Study Material for B.tech

8/13/2019 Study Material for B.tech

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Karnataka State Open University

Study Material for B.Tech

Mathematics - Code - BTC 11

by

K.S. SrinivasaRetd. Principal &

Professor of Mathematics

Bangalore

Published by

Sharada Vikas Trust (R)

Bangalore


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BTC 11 MATHEMATICS

Syllabus

1. Complex Trigonometry

Revision of Plane Trigonometry - trigonometric ratios, expressions for relation between allied angles and trigonometricalratios. Addition formulae for trigonometrical ratios and simple problems. Complex numbers and functions, definition,

properties, De Moivre's Theorem (without proof), Roots of a complex number, expansions of sin ( n ), cos ( n ) inpowers of sin & cos , addition formulae for any number of angles, simple problems.

2. Matrix Theory :

Review of the fundamentals. Solution of linear equations by Cramers' Rule and by Matrix method, Eigen values and Eigen vectors, Cayley Hamilton's Theorem, Diagonalization of matrices, simple problems.

3. Algebraic Structures

Definition of a group, properties of groups, sub groups, permutation groups, simple problems, scalars & vectors, algebra

of vectors, scalar & vector products, scalar triple product, simple problems.

4. Differential Calculus

Limits, continuity and differentiability (definition only), s tandard derivatives, rules for differentiation, derivatives of

function of a function and parametric functions, problems. Successive differentiation,nderivative of standard functions,t hstatement of Leibnitz's Theorem, problems, polar forms, angle between the radius vector and the tangent to a polar curve, (no derivation) angle between curves, pedal equation, simple problems, indeterminate forms, L' Hospital's rule,

partial derivatives, definition and simple problems.

5. Integral Calculus

Introduction, standard integrals, integration by substitution and by parts, integration of rational, irrational and trigonometric

functions, definite integrals, properties (no proof), simple problems, reduction formulae and simple problems.

6. Differential Equations of first order

Introduction, solution by separation of variables, homogeneous equations, reducible to homogeneous linear equation,

Bernoulli's equation, exact differential equations and simple problems.

Text Books

1. Elementary Engineering Mathematics by Dr. B.S. Grewal, Khanna Publications

2. Higher Engineering Mathematics by B.S. Grewal, Khanna Publications

Reference Books

1. Differential Calculusby Shanti Narayan, Publishers S. Chand & Co.

2. Integral Calculusby Shanti Narayan, Publishers S. Chand & Co.

3. Modern Abstract Algebra by Shanti Narayan, Publishers S. Chand & Co.


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CONTENTS

Page Nos.

1. Complex Trigonometry 01

2. Matrix Theory 27

3. Algebraic Structures 47

4. Differential Calculus 69

5. Integral Calculus 101

6. Differential Equations 123


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COMPLEX TRIGONOMETRY

Trigonometric ratios of acute angles

AConsider a right-angled triangle ABC, right angled at C. Let The side ACABC= .opposite to angle is called opposite side. The side BCis called adjacent side. ThesideABis hypotenuse.

opposite side ACThe ratio is defined as sine of & written as sin .ie

hypotenuse AB

adjacent side BCThe ratio is defined as cosine of & written as cos .ie

hypotenuse AB

opposite side AC 90The ratio is defined as tangent of & written as tan .ieBCadjacent side BC

sin tanIt can be seen by the above definition that =

cos

1 ABThe reciprocal of sin ie is defined as cosecant of and written as cosec .iesin AC

1 ABThe reciprocal of cos ie is defined as secant of and written as sec .ie

cos BC

BCThe Reciprocal of tan ie is defined as cotangent of and written as cot .

AC

Identitie s:- (1) s in + cos = 12 2

(2) sec = 1 + tan2 2

(3) cosec = 1 + cot2 2

Proof:- From the right angled triangle ABC

AC+ BC= AB2 2 2

AC BC2 2ie sin + cos = 12divide byAB ,we have + = 1 2 2

2 2AB AB

2 2AC ABie tan + 1 = secBC , + =2divide by we have 1 2 2

BC BC2 2

2 2BC ABie 1 + cot = cosecAC , + =2divide by then 1 2 2

AC AC2 2

Trigonometric ratios of 30 & 60

Consider an equilateral triangle of side 2 units AB = BC = AC = 2. , thenBD = DC = 1. Then in theDraw AD toBCr

triangleABD, , furtherAD= ABBD= 4 1 = 3, AD= 3ADB= 9 0 , ABD= 60 & BAD=30 2 2 2


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2 KSOU Complex Trigonometry

In the triangleABD A

AD 3sin 60 = = 30take ABD

AB 2

BD 1cos 60= =

AB 22 2

3AD 3tan60= =

BD 1

2 19060also cosec 60= , sec 60 = 2 & cot 60 =

3 3BDC 11

BD 1take 2BAD sin30 = =

AB

AD 3cos30= =

AB 2

BD 1tan 30= =

AD 3

2also cosec 30= 2,se c 30 = & cot 30 = 3

3From the above results, it can seen that sin60 = cos30, cos60 = sin30 and tan60 = cot30.

Trigonometric ratios of 45

AConsider a right-angled isosceles triangle ABCwhere ACB= 90

ABC=45 =BAC

Let AC = BC = 1 unit then unitsAB 2=2

1AC 1\ sin 45 = =

AB 2

BC 1cos 45= = B C1AB 2

AC 1tan 45= = = 1

BC 1

also, cosec45 = 2 , sec 45 = 2 and cot 45 = 1

Note :- Trigonometric ratios of 30, 45 and 60 are called Standard Trigonometric ratios which are always useful,hence these values have to be always remembered.

Trigonometric ratios of any angle (from 0 to 360 )

LetXOX'& YOY'be co-ordinate axes where Ois the origin. Consider a circle of radius rwith centreO. LetPbe any pointon the cirlce whose co-ordinates are ( x, ).

DrawPMperpendicular to OX.


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BTC11

ThenOM = x& PM = y Y

Let MOP=

xsin = , cos = , tan =

r r x (, +) (+, +) P(x, )r r x

also cosec = , sec = ,cot = xx X' XO M

When ,MOPsatisfy 0 < < 90 the printPwill be in first quadrant(, ) (+, )of the circle, when it satisfy 90 < < 180, Pwill in second quadrant,

when 180 < < 270, Pwill be in third quadrant & finally when 270< < 360 the point Pwill be in fourth quadrant, because of thesepositions, signs of the Trigonometric ratios changes.

Y'In the I quadrant bothx& are +ve andris always +ve.

Therefore sin , cos & tan are +ve, their reciprocals are also +ve.

In the II quadrantxis ve, is +ve

Therefore sin & cosec are +ve cos , tan , sec & cot are ve.

In the III quadrant bothx& are ve

Therefore, tan & cot are +ve and sin , cos , cosec & sec are ve.

In the IV quadrantxis +ve & is ve

Therefore, cos & sec are +ve and sin , tan , cosec & cot are ve.

Note :- The signs of the trigonometric ratios can be easily remembered with the help of the following diagram

Sine is +ve All are +veSA

in short

TC

tan is +ve cos is +ve

PTrigonometric ratios of angles 0 , 90 , 180, 270and 360 (can be called border angles).

xO M

Let be an angle whose measure is very close to zero (as in fig. 1)MOP

As 0, 0, x r

0\ si n 0 = = 0

rfig.1


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Pr

cos 0= =1r

0 tan 0= =0

r

xO MIf ,MOPis very close to 90 as in fig. 2.

