the rows and columns of a

matrix A of order m x n is called the transpose matrix of A or transpose of the matrix and is denoted by

A’ or At.

Representation of Points: A point is represented in two dimensions by its coordinates. These two values

are specified as the elements of a 1 x 2 matrix i.e.

In three-dimensions a 1 x 3 matrix

Row matrix or column matrix like are frequently called position vectors.

Elementary Transformations:

An elementary transformation, which is also known as E-transformation, is an operation of any one of

the following types.

i) Interchange of two rows (columns)

ii) The multiplication of the elements of rows (columns) by a non-zero number.

iii) The addition to the elements of a row (column), the corresponding elements of a row ( column)

multiplied by any number

An elementary transformation is said to be row-transformation or column transformation according

as it applies to rows or columns.

ngular upper tria

600

530

421

ngularlower tria

654

032

001

[ ]12y

xor ,

xyx

[ ]

13z

y

x

or used is

x

zyx

[ ]

y

xor yx



Symbols to be used for elementary transformation:

i) Rij for the interchange of ith

and jth

rows

ii) Ri(k) for the multiplication of ith

row by k ≠ 0

iii) Rij(k) for the addition to the ith

row, the products of the jth

row by k. Similarly we use the symbols

Cij, Ci(k), Cij(k) for the corresponding column operations.

Equivalent Matrices:

Two matrices A and B are said to be equivalent if one can be obtained from the other by a

sequence of elementary transformations. Two equivalent matrices have the same order. The

symbol ~ is used for equivalence.

Minor : Let A be a matrix square or rectangular, from it delete all rows leaving a certain t-rows

and all columns leaving a certain t-column. Now if t > 1 then the elements that are left,

constitute a square matrix of order t and the determinant of this matrix is called a minor of A of

order t

Rank of a Matrix:

Definition: A matrix is said to be rank r when

I ) It has at least one non-zero minor of order r and

ii) Every minor of order higher than r vanishes.

Briefly, the rank of a matrix is the largest order of any non-vanishing minor of the matrix

For example:1) , ρ(A) = 1

2) , ρ(A) = 1

Note : i) If a matrix has a non-zero minor or order r, its rank is ≥ r

ii) If all minors of a matrix of order of r+1 are zero, its rank is ≤ r

Example:

=642

321A

=

12963

8642

4321

A

3(A) 02A

220

432

321

=≠=

= ρA

3)( 0

100

010

001

0000

1000

0100

0010

=≠=

AA ρ



Example:. Find the rank of the matrix

Example:. Reduce the matrix to a unit matrix by only elementary

row transformations

Solution:

Example:. Find the rank of the matrix

2)(

0000

0000

1310

0211

12

0000

0000

0211

1310

244

;2

R-3

R3

R

0211

0211

0211

1310

operate

0211

0633

0211

1310

1

2R3

R3

R;1

R2

R2

R

0211

2013

1101

1310

=−−

−

≈

−

−−

≈−==

−

−

−

−−

≈

−

−

−

−−

≈+=+=

−

−−

=

A

RRRR

ρ

13165

452

562

=A

3(A)

100

010

001

~2

411

100

010

041

~

32233

11100

110

341

~2

233

120

110

341

~

12

33562

110

341

~ 13

R

341

110

562

~

12

33:

122R

13165

452

562

=−+=

−=−=−−−=

−−

−−

−=−−−−

−=−==

ρRRR

RRRRRRRRR

RRR

RRRRRA

−=−=−=

==

00

01~

00

000

001

~13

00

000

100

~

13

33;

12

2200

200

100

~3

222

c

60

240

120

~

2

c-3

c3

c

960

640

320

00

33

Ic

RRRRRRcc

A



Normal form of a Matrix:

By means of a sequence of elementary transformations every matrix A of order m x n can be reduced to

one of the following form

Example: Reduce the matrix A to its normal form

Solution:

Consistency of a system of linear equations:

Consider the following system of m linear equations in n-unknowns x1, x2, x3,…..xn

a11 x1 + a12 x2 + a13 x3 + …….a1n xn = b1

a21x1 + a22 x2 + a23 x3 + …….a2n xn = b2

…………………………………………

am1x1 + am2 x2 + am3 x3 + …….amn xn = bn The system of equations can be written as

00

0rI

−

=

1312

6204

2210

A

2)(

0000

0010

0001

~23

0000

0100

0001

33

44;

32

220000

3120

0001

~233

3120

3120

0001

~

2

2

3120

6240

0001

~1

244

;1

233

3120

6240

2201

~

1331321

6240

2201

~12

1312

6204

2210

~ =

−=−=−=

+=−=

−

−=

−−

=

Ac

ccccccRRR

Rcccccc

RRRcA

ρ



Conditions for Consistency:

� Here the system of equations AX = B is called the non homogeneous set of equations and the

system of equations A X = 0 is called the homogeneous set of equations.

� A system of linear equations is said to be consistent if it possesses a solution and inconsistent (

not consistent) if it does not possess a solution

� A homogeneous system always possesses at least one solution (namely the trivial solution ), a

homogeneous system is always consistent. But a non–homogeneous system may or may not be

consistent.

� The equation A X = B is consistent i.e. has a solution (unique or infinite ) iff the matrices A and D

are of the same rank i.e. ρ(A) = ρ(D).

� i) If ρ(A) = ρ(D)= n, the number of unknowns, then only the given system of equations will have a

unique solution

� ii) If ρ(A) = ρ(D) < n, then the given system will have infinite solutions.

� iii) If ρ(A) ≠ ρ(D) then the given system of equations will be inconsistent or will not have any

solution.

Example: Test the consistency of the following equations

4x – 2y + 6z = 8

x + y – 3z = -1

15x – 3y + 9z = 21

Solution:

..21

..........33

..3231

22..

1221

11..

1211

..3

2

1

B

nx

..3

x2

x1

x

X

21

33231

21221

1,,..

1211

....

..........

....

,,..

=

===

nbmnam

am

a

bn

aaa

bn

aaa

bn

aaa

D

nb

b

b

b

mnam

am

a

naaa

naaa

naaa

A

33

3,

12R

219315

1311

8624

:

21

1-

8

B ,

z

y

x

X ,

9315

311

624

RRDAD

A

=

−

−−

−

==

==

−

−

−

=



If ρ(A) = ρ(D)= 2< number of unknowns, the system is consistent and has an infinite number of solutions

Example: Show that the following equations are inconsistent

2x + y + 5z = 4

3x – 2y+2z = 2

5x – 8y -4z =1

Soln.: Ax = B i.e.

D=[A/B] =

−−⇒−=−−

−−

−=−

−−

−=

−

−

−−

−=−=

−

−

−−

0000

2310

1001

2R

1R

1R

0000

2310

1311

2

R6

12

R

0000

121860

1311

2

R3

R3

R

121860

121860

1311

1R5

3R

3R ;

1R4

2R

2R

7315

8624

1311

=

−−

−

1

2

4

485

223

512

z

y

x

233;

1211485

2223

4512

RRRRRR −=−=

−−

−

−−−

−=−=

−−−

−

−−−

3000

81170

2331

12

33;

13

221662

2223

2331

RRRRRR



ρ(A) = 2, ρ(D)=3 ρ(A) ≠ ρ(D) Hence the given equations are inconsistent

Example: Show that the system of equations

2x –y +z = 4

3x –y +z = 6

4x – y +2z =7

-x + y -z = 9 is consistent.