As 90, x 0, r

r\ sin 90= =1

r fig.20

cos 90 = = 0r

rtan 90 = = 8

P0

If is very close to 180 as in fig. 3 xM OAs 180, x -r, 0

0\ sin 180 = =0

r

- r fig.3cos 180 = =- 1r

0tan 180 = =0

r

If is very close to 270 as in fig. 4M

As 270 , 0x , -r x O

- r\ si n 270 = =- 1 r

r

0cos 270= = 0r P fig.4

- rtan 270 = =-8

0

If is very close to 360 as in fig. 5

When 360 , x r, 0x M

0\ sin 360 = = 0

rr

Prcos 360= = 1

r

0tan360 = = 0

fig.5r


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BTC11

Thus we have the following

s in 0 = 0, cos0=1, tan 0 = 0sin 90=1, cos 9 0 =0, tan 90 = 8sin 180 = 0, cos 1 80 =- 1, tan 180 =0sin 270 =- 1, cos 270 =0, tan 270 =-8

sin 360 = 0, cos360 = 1, tan 360 = 0

Note :- From the above derivations it can be seen that the values of sin & cos always be between 1 & 1. Whereas thevalue of tan will be between & and hence the graphs of the trigonometric funct ions-8 8

are as follows.= s inx, =cosx& = t anx

=sin x

O x 90 270180 360

fig.6

=cos x

O x 90 270 180 360

fig.7

= tanx

O x 90 270180 360

fig.8


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Rules for allied angles

sin( 90 - ) =cos sin( 270 - ) = - cos

cos( 90 - ) =sin cos( 270 - ) =- sin

tan(90 - ) = cot tan(270 - ) = cot

sin( 90 + ) =cos sin( 270 + ) = - cos

cos( 90 + ) =- sin cos( 270 + ) =sin

tan(90 + ) = - cot tan(270 + ) = - cot

sin(180- ) = sin sin( 360 - ) =- sin

cos(180 - ) =- cos cos( 360 - ) = cos

tan(180- ) = - tan tan(360 - ) = - tan

sin(180+ ) = - sin sin(360 + ) =sin

cos(180 + ) =- cos cos( 360 + ) = cos

tan(180+ ) = tan tan(360 + ) = tan

Using the Trigonometric ratios of standard angles 30, 45 & 60 and using the above rules for allied angles, we can

find the trigonometric ratios of other angles as follows.

3Eg.1 2sin 120 = sin(90 +30) = cos30 =

3or 2sin120= sin(180 - 60) = sin 60 =

3Eg.2 2cos150 = cos(90 + 60) = - sin 60 = -

3or 2cos 150= cos(180- 30 ) = - cos 30= -

1Eg.3 si n315 = si n( 360 - 45 ) =- sin 45 = -

2

1or si n315 = si n( 270+ 45 ) =- cos 45 =-2

B

Radian Measure r rConsider a circle of radius rwith centre O. LetABbe a arc such that arc AB = r.

1 radianAO rMeasure of the is cal led a 'radian' denoted as 1 .AOB C

To prove that radian is a Constant angle

Consider a circle of radiusrwith centre O. Let arcAB = rso that is 1 radian.AOBLetCbe a point on the circle such that AOC= 90 .

Since the angle subtended at the centre of a circle is proportional to the corresponding arc.


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MCA 11 - Mathematics SVT 7

CAOB arcAB=

BarcACAOC

1 radian r 2 r= = r190

(2 r)4

AO r1 1= Q AC= Circumfere nce ( 2r)

4 4

radians =180 ie = 180 imp. result.

=60 , = 30 .=90 , =45 ,

2 4 3 6

Using the relation = 180 measurement of an angle in degrees can be converted to radians and vice-versa.

Eg. (1) Qn. : Convert 40 to radians

2Solution : 40= 40 = radius

180 9

2Eg. (2) Qn. : Convert radians in to degrees

3

2 180Solution : = 120

3

Some Problems

1. Show that

2 2 2(1+cotA) + (1 - cotA) = 2 cosec A

2 2Sol ution : LHS = 1+ cot 2A+ 2cot A+1+ cot A- cot A2= 2 cot+2 A

= 2(1+cot A)2

2= 2cosec A= RHS.

2. Show that1+ sin 4A 1- sin A

- = sec AtanA1- si n A 1+ si nA

(1+sinA) - (1- sinA)2 2(taking LCM)Solution: LHS=

(1- sinA)(1+sinA)

2 2(1 +sin A+2 sin A) - (1 + sin A- 2 sin A)=

21 - sin A

2 21+ si n A+2sinA- 1 - sin A+2sin A=

2cos A

4 sinA 1 sin A= =4

cosA cosAcos A2

= 4 s ec A tanA= RHS


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2a - 13. If sec A+ tan A = a, then prove that = si nA

a + 12

2 2 2a - 1 sec A+tan A+2sec Atan A- 1Solution : LHS = =

a +1 sec A+tan A+2sec Atan A+ 12 2 2

2 2

sec A- 1+tan A+ 2 sec Atan A=sec A+ tan A+1+ 2 sec Atan A2 2

22tan A+2sec AtanA= (using the identity)

2sec A+2 sec Atan A2

2tanA( tanA+ sec A)=

2sec A(sec A+tanA)

tan A sin A cos A= = =sin A= RHS

sec A cos A 1

4 sin +cos4. If find the value of sin 5 = ,

tan - cot

4then opposite side is 4 hypotenusis 5.Solution : If sin =

53 4

cos = & tan =adjacent side = 25 - 16 = 9=35 3

4 3 7+

sin +cos 125 5 5= = =

4 3 7tan - cot 5-

3 4 12

3 5 cos + 8tan5. If tan =- , < < find

4 2 8sec - 3cosec

Solution : Since lies in II Quadrant sine is +ve cosine & tangant are ve.

3 4sin =

and cos =

-

5 5

- 4 - 35 + 8

5cos +8 tan 5 4=

8sec - 3cosec - 5 58 - 3

4 3

- 4- 6 10 2= = =

- 10 - 5 15 3

6. Prove that

tan(180 sec+ )sec(180 - ) cosec (90 + )= 2

sec(360 - ) cot( 90 + ) sin( 90- )


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tan (- sec ) sec= = 2Solution sec: LHS

sec ( - tan ) cos

7. Prove that

-sin( + ) cos( 2 - ) cot2 =sin 3+ +

tan cot sin( - )2 2

(- sin ) cos tanSol ution : LHS =

( - cot ( - tan )( - sin )

= cos tan

sin= cos = sin = RHS

cos

Addition Formulae

sin(A+B) = sin AcosB+cosAsin B

cos( A+B) =cos Acos B- sinAsinBtanA+tanB

tan(A+B) =1- tanAtanB

replacingBby B

sin(A- B) = sin AcosB- cos Asin B

cos( A- B) =cos Acos B+sinAsinB

tanA- tanBtan(A- B) =

1+tanAtanB

Using the above formulae we can find the trigonometric ratios of 15, 75, 105 etc.

s in 75=sin(45 +30)

= + sin 45 sincos 30 cos 45 30

1 3 1 1 3 +1= + =2

2 22 2 2

cos 75 =cos ( 45 + 30)