Solution:

ρ(A) = 3, ρ(D)=4 ρ(A) ≠ ρ(D) The system is inconsistent

Example : Investigate the values of λ and µ so that the equations

2x + 3y + 5z = 9

7x + 3y – 2z = 8

2x + 3y + λ z = µ

have i) Unique solution ii) no solution iii) an infinite number of solutions

+=

+=

+=

−−

−

−

−

==

==

−−

−

−

−

=

4R4

3R

3R

4R3

2R

2R

4R2

1R

1R

9111

7214

6113

4112

:

9

7

6

4

B ,y

x

X

111

214

113

112

BAD

z

A

14

11010

23100

1010

13001

24

1010

23100

11000

13001

14

13001

23100

11000

1010

4R)1(

4R

3R

1R

1R

13001

23100

11000

22110

1R

4R

4R

1R3

3R

3R

1R2

2R

2R

9111

43230

33220

22110

RRR

−

−

−⇒

−

−

−⇒

−

−

−

−=

+=

−−

−

−

−

−=

−=

−=

−−

−

−

−



Solution:

(1)

i) If equation (1) has a unique solution then the determinant of the coefficient matrix A≠0

i.e the given system of equations will have a unique solution if λ≠ 5 whatever may the value of µ.

If λ = 5 and µ≠ 9, then ρ(A) = 2, ρ(D) = 3 ρ(A) ≠ ρ(D) Hence the given system of equations has no

solution

c) If in the augmented D, λ = 5 and µ= 9 then

It is clear that ρ(A) = ρ(D) = 2 (< 3 the number of unknowns ) Therefore their exist infinite number of

solutions to the given set of equations

=−

µλ

8

9

32

237

532

z

y

x

5 .0A -515

32

237

532

≠≠∴=−=

λλ

λ

A

−=−==

9-5-00

82-37

9532

133R

32

8237

9532

/

µλ

µλ

RRBAD

−=

0000

8237

9582

D



Basically, there are TWO types of methods available, viz, Direct methods and Indirect methods or

Iterative methods.

Direct Method:

i) Cramer’s rule

ii) Gaussian Elimination

iii) Gauss-Jordan method

iv) Choleskey Method

Indirect Method:

i) Jacobi Method

ii) Gauss seidal method

iii) Successive Over-Relaxation Technique

Gauss Elimination method:

Gauss Elimination method applied to three linear equations. First we explain this method applied to a

particular systems of order three given by

a11 x1 + a12 x2 + a13 x3 = b1

a21x1 + a22 x2 + a23 x3 = b2 (1)

a31x1 + a32 x2 + a33 x3 = b3

The augmented matrix of the above system

To eliminate x1 from the second equation multiplying the first equation by and add it to

the second equation.

Similarly to eliminate x1 from the third equation multiply the first equation by and add it to the

third equation . Thus we get the matrix.

~

3333231

2232221

1131211

baaa

baaa

baaa

11

21a

a−

11

31a

a−

−−−−

−−−−

111

31313

11

313312

11

313211

11

3131

111

21213

11

212312

11

212211

11

2121

1131211

ba

aba

a

aaa

a

aaa

a

aa

ba

aba

a

aaa

a

aaa

a

aa

baaa

′′′

′′′

333320

223220

1131211

baa

baa

baaa



Where , are called the multipliers for the first stage of elimination. The first

equation is called the pivotal equation and is called the first pivot is the new pivot

and the multiplier is

i.e. multiply the second row by and add it to the third row.

~

~

The values of x1, x2 and x3 can be obtained by the substitution

Remark: This method fails if one of the pivots say , or

Vanishes. We can modify by rearranging the rows so that the pivot is non-zero.