=cos 45 sin cos30 - sin45 30

1 3 1 1 3- 1= - =2

2 22 2 2

sin75 3+1tan75 = =

cos 75 3- 1

sin 1 5 =sin( 45- 30 )

=sin 45 sincos 30 - cos 45 30


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1 3 1 1 3- 1= - =2 2

22 2 2

cos15 = c os( 45- 30 )

= cos 45 sincos30 +sin 45 30

1 3 1 1 3+1= + =2 2

22 2 2

sin15 3 - 1tan15= =

cos15 3 +1

sin 105 = si n( 60+ 45 )

= sin 60 sincos 45 + c os 60 45

3 1 1 1 3+1= + =2 2

2 2 2 2

cos105 =cos( 60 + 45)

= cos 60 sinsi n 4 5 - sin 6 0 45

1 1 3 1 1 - 3= - =2 2

2 22 2

3 +1tan105 =

1- 3

Alternate method

sin 15 cos = si n( 9 0 - 75 ) = 75

3 - 1=

2 2

cos15 sin= cos( 90- 75 ) = 75

3 +1=

2 2

sin 105 sin= sin(180 - 75 ) = 75

3 +1=

2 2

cos105 cos =cos(180 - 75 ) = - 75 ()

- 3 - 1 1 - 3= =

2 2 2 2


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To find sin2 , cos2 & tan2 .

sin(A+B) = sin AcosB+cosAsin Bput A= =B

s in 2 = sin cos + cos sin= 2 sin cos

cos( A+B) =cos Acos B- sinAsinB

put A=B=

cos 2 = cos cos - sin sin2 2= cos - sin

using cossin = 1-2 2

cos 2 =2 cos - 12

2 2also sinusing cos = 1-

cos 2 sin=1 - 2 2

tanA+tanBtan(A+B) =

1- tanAtanB

put A= =B

tan +tantan 2 =

1 - tan tan

2 tan=

1 - tan2

To find sin3 , cos3 & tan3 .

s in 3 = sin( 2 + )

=sin 22 cos +cos sin

2= 2 sin cos cos + (1- 2 sin )sin

2 3= 2 sinsin (1- sin ) + sin - 2

3 3=2 sinsin - 2 sin +sin - 2

3=3 sinsin - 4

cos 3 =cos( 2 + )

=cos 22 cos - sin sin

= (2cos - 1) cos - 2sin cos2 2

3 2= 2cos - cos - 2(1 - cos )cos

3 3=2 coscos - cos - 2 cos +2

3=4cos 3 - cos


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tan 3 =tan(2 + )

tan 2 + tan=

1- tan 2 tan

2 tan+ tan

1- tan2

= 2tan1 - tan21 - tan

22tan + tan (1- tan )21- tan=

2 21- tan - 2tan2(1- tan )

tan +tan - tan32=

21 - 3 tan

tan - tan33=

21- 3 tan

Thus we have,

sin 22 = sin cos

2 2 2 2cos2 sin= cos - = 2 sincos - 1 =1- 2

2tantan 2 =

21- tan

3sin 3 sin= 3sin - 4

3c os 33 =4cos - cos

33tan - tantan 3 =

21- 3tan

Problems

sin3A cos 3A1. Show that - =2sin A cosA

3 33sin 3A- 4sin A 4 cos A- cos ASol ution : LHS = -

sin A cos A

2 2=3 - 4 sin A- 4 cos A+3

) = 6- 4(sin A+cos A2 2

=6 - 4 =2=RHS

22. If find sin2 A& cos3 A.cosA=

3

2 9- 4 5Solution : Given cosA= , sin A= =

3 3 3


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5 2 4 5sin 22A= sinAcos A =2 =

3 3 93cos 33A= 4 c os A- cos A

8 2 32 32 - 54 - 22= 4 - 3 = - 2 = =

27 3 27 27 27

4 5 - 22& cos 3A=sin2A=

9 27

3. Prove that

1+ sin 2= tan ( 45 + )2

1- sin 2

1+ 2sin cosSol ution : LHS =

1- 2sin cos

2(cos +sin )=

(cos - sin )2

2cos +sin

= (dividing Nr & Dr inside the bracket bycos )cos - sin

21+ tan

=1- ta n

= tan(45+ ) =RHS

1- cos4. Prove that and hence prove that 32 tan15 = 2-= tan

1+cos 2

- 21- 1 2sin1- cos 2

using cos2AformulaSolution : =1+cos - 21 + 2 cos 1

2

2 21- 1+ 2 sin 2sin2 2

= = =tan2

21 + 2cos - 1 2 cos2 2

2 2

1- cos2= tan

1+cos 2

31-

1 - cos30 2 - 32put =30 . tan 15 = = =2

1 + cos30 3 2 + 31+

2

()()()() 2-

()()2

- 3 2 - 3 2 3 2= = = 2- 3

4 - 32 + 3 2 - 3

tan = 2- 32


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Complex Number

Definition :A number of the form is defined as a Complex Number and usually denotedx+iywhere x R, R& i= - 1asZ. xis called Real part & is called Imaginary part.x iy is called Conjugate. Complex number denoted as .ZAorzcomplex number can be represented by a point on a plane by taking real part onx-axis & imaginary part on -axis. The planeon which complex numbers are represented is called aComplex Plane. For every point in a plane there is a complex number

& for every complex number there is a point in the plane. In a complexx-axis is called Real axis & -axis Imaginary axis.Properties

(1) Equality.Two complex numbers are said to be equal if z =x +iy , z =x +iy x =x , =1 1 1 2 2 2 1 2 1 2

(2) Addition. If z =x +iy , z =x +iy1 1 1 2 2 2

z +z = (x +x ) +i( + )1 2 1 2 1 2

(3) Subtraction. z - z =(x - x ) +i( - )1 2 1 2 1 2

(4) Multiplication. zz = (x+iy )(x +iy )1 2 1 1 2 2

2=xx + ix + ix +i1 2 2 1 1 2 1 2

2=xx - + i(x +x ) Qi = - 11 2 1 2 1 2 2 1

Note :i= - 1,i = - 1, i =-i,i =1 etc2 3 4

z (x +iy )=1 1 1(5) Division.

z (x +iy )2 2 2

multiply numerator & denominator by the conjugate ofx + iy ie x iy, then2 2 2 2

z (x +iy )(x - iy )=1 1 1 2 2

z (x +iy )(x - iy )2 2 2 2 2

xx + +i(x - x )= 1 2 1 2 2 1 1 2

x +2 22 2

xx + x - x= +i1 2 1 2 2 1 1 2

2 2 2 2x + x +2 2 2 2

which is a complex number.

Note :- Product of a complex number with its conjugate is always a positive real number.

2 2iez z= (x+iy)(x-iy) =x + & z+z= 2x, z- z= 2iy

z+ z z- zx= & =

2 2i

Polar form the complex number

z = x + iy is called the Cartesian form.P(x, )

LetP(x, ) be any point in the plane which represents a complex number. r yrDraw PM tox-axis & joinPM. Then OM = x,MP = y. Let &MOP =

xOP = r. O x MFrom the triangleOPM,


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OM xcos = = x=rcos

OP

PMsin = = =rsin

OP rz= x+iy=rcos +irsin =r(cos +isin )

This form of the complex number is called 'Polar form'. Where ris called Modulus & is called argument which aregiven by

2 2 - 1r= x + & =tan

xris always positive and argument varies from 0 to 360. The value of argument satisfying < = is defined

as amplitude which is unique for a complex number.