Example: Solve by Gauss elimination Method

11

21a

a−

11

31a

a−

11a

≠0

11a

22a′

22

32

a

a

′

′−

22

32

a

a

′

′−

′′

′−′′

′

′−′′

′

′−′

′′′

222

32323

22

323322

22

3232

0

223220

1131211

ba

aba

a

aaa

a

aa

baa

baaa

′′′′

′′′

33300

223220

1131211

ba

baa

baaa

33

33

3333

2323222

1313212111

equation xlast theFrom

a

b

bxa

bxaxa

bxaxaxa

′′

′′=

′′=′′

′=′+′

=++

11a

22a′

33 a ′′

1694

18323

102

=++

=++

=++

zyx

zyx

zyx



Solution: The augmented matrix is

Eliminating x from second and third equation

The augmented matrix is

~

~

Eliminating y from the third equation R3=R3-7R2

z = 5, y = -9, x = 7

112

17

2

70

32

3

2

10

10112

12

133

R;12

322

R

16941

18323

10112

RRRR −=−=

102

1161

2

191

2

142

2

11

102

3181

2

331

2

322

2

33

10112

−−−−

−−−−

10200

32

3

2

10

10112

−−

102

32

32

1

102

=−

=+

=++

z

zy

zyx



Gauss-Jordan Method:

Instead of eliminating x2 only in the third equation, we could have eliminated it from the first

equation also, so that at the end of the second stage, the augmented matrix becomes.

Gauss-Jordan Method: consider

a11 x1 + a12 x2 + a13 x3 = b1

a21x1 + a22 x2 + a23 x3 = b2 (1)

a31x1 + a32 x2 + a33 x3 = b3

The augmented matrix of the above system

(2)

This modification of Gaussian elimination is called the Gauss-Jordan method

Example: Solve by Gauss-Jordan method

Solution: The augmented matrix is

1694

18323

102

=++

=++

=++

zyx

zyx

zyx

3333231

2232221

1131211

baaa

baaa

baaa

′′

′−′′

′

′−′′

′

′−′

′′′

222

32323

22

323322

22

3232

0

223220

1131211

ba

aba

a

aaa

a

aa

baa

baaa

′′

′′′

′′′

33300

223220

1130

11

ba

baa

baa

12

133

R;12

322

R

16941

18323

10112

RRRR −=−=



Eliminating x from second and third equation

2x -2z = 4

(½) y +(3/2) z = 3

-2z = -10 z = 5, y = -9, x = 7

Example :Solve the following system of linear equations

x + 2y + z = 8

2x + 3y + 4z = 20

4x + 3y + 2z = 16 Using Gauss-Jordan method.

Solution: Let the given system of equations be written in the matrix form AX= B where

The augmented matrix is

12

13

R3

R;12

32

R2

R i.e

102

1161

2

191

2

142

2

11

102

3181

2

331

2

322

2

33

10112

RR −=−=

−−−−

−−−−

27

33R;

22

11R

112

17

2

70

32

3

2

10

10112

RRRR −=−=

10200

32

3

2

10

4202

−−

−

===

16

20

8

B and

z

y

x

X,

234

432

121

A

16

20

8

234

432

121

:

=BA

16

20

8

234

432

121

:

=BA 1

43

R3

R 1

2R-2

R2

R R−==



We get the following system of equations X = 1, y = 2 and 12z = 36

Then required solution is x =1 , y = 2, z =3

−−

−

16-

4

8

250

210

1212

53

R3

R R−=

36-

4

8

1200

210

121

−

− 36/1

2312/1

1R RRR ++

36-

2-

5

1200

010

021

−

−2

21

R1

R R+=

36-

2-

1

1200

010

001

−

−



Unit-VIII: Linear Algebra-II

Linear Transformations: Definition

A transformation ( or function or mapping):

A transformation T from Rn to R

m is a rule that assigns to each vector x in R

The set Rn is called the domain of T, and R

that the domain of T in Rn and the co

For x in Rn, the vector T(x) in R

m is called the image of x (under the action of T).

The set of all images T(x) is called the range of T.

For example: AX = b and Au = o

Say that multiplication by A transforms X into b and Similarly multiplication by A transforms u into the o

(zero) vector.

Matrix Transformation : T:Rn R

matrix, For simplicity, we sometimes denote such a matrix transformation by x Ax.