Thus we have

2 2Mod. z=z =r= x +

and amp. z = where < = -1arg z=tanx

xwhi le finding the ampli tude of acomplex number, w e have t o find satisfying ie cos = andsin =

x

Examples

2 + 3i1. Express in the form x + iy

1 +i

22+ 3i- 2i- 3i2 + 3i ( 2 +3i)(1- i)=Solution : =

1 +11 +i (1 +i)(1 - i)

2+i+ 3 5+i 5 1= = = +i

2 2 2 2

(1 +i)22. Express in the formx + iy

3- i

2 2(1 + i) 1+i + 2i 2i( 13+i)Solution : = 2= (using i = - )

3-i (3 -i) (3- i)(3 +i)

26i+2i - 2 +6 i - 2 6 i 1 3i= = = + =- +

9 + 1 10 10 10 5 5

3. Find the modulus and amplitude of 1 + i.

Solution : z=r= x + = 1 + 1= 22 2

x 1 1cos = = and sin = = =45 =

r 2 r 42

Modulus is 2 and amplitude is

4


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4. Find the modulus & amplitude of - 3 + i

Solution : z= 3 +1 = 4=2

- 3cos =

5 2= 61

sin =2

5Modulus is 2 & ampli tude is

6

5. Find the modulus & amplitude of 3 - 1-i

Solution : z= 1 + 3= 4=2

1cos = -

2 2 = -

33sin = -

2

2Modulus is 2 & ampli tude is-

3

6. Find the modulus & amplitude of 1 i.

Solution : z= 1 +1= 2

x 1cos = =

r 2=-

1 4sin = =-

r 2

Modulus is 2& amplitude is -

4

1a a2 2 2 2If +i = ,then prove that ( + )(a +b ) = 17.

a+ib

1 (a- ib)aSolution : +i = =

a+ib (a+ib)(a- ib)

a- ib a b= = - i

2 2 2 2 2 2a +b a +b a +b

Equating real and imaginary parts separately

a - ba = and =

a +b a +b2 2 2 2


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Squaring and adding

2 2 2 2a b (a +b ) 1a + = ()() + = () =

2 2

a +b2 2 2 2 22 2 2 2 2 2a +b a +b a +b

cross - multiplyin g

2 2 2 2(a + )(a +b ) = 1

DeMoivre's Theorem

Statement : Ifnis a +ve or ve integer, then (cos + isin )n = cos n +isin n

If n is a +ve or ve fraction, one of the values of is(cos + isin )n cosn +isinn .

Proof : Case (i) when nis a +ve integer proof by Mathematical Induction.

Whenn= 1,1(cos +isin ) =cos +isin =cos 11. +isin .

result is true for n= 1.

Let us assume that the result is true for n = mmie ( cos +i

sin ) = cosm

+ i

sinm

multiply both sides by cos +isin

(cos +isin ) (cos + isin )m =(cos m + isinm )(cos +isin )

+ie ( cos +isin )m1

=cosm cos +isinm cos +icos m sin - sinm sin=cosm cos - sinm sin +i[sinm cos +cos m sin ]

=cos( 1m+ 1) +isin(m+ )

the result is true for n = m +1. Thus if the result is true for n = m then it is true for n = m+ 1.

ie If it is true for one integer it is true for next integer, hence by Induction the result is true for all +ve integers.

Case (ii) When where & qare +ve integers.n=q

q+ Consider cos isinq q

= (cos + isin )p= cos q +isin q =cosq +isin q q

q+

pie (cos + isin ) = cos isinq q

takingqth roots on both sides. One of the values of (cos +isin ) = cos +isinqq q

Case (iii) whennis ve integer or ve fraction. Let n = mwhere m is a +ve integer or +ve fraction

(cos +isin ) = (cos + isin )n -m


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1 1= =

cos m +isinmm(cos +isin )

(cos m - isin m ) cos m - isin m= =

(cos m +isinm )(cos m - isinm ) 2 2cos m +sin m

cosm - isinm= =cos( -m) +isin(- m) = cos n +isin n .1

Important Results

1(i) If then =cos - isinx=cos +isin

x1 1

x+ = 2cos and x- = 2isin .x x

(ii) If x=cos +isin

x = (cos + isin )n n = cos n +isin n

1 1&= cos - isin = (cos - isin )n

x xn= cos n - isin n

1 1&n nx + =2cos n x - =2isinn

x xn n

Note :- For convenience can be written ascos +isin cis .

Roots of a complex number

Letz = x + iy , express the complex number in the polar form.ie z=r(cos +isin )

[] = rcos( 2k + ) +isin( 2k + ) where k I

[] 1 1 1z =r cos( 2 )k + ) +isin( 2k +n n n

2k + ( 2k )+= r cos +isin1n

n n

wherek =0, 1, 2, ......... n 1. Let us denote the nth roots of the complex number by z,z, z, .............. z0 1 2 n 1Then,

+k= 0, z =r cos isin =r cisfor 1 1n n0 n n n

2 + 2 2 ++k= 1, z =r cos +isin =r cis1 1n n

1 n n n

4 + 4 4 ++k= 2, z =r cos +isin =r cis1 1n n

2 n n n

2(n- 1) + 2(n- 1) 2(n- 1) ++k=n- 1, z =r cos +isin =r cis1 1n n

n-1 n n nthe above nvalues gives nth roots ofz = x + iy


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Note :- If k = n,n + 1,n+ 2 etc. The values will repeat. Hence these will be only nvalues of which are distinct.1znUsing the polar form of the complex number we can plot the nth roots of the complex in the following way.

z2

Draw a circle of radius whose centre is O. Mark a point on the circle and taker1nz

C 1OAas intial line. Take a point Bsuch that . ThenBrepresentz. Take aAOB=0n

z2 +pointCsuch that then Crepresent z, like this all the nth roots can 0AOC = Bn 1be represented. This diagram is called 'Argand Diagram'. /nO A

1rn

Problems :

z2 - 3cos5 - isin 5 ) (cos 7 +isin 7 )(1) Simplify n

9 5(cos 4 - isin 4 ) (cos +isin )

Solution : G.E. (given expression)

(cos +isin ) (cos +isin )- 1 0 - 2 1=

-3 6 5(cos +isin ) (cos +isin )

- - + -=(cos + isin ) = (cos + isin ) = 110 21 36 5 0

- 5 -3(cos 2 +isin 2 ) (cos 3 - isin 3 )(2) Simplify

4 10(cos 4 -is in 4 ) (cos +isin )

(cos +isin ) (cos +isin )-10 9Solution : G.E. =

- 16 10(cos +isin ) (cos +isin )

- + + -= (cos +isin ) = (cos +isin )10 916 10 5= cos 5 sin+i 5

1 1(3) If 2 cos = x+ & 2cos = +

xShow that

m n1 xm m &(i) x + = 2cos( m +n ) (ii) + =2 cos(m - n )m m n mx x

21 x + 1Solution : 2 cos =x+ =

x x

x + 1= 2cos x0 ie x - 2 cos x+1 =2 2

2cos (- 4)(1- cos )22 cos 4 cos - 422=x=

2

22 cos - 4 sin 2 cos 2isin= = = cos isin

2 2

1If x=cos +isin ,then = cos - isin

x1 1Similarly i 2cos = + , = cos +isin & = cos - isin


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(i) = (cos + sin )m mx i =cosm +isinm

= (cos + isin ) n n=cos n +isinn

x = (cosm + isinm )(cosn +isinn )m n

(1)=cos( m +n ) +isin(m +n )

1 = 2cos(m +n ) - isin(m +n ) (2)m nxadding (1) & (2)

1x + = 2 cos(m +n )m n

xm n

(ii) x =cosm +isin mm

1 = cos n =isin nn

xm n& = cos(m - n ) +isin(m - n ) =cos(m - n ) - isin(m - n )

n mx

addingm n

x+ = 2c os(m - n )

n mx

(4) Prove that

m b2 2 - 1(a+ib) +(a- ib) = 2 tan(a +b ) cosm m mn n 2n

n a

b2 2 -1Solution : Leta+ib=r(cos +isin ) wher e r= a +b & = t an

a

m m+ (1)( + )m = m (cos + sin )ma ib r i =r cos isinmn n n nn n

a- ib= r(cos - isin )

m m-(a- ib) =r cos isinm m (2)n nn n

adding (1) & (2)

m( 2a+ib) + (a- ib) =r cosm m mn n n

n

mm b+ 2n

(a+ib)m +(a- ib)m = a b cos tan2 2 - 12n nn a

b bQ r= a +b & tan = = tan2 2 -1

a a

() m bm2 2 - 1(a+ib) +(a- ib) =2 a +b 2cos tanm m 2nn nn a


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(5) Find the cube roots if 1 + iand represent them on Argand diagram.

Solution ): Letz= 1+i= r(cos +isin say

rcos =1,rsin =1

Squaring& addingr (cos + sin ) =1+12 2 2

r =2 r= 221

cos =2

=1 4

sin =2

+z= 2 cos isin

4 4

+ += 2 cos 2k +isin 2k for k I

4 4

+ z= 1+i= 2 cis 2k4

()1

8k+ 3 8k +1z = (1 + i) = 2 cis1 1 3 1= 2 cis fork=0,1, 23 3 6

4 12

1 1

when k= 0, z = 2 cis =2 cis15 6 60 12

9 31 1 1k=1, z =2 cis =2 cis = 2 cis 135 6 6 6

112 4

17 171 1 1k=2, z = 2 cis k=2, z =2 cis = 2 cis255 6 6 6

2 212 12

To represent them on Argand diagram.

Draw a circle of radius with centre O. LetOAbe the initial line.1 C26

Take a pointBon the circle such that ,AOB= 15 take a pointCon the circleB135such that & take a point Don the circle such that thenAOC = 135 AOD=255

the points B,C,Drepresentz,z,z. A15255 O0 1 2 126

1

- 1 3 4i(6) Find all the values of & find their continued product.

2 2

1 3Solution: Letz= -i =r(cos +isin ) D

2 2

1 3rcos = , rsin = -

2 2Squaring & adding

1 3 4r = + = =12

4 4 4


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1cos =

2 = -

3 3sin = -

2

- -z=1 cos +isin

3 3

- z= cis 2k for k I3

1-

1 1

1 3 - - 6k4 4 4z = i = cis 2k = cis1 4

2 2 3 3

6k -= cis , for k= 0, 1,2 3

12

5-for k=0, z = cis k=1, z = cis

0 112 12

11 17k= 2, z = cis k=3, z =cis2 312 12

Their continued product

5 11 17-= ci s cis cis cis

12 12 12 12

+ 5 11 17 32 8= cis - + + = cis =cis

12 12 12 12 12 3

1 3+ 2 2 2 2= - + i= cis 2 = cis =cos +isin

2 23 3 3 3

Expansion of sin ( n ) and cos ( n ) in powers of sin & cos

nConsider (cos +isin ) = cos n +isinnnExpand (cosn + isinn )

Using Binomial Theorem

( + ) = + + + ........ +n n n- 1 n- 2 2 nx a x nCx a nCx a nCa1 2 n

Equating real and imaginary parts separately we get the expressions of cosn & sinn

Eg. (1) Express in powers of & .cos5 sin+i 5 sin cos5Solution ): cos 5 + isin 5 = (cos +isin

= cos + C cos isin + C cos (isin ) + C cos (isin ) + C cos ( isin ) + C(isin )5 4 3 2 2 3 4 55 5 5 5 51 2 3 4 5

5 4 3 2 2 3 4 5= cos + 5icos sin - 10 cos sin - 10 icos sin + 5 cos sin +isin

= cos - 10 cos sin + 5 cos sin + i[5cos sin - 10 cos sin + sin ]5 3 2 4 4 2 3 5


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Equating real & imaginary parts separately

and5 3 2 4cos 5 sin=cos - 10 cos sin + 5 cos4 2 3 5sin 5 sin= 5 cos sin - 10 cos sin +

Eq. (2) Express cos6 & sin 6 in powers of cos +isin

cos +isin = (cos + isin 6Solution ): 6 6= cos + 6C cos (isin ) + 6C cos (isin ) + 6C cos (isin ) + 6C cos (isin )6 5 4 2 3 3 2 4

1 2 3 4+ 6 )C cos (isin ) + 6C (isin5 6

5 6

=cos + icos sin - cos sin - icos sin + cos sin + icos sin - sin6 5 4 2 3 3 2 4 5 66 15 20 15 66 4 2 2 4 6 5 3 3 5= cos - 15 cos sin + 15 cos sin - sin +i[6cos sin - 20 cos sin + 6 cos sin ]


cos 6 = cos - 15 cos sin + 15 cos sin - sin and6 4 2 2 4 6

5 3 3 5sin 6 sin= 6 cos sin - 20 cos sin + 6cos

Addtion formulae for any number of angles

We have,

cis cis cis ......... cis =cis( + + +.......... ...+ )1 2 3 n 1 2 3 n

Now cis cis cis ......... cis1 2 3 n

=cos (1+itan ) cos (1 +itan )......... cos (1+itan )1 1 2 2 n n

=cos cos ......... cos (1+itan )(1 +itan ) .. .. ..... ... .. (1+ itan )1 2 n 1 2 n

( + + +.......... ..+ )i e cis1 2 3 n

2 3= cos cos cos .......... cos (1+ iS + i S +iS + .......... ....)1 2 3 n 1 2 3

where S = tan1 1

S = tan tan2 1 2

S = tan tan tan3 1 2 3

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

cis ( + + + )............1 2 n

=cos cos .......... cos ( +iS - S - iS +S .......... ....)11 2 n 1 2 3 4

=cos cos .......... cos ( - S +S .......... ....) +icos cos .......... cos ( S - S .......... ..)11 2 n 2 4 1 2 n 1 3


cos( + + +.......... ... + ) = cos cos .......... cos (1 - S +S .......)1 2 3 n 1 2 n 2 4and

sin( + + +.......... ... + ) = c os cos .......... cos (S - S .......)1 2 3 n 1 2 n 1 3

S - S................. + + +.............+ = 1 3tan( )

1 2 3 n 1- S +S.................2 4


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Exercise

1. Findthe continued product of fourth roots of unity.

2 32. If ' ' isthe cube root of unity,then findthevalue of (4- 3 - 3 ) .