Example: Let

And define a transformation T:R2

Find T(u), the image of u under the transformation T.

=−

8

5

1

1

1

1

1502

3134

−=

−

−

= ,1

2u,

71

53

31

A

Linear Transformations: Definition:

nsformation ( or function or mapping):

is a rule that assigns to each vector x in Rn a vector T(x) in R

is called the domain of T, and Rm

is called the co-domain of T. The notation T:R

and the co-domain is Rm

.

is called the image of x (under the action of T).

The set of all images T(x) is called the range of T.

and


Rm

for each x in Rn, T(x) is computed as A(x), where A is an m x n


R3 by T(x) = Ax so that T(x)= A(x)=

the transformation T.

=−

0

0

3

1-

4

1

1502

3134

−

=

5

2

3

b ,

−

−

71

53

31

a vector T(x) in Rm

.

domain of T. The notation T:Rn R

m indicates


, T(x) is computed as A(x), where A is an m x n


by T(x) = Ax so that T(x)= A(x)=

+−

+

−

=

27

1

25

13

23

1

2

1

xx

xx

xx

x

x



Definition: A transformation ( or mapping ) T is linear if

i) T(u + v) = T(u) + T(v) for all u, in the domain of T

ii) T(cu)=c T(u) for all u and all scalars c.

Characteristic Equation of a Matrix :

If A = (aij) is any square matrix of order n and λ be any variable, then the matrix [ A- λI] is called the

characteristic matrix of A. The determinant of this matrix i.e. Is called the characteristic function.

The equation =0 is the characteristic equation of A and of degree n. The roots of the equation

= 0 are called the characteristics or Eigen values of A.

Example: Determine the characteristic roots of

Solution: The characteristic equation is =0

iE

By synthetic division method

1) -1 7 -11 5

-1 6 -5

-1 6 5 0

-λ2+6 λ-5 = 0 (-λ+1)(λ-5)=0

λ1= 1, λ2 =1 λ3 = 5 are the require roots or Eigen values

Example: Find the eigen values and the eigenvector of the matrix

Solution: The characteristic equation of A is = 0

i.e expanding we get

The roots of this equation are, λ = 6 λ = 1 which are the eigen values of A.

To find the eigen vectors, consider the matrix equation is

−

=−

−

−

==

9

1

5

1

2

71

53

31

)( AuuT

0511273 =+−+− λλλ

=

221

131

122

A

IA λ−

0

221

131

122

=

−

−

−

λ

λ

λ

21

45=A

021

45=

−

−

λ

λ

IA λ−

0672 =+− λλ



This gives 2 homogeneous linear equations

(5-λ) x1 +4x2 = 0

x1+(2-λ)x2 =0

i) For λ = 6 above equations becomes

-x1 + 4x2 = 0

x1- 4x2 =0 Both represent the same equation x1-4x2 =0

giving the eigenvector

ii) Again for λ = 1,

4x1 + 4x2 = 0

x1 + x2 = 0 Both represent the same x1 + x2 = 0

giving the eigenvector

Example: Find the eigen value and the eigen vectors of the matrix

Solution:

The characteristic equation of the matrix A is

12

11

−=∴

xx

=−

−

==−

0

0

2x

1x

21

45.

2x

1x

X where0

λ

λ

λ

ei

XIA

12

41

xx=∴

=1

4

1X

−=

1

1

2X

−

−−

−

342

476

268

−

−−

−

=

342

476

268

ALet

0=− IA λ



i.e

Expanding, we get λ3-18λ2

+45λ=0 : λ3-18λ2

+45λ=0 λ(λ2-18λ+45) = 0

i.e λ = 0 or (λ2-18λ+45) = 0

i.e λ = 0 or (λ -3) (λ -15)=0

i.e λ1 = 0 , λ2 = 3, λ3 = 15 These are the eigen values of A

To find the eigen vector consider the matrix equation is

i.e

This gives the three homogeneous equations

(8-λ)x1 - 6x2+ 2x3 = 0

-6x1 + (7-λ)x2-4x3 = 0 (1)