8sin( /8) +icos( / 8)

3. Findthe value of sin( /8) - icos( / 8)

4 . Write the conjugate o the multiplica tive inverse of t he complex number ( sin +icos ).

5. If z=x+ iythen wha t does z+ 1 =1represents?

26. Simplify (sin 2x+ icos2x) using DeMoivre'sTheorem.

7. Find the realvalues o xand which sati sfy th e equation ( 02x- ) + i(3x+2 +1) = .

8. Find the realvalues o xand which sati sfy th e equation ( 02+i)x+ (i- 3) - 4 = .

9. If 'n'is anyinteger, find i +i + i + i .-n - n+1 -n+ 2 -n+ 3

10. What i sthe multiplicative inverseof i ?101

3 2 311. If ' ' in the cube root of '1',findthevalue of (1+ ) - (1+ ) .

2 4 2 4. +

+

+ .

12 Simplify i i i i2 3 4

+ 3 313 . Express 2 cos isin in theCartesian form.

4 4

- 1+

cos isin .14. Find the real part of 1 +5 5

a+ib a +b2 22 2 215 . If x+iy= , prov e t hat (x + ) = .

c+id c +d2 2

n1 +i

16 . Find the least positive integer 'n'for which = 1.1 - i

2 217 . If z= x+iy and z+6 = 2z+3 , showthat x + =9.

2 218 . If z=a+ib and z- 2 = 2z- 1 , show that a +b = 1.

z- 12 219 . If z= x+iy be such that amp = , showthat x + - 2 = 1 .

z+ 1 4

20 . Express 1 - i in the polarform.

3-i21. Express in thepolar form.

2

22 +i

22. Find the modulus and amplitude o .3 - i

(sin + icos )323 ). Simplify

(cosa+ isina 2


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(cos 7 +isin7 ) (cos3 + isin 3 )-2 424. Simplify

-4(cos +isin )

25. If is a cube rootof unity,showthat (1- + )(1- + )(1- + )(1- + ) =16.2 2 4 4 8 8 16

1526. If x=cis ,t hen find x + .

x527. If i = - 1,then find i +i +i + L + (2n)terms.2 2 4 6

ax- -a28. If x= cis and = cis , provethat =itan

x+ 2

na a2 n n n+ 129 . If & arethe roots of x - 2x+ 4= 0,then prove that + = 2 cos .

3

30. Provethe followi ng

()() 6 6i) (1+ i) + (1 -i) = - 4 ii) 1- i 2 + 1+ 2 = 46.3 3

1 1n n31 . If Z= cos +isin , show that Z + =2 cosn and Z - = 2isinn .

Z Zn n

a32 . If x=cis , = cis , provethat1 1a ai) xy+ = 2 cos( + ) ii) xy- = 2isin( + ).

xy xy

a33 . If x=cis , = cis , prove that1 1

a a2 3 2 3i) x + = 2 cos( 2 +3 ) ii) x - = 2isin( 2 +3 ).x x2 3 2 3

x - 12n34 . If x= cis , prove that =itann .

2nx + 18 8 9 835. Provethat (1+ cos +isin ) + (1+ cos -isin ) = 2 cos(4 ) cos ( / 2).

36. If cosa+ 2cos + 3cos = 0= sina+ 2sin +3 sin , provethati) cos 3a+8 cos 3 + 27cos 3 = 18cos(a+ + ) ii) sin3a+8sin3 + 27sin3 =18sin(a+ + ).

1 1 137. If 2 cosa =x+ , 2cos = + and 2cos =z+ , provethatx z1 1

i) xyz+ = 2cos(a+ + ) ii ) xyz- = 2isin(a+ + ).xyz xyz

[]n n n -38. Showthat (a+ib) - (a- ib) = 2i(a +b ) sin 2n tan (b/a)2 2 2 2 1

n1 + sin +icos -n -n

39 . Prove that = cos n +isin n1 + sin - icos 2 2

71 - cos - isin

740 . Prove that =itan cis 7 .1 + cos - isin 2

a n a1 +itan 1+itan41 . Prove that = .

a a1 - itan 1 -itan


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. Z = ( / );r= , , L Z Z Z L L 8 =i.r42 If cis 3 12 3 provethatr 1 2 3

43. Findthe cube roots of - 1+ i 3 ./ 344. Find allthe values of (1+ i) .2

2/ 345. Find allthe values of (1- i) .

46 . Find the fourth roots of 1 +i 3 andrepresent them in the Argand diagram.

747. Solve x - x= 0.548 . 0 Solve x + 1= .

49. Express sin 7 & cos 7 in powers of sin & cos .

50 . Express cos 8 & sin 8 in powers of cos & sin .


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MATRIX THEORY

Review of the fundamentalsA rectangular array ofmnelements arranged inmrows &ncolumns is called a'Matrix' of a ordermnmatrices are denotedby capital letters of The English Alphabet.

Examples

a b1 1

a bMatrix of order 32 is 2 2a b

3 3

a a a1 2 3

b b bMatrix of order 43 is 1 2 3

c c c1 2 3

d d d1

2 3

a b c1 1 1

a b cMatrix of order 33 is 2 2 2a b c

3 3 3

Note :- Elements of Matrices are written in rows and columns with in the bracket ( ) or [ ].

Types of Matrices

(1) Equivalent Matrices : Two matrices are said to be equivalent if the order is the same.

(2) Equal Matrices : Two matrices are said to be equal if the corresponding elements are equal.

(3) Rectangular & Square Matrices : A matrix of ordermn is said to be rectangular ifm n, square ifm = n.

(4) Row Matrix :

A matrix having only one row is called Row Matrix.

(5) Column Matrix : A matrix having only one column is called Column Matrix.

(6) Null Matrix or Zero Matrix : A matrix in which all the elements are zeros is called Null Matrix or Zero Matrixdenoted asO. [English alphabetOnot zero where as elements are zeros]

(7) Diagonal Matrix : A diagonal matrix is a square matrix in which all elements except the elements in the principal

diagonal are zeros.

2 0 04 0

Example 0 1 00 6

0 0 4

are diagonal matrices of order 2 & 3.

(8) Scalar Matrix : A diagonal matrix in which all the elements in the principal diagonal are same.


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28 KSOU Matrix Theory

8 0 04 0

Example 0 8 00 4

0 0 8

are Scalar Matrices of order 2 & 3.

(9) Unit Matrix or Identity Matrix : A diagonal matrix in which all the elements in the principal diagonal is 1 iscalled Unit Matrix or Identity Matrix denoted by I.

1 0 0 01 0 0 1 0 0

Example , :0 1 0 0 1 0

0 0 0 1

are unit matrices of order 2 & 4.

(10) Transpose of a Matrix : IfAis any matrix then the matrix obtained by interchanging the rows & columns ofAiscalled 'Transpose ofAand it is written as A'or A.T

a ba c e

Example: If A= c d then A' isb d

e

Ais of order 3 2 but A'is of order 2 3.

Matrix addition

Two matrices can be added or subtracted if their orders are same.

a b c c d e1 1 1 1 1 1Example : If A= & B=

a b c c d e2 2 2 2 2 2

a +c b +d c +eA+B= 1 1 1 1 1 1

a +c b +d c +e2 2 2 2 2 2

a - c b - d c - e1 1 1 1 1 1A- B=

a - c b - d c - e2 2 2 2 2 2

Matrix Multiplication

IfAis a matrix of orderm andBis matrix of order n, then the productABis defined and its order ismn. (ie. forABto be defined number of columns of Amust be same as number of rows ofB)

a1 1a b c

A= & B= a1 1 1Example: Let2 2a b c

2 2 2 a3 3

a a aa +b +c a +b +c1 1 1 2 1 3 1 1 1 2 1 3then AB=a a aa +b +c a +b +c

2 1 2 2 2 3 2 1 2 2 2 3which is of order 2 2.

Note :- IfAis multiplied byAthen AAis denoted as A,AAA.... asAetc.2 3


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Scalar Multiplication of a Matrix

IfAis a matrix of any order andKis a scalar (a constant), thenKArepresent a matrix in which every element ofAis multipliedbyK.

a b c Ka Kb Kc1 1 1 1 1 1

Example: IfA= a b c thenKA= Ka Kb Kc2 2 2 2 2 2

a b c Ka Kb Kc3 3 3 3 3 3

Symmetric and Skew Symmetric Matrices

LetAbe a matrix of ordernnan element in irow and column can be denoted as aij. Hence a matrix of ordernncant h thbe denoted as (aij) or [aij] wherei= 1, 2, .......n, = 1, 2, .......n

A matrix of ordernnis said to be Symmetric ifaij = aji and Skew Symmetric ifaij = ajiorAis symmetric ifA = ATorA=A', skew symmetric ifA = A orA = A'also A + A'is symmetric & A A'is skew symmetric.T

Note :- In a skew symmetric matrix the elements in principal diagonal are all zeros.

2 3 5'Example :A= 3 7 6 is symmetricwhere A=A

5 6 8

0 - 2 7

B= 2 0 6 is skew symmetricwhere B= -B'- 7 - 6 0

Determinant

A determinant is defined as a mapping (function) from the set of square matrices to the set of real numbers.

AIf A is a square matrix its determinant is denoted as .

a b c a b c1 1 1 1 1 1

Example: LetA= a b c then det.AorA=a b c2 2 2 2 2 2

a b c a b c3 3 3 3 3 3

Minors and Co-factors

Let ,A= (aij) i=1, 2, 3 = 1, 2 3

a a a11 12 13

ie A= a a a21 22 23

a a a31 32 33

a a a11 12 13

A= a a a21 22 23

a a a31 32 33

a a2 2 2 3Consider which is a determinant formed by leaning all the elements of row and column in which all lies. This

a a3 2 3 3

determinant is called Minor ofa . Thus we can form nine minors. In general ifAis matrix of ordernnthen minor ofaijis11


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obtained by leaning all the elements in the row and column in whichaijlies in .AThe order of this minor isn 1where as theorder of given determinant isnif this minor is multiplied by (1) then it is called Co-factors ofai j.i + j

a a a11 12 13

Example: LetA= a a a21 22 23

a a a31 32 33

a a22 23a =Mi nor o

1 1 a a32 33

a a a a22 2 3 2 2 2 3Co -factor of a =( - 1) =1+1

1 1 a a a a3 2 3 3 3 2 3 3

a a1 2 13Mi nor o a is

21 a a3 2 3 3

a a a a1 2 13 1 2 13Co -factor of a is (- 1) = -2+1

21 a a a a3 2 3 3 3 2 3 3

Value of a determinant

Consider a matrixAof ordernn. Consider all the elements of any row or column and multiply each element by its correspondingco-factor. Then the algebraic sum of the product is the value of the determinant.

a bA= 1 1Example : Let

a b2 2

Co-factor ofais b1 2

Co-factor ofbis a1 2

A=ab - ba1 2 1 2

a b c1 1 1

Let A=a b c2 2 2

a b c3 3 3

b c b ca ( - ) =1+1 2 2 2 2Co - factor of is 11 b c b c3 3 3 3

a c a c2 2 2 2Co -factor of b is (- 1) = -1+ 2

1 a c a c3 3 3 3

a b a b1+ 3 2 2 2 2Co -factor of c is (- 1) =

1 a b a b3 3 3 3

b c a c a b2 2 2 2 2 2A=a - b +c

1 1 1b c a c a b3 3 3 3 3 3

=a (bc - bc ) - b (ac - ac ) +c(ab - ab )1 2 3 3 2 1 2 3 3 2 1 2 3 3 2

=abc - abc ) - abc - abc +abc - abc1 2 3 1 3 2 2 1 3 3 1 2 2 3 1 3 2 1


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Properties of determinants

(1) If the elements of any two rows or columns are interchanged then value of the determinant changes only in sign.

(2) If the elements of two rows or columns are identical then the value of the determinant is zero.

(3) If all the elements of any row or column is multipled by a constant K, then the value of the determinant is multipledbyK.

(4) If all the elements of any row or column are written as sum of two elements then the determinant can be written as

sum of two determinants.

(5) If all the elements of any row or column are multiplied by a constant and added to the corresponding elements of

any other row or column then the value of the determinant donot alter.

Adjoint of a Matrix

a b c1 1 1

Let A= a b c2 2 2

a b c3 3 3

Let us denoted the co-factors ofa,b,c,a,b,c,a,b,cas A,B,C,A,B,C,A,B,Ctranspose of matrix of1 1 1 2 2 2 3 3 3 1 1 1 2 2 2 3 3 3

co-factors is called Adjoint of the Matrix.

A

B

C

1 1 1MatrixofCo- factors = A B C2 2 2

A B C3 3 3

A A A1 2 3

Adjoint ofA= B B B1 2 3

C C C1 2 3

Theorem A. adjadj.A= AI= .A.A

a b c A A A1 1 1 1 2 3

A.adj.A= a b c B B B2 2 2 1 2 3

a b c C C C3 3 3 1 2 3

aA+bB+cC aA +bB +cC aA+bB +cC1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3

= aA+bB+cC aA +bB +cC aA+bB +cC2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3aA+bB+cC aA +bB +cC aA +bB +cC

3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3

b c a c a bThe value of the det.A.2 2 2 2 2 2Now aA +bB +cC = a - b +c =

1 1 1 1 1 1 1 1 1b c a c a b3 3 3 3 3 3

aA +bB +c C =Similarly2 2 2 2 2 2

aA +bB +cC =3 3 3 3 3 3

b c a c a b1 1 1 1 1 1aA +bB +cC = - a +b - c

1 2 1 2 1 2 1 1 1b c a c a b3 3 3 3 3 3

=- a (bc - bc) +b(ac - ac) - c(ab - ab)1 1 3 31 1 1 3 3 1 1 1 3 3 1

=- abc +abc +abc - abc - abc +abc = 01 1 3

1 3 1 1

1

3 3 1 1

1 3 1 3 1 1


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Similarly the other five elements ofAadj.Ais zero.