2x1 - 4x2 + (3-λ)x3 = 0

i) λ1 = 0 equation (1) become

8x1 - 6x2+ 2x3 = 0

-6x1 + 7x2-4x3 = 0 (2)

2x1 - 4x2 + 3x3 = 0

Solving any two non-identical equations of (2) for x1, x2, x3 by the rule of cross multiplication. We get

0

342

476

268

=

−−

−−−

−−

λ

λ

λ

==−

3

2

1

.0

x

x

x

whereXXIA λ

=

−−

−−−

−−

0

0

0

x

x

x

342

476

268

3

2

1

λ

λ

λ

36563

12322

14241

−=

+−

−=

−

xxx

203

202

101

xxx=

−=

23

22

11

xxx=

−=



Therefore the eigenvector corresponding to λ1 = 0 is

ii)λ2 = 3 equation (1) become

5x1 - 6x2+ 2x3 = 0

-6x1 + 4x2-4x3 = 0 (3)

2x1 - 4x2 + 0x3 = 0

Solving by the rule of cross multiplication We get

Therefore the eigenvector corresponding to λ2= 3 is

iii) λ3 = 15 equation (1) become

-7x1 - 6x2+ 2x3 = 0

-6x1 - 8x2-4x3 = 0 (3)

2x1 - 4x2 - 12x3 = 0

Solving by the rule of cross multiplication We get

Therefore the eigenvector corresponding to λ3= 15 is

� Orthogonal Matrix:

A square matrix A is said to be orthogonal if AA’=I=A’A

� Singular matrix and Nonsingular matrix:

A square matrix A is said to be singular if its determinant is zero and A is said to be nonsingular if its

determinant is not equal to zero.

� Inverse of a square matrix:

If A and B are two square matrices of the same order such that AB-BA = I then B is called the inverse of A

denoted by A-1

.

� Similarity of Matrices:

Two square matrices A and B of the same order are said to be similar if there exists a non singular

matrix P such that B= P-1

AP

Here B is said to be similar to A.

=

2

2

1

1x

13

22

21

xxx=

−=

23

12

21

−==

xxx

−

=

2

1

2

2x

−=

1

2

2

3x



Example:

Reduce the matrix to the diagonal form .

Solution: The characteristic equation of A is

Are the eigen values of A.

Consider

Casei) let λ=1, we get -2x+3y=0 or 2x=3y or x/3 = y/2

X1 = (3,2)’ is the eigen vector corresponding to λ=1

Caseii) let λ=2,

we get -3x +3y = 0 or x = y or x/1 = y/1

X2 = (1,1)’ is the eigen vector corresponding to λ=2.

=− 0XIA λ

0)y-(42x- .

0

0

42

31 .

=+

=−−

−−

λλ

λ

ei

y

xei

−

−=

42

31A

0=− IA λ

2 10232042

31 . andei =⇒=+−⇒=

−−

−−

λλλλ

λ

==

==

AdjPp

andwe

XXP

11-p 1P have

12

13

21

matrix diagonal theis 20

011-p

20

01

22

23

32

11

12

13

42

31

32

111-p

32

111

==

=−

−=

−

−

−

−=

−

−=−

DAPThus

AP

P



Reduce the following matrices into diagonal matrices

1500

030

000

:Ans.

342

476

268

A

200

010

000

:Ans.

6410

527

7411

−

−−

−

=

−−

−−

−−

=A



Unit-VII: Linear Algebra-I - BookSpar · Unit-VII: Linear Algebra-I ... row matrix from the Table 1...

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Transcript of Unit-VII: Linear Algebra-I - BookSpar · Unit-VII: Linear Algebra-I ... row matrix from the Table 1...