0 0A.adj.A= 0 0 where = A

0 0

1 0 0

= 0 1 0 = .I0 0 1

A. adjadj.A= AI= .A

Singular and Non-singular Matrices

A square matrixAis said to be singular if 0A= and non-singular if .A 0

Inverse of a Matrix

Two non-singular matricesA& Bof the same order is said to be inverse of each other ifAB = I = BA. Inverse ofAis denotedasA . Inverse ofBis denoted asB and further (AB) =BA . 1 1 1 1 1

To find the inverse of A

A.adj.A= AI

A ,AA . .A= AA- 1 - 1 - 1multiply by adj

.Aie adj. adjA=AA A =-1 - 1

A

1 4 - 2- 2 - 5 4Example : Find the inverse of1 - 2 1

1 4 - 2Let A= - 2 - 5 4

1 - 2 1

- 5 4 - 2 4 - 2 - 5-

- 2 1 1 1 1 - 24 - 2 1 - 2 1 4

Matrix of Co- factors = - -- 2 1 1 1 1 - 24 - 2 1 - 2 1 4

-- 5 4 - 2 4 - 2 - 5

(- 5+ 8) - (- 2 - 4) (4+ 5)= - (4- 4) (1 + 2) - ( - 2 - 4)

(16 - 10) - (4- 4) (- 5 +8)

3 6 9= 0 3 6

6 0 3


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3 0 6adj.A= 6 3 0

9 6 3

1 4 - 2A= - 2 - 5 4 = 1( 5- 5+ 8) - 4( - 2- 4) - 2( 4+ ) =3+ 24 - 18 = 9

1 - 2 1

3 0 6 03 69 9

11= 6 3 0 = 0- 1A = adj.A 6 39 99A

9 6 3 9 6 39 9 9

01 23 3

A = 0- 1 2 13 31 2 1

3 3

Solutions of Linear equations

Cramer's Rule

To solve the equations

ax+b +cz=d1 1 1 1

ax+b +cz= d2 2 2 2

ax+b +cz= d3 3 3 3

a b c1 1 1

Consider =a b c(1)2 2 2

a b c3 3 3

first evaluate & if it is not zero then multiply both sides of (1) by x.

a b c ax b c1 1 1 1 1 1

x=xa b c = ax b c2 2 2 2 2 2

a b c a b c3 3 3 3x 3 3

multiply the elements of columns 2 & 3 by & zand add to elements of column 1.

ax+b +cz b c1 1 1 1 1

then x= ax+b +cz b c2 2 2 2 2

ax+b +cz b c3 3 3 3 3

d b c1 1 1

= d b c = (say) (2)2 2 2 1

d b c3 3 3

multiply both sides of (1) by

a b c a b c1 1 1 1 1 1

= a b c = a b c2 2 2 2 2 2

a b c a b c3 3 3 3 3 3


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multiply the elements of columns 1 & 3 byx&zand add to the elements of column 2.

a ax+b +cz c a d c1 1 1 1 1 1 1 1

= a ax+b +cz c = a d c = (say) (3)2 2 2 2 2 2 2 2 2

a ax+b +cz c a d c3 3 3 3 3 3 3 3

multiply both sides of (1) by z

a b c a b cz1 1 1 1 1 1

z=za b c =a b cz2 2 2 2 2 2

a b c a b cz3 3 3 3 3 3

multiply the elements of columns 1 & 2 byx& and add to the elements of column 3.

a b ax+b +cz1 1 1 1 1

= a b ax+b +cz2 2 2 2 2

a b ax+b +cz3 3 3 3 3

a b d1 1 1

= a b d = (say) (4)2 2 2 3

a b d3 3 3

then 1x= from (2)

= from (3)2

z= from (4)3

Note :- Verification of values ofx, ,zcan be done by substituting in the given equations.

Example - 1

Solve 2x+ - z=3

1x+ +z=

4x- 2 - 3 =

2 1 -1Let = 1 1 1

(1)1 - 2 - 3

= 2( 1- 3+2) - 1(- 3- 1) - 1( - 2 - ) =- 2 +4 +3 = 5

multiply both sides of (1) by x

2 1 -1 2x 1 -1then x=x1 1 1 = x 1 1

1 - 2 - 3 x - 2 - 3

multiply the elements of columns 2 & 3 by andzand add to the elements of column 1.

2x+ - z 1 -1x= x+ +z 1 1

x- 2 - 3 - 2 - 3


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3 1 -1= 1 1 1 =3( 4- 3+2) - 1( - 3- 4) - 1(- 2 - ) = - 3+7 +6 =10

4 - 2 - 3

10 10x= = = 2

5

multiply both sides of (1) by 2 3 -1 2 -1

= 1 1 1 == 1 11 4 - 3 1 - 2 - 3

multiply the elements of column 1 byx& 3 by zand to the corresponding elements of column 2.

2 2x+ -z -1then = 1 x+ +z 1

1 x- 2 - 3z - 3

2 3 -1= 1 1 1 = 2 ( 1- 3 - 4) - 3( - 3- 1) - 1(4 - ) =- 14 + 12 - 3 = - 5

1 4 - 3

- 5 - 5= = =- 15

multiply both sides of (1) by z

2 3 -1 2 3 -zz=z1 1 1 = 1 1 z

1 4 - 3 1 4 - 3z

multiply the elements of column 1 byx& column 2 by and to the corresponding elements of column 3.

2 1 2x+ -zthen z= 1 1 x+ +z

1 - 2 x- 2 - 3z

2 1 3= 1 1 1 = 2 ( 14 +2 ) - 1(4- 1) +3( - 2- ) = 12 - 3 - 9 = 0

1 - 2 4

0 0z= = = 0

5

Thus solution is x= 2, = 1 &z= 0 which can be verified by substituting in the given equations.

Example - 2

Solve 4x+ =7

53 + 4z=

5x+ 3z= 2

4 1 0= 5 3 4 = 4( 159 - 0) - 1(0- 20 ) +0( 0 - ) = 36 + 20 = 56

2 0 3


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7 1 0= 5 3 4 =7( 69 - 0) - 1(15 - 8) + 0(0 - ) =63 - 7 =56

1

2 0 3

4 7 0

= 0 5 4 = 4(15 - 8) - 7(0 - 20 ) +0( 0 - 25) =28 + 140 =16825 2 3

4 1 7= 0 3 5 =4( 156 - 0) - 1(0 - 25 ) +7( 0- ) = 24 +25 - 105 = - 56

3

5 0 2

561x= = = 1

56

168= = =3256

- 56z= = = - 13

56

Solution of Linear equations by Matrix Method

Given ax+b +cz=d1 1 1 1

ax+b +cz=d2 2 2 2

ax+b +cz=d3 3 3 3

a b c x d1 1 1 1

ConsiderA= a b c X= &B= d2 2 2 2

a b c z d3 3 3 3

then given equations can be written in Matrix form asAX = B. If 0A solution exists multiply both sides by A1

-1 -1A (AX) =A B

A AX=AB-1 -1

ie IX = A B-1

-1X=A B

Example

Solve 3x- + 2z=13

32x+ - z=

x+ 3 - 5z= -8

3 - 1 2 x 13Let A= 2 1 - 1 X= , B= 3

1 3 - 5 z - 8

then given equations can be written asAX = B


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X A B (1)-1=

To findA1

3 - 1 2A= 2 1 - 1 =3(- 5+3) +1 (- 10 + 1) + 2 ( 6- 1) = - 6 - 9 + 10 =- 5

1 3 - 5

1 - 1 2 - 1 2 1-

3 - 5 1 - 5 1 3- 1 2 3 2 3 - 1

Matrix o Co - factors = - -3 - 5 1 - 5 1 3

- 1 2 3 2 3 - 1-

1 - 1 2 - 1 2 1

(- 5+ 3) - ( - 10 +1) (6- 1) - 2 9 5= - (5 - 6) (- 15 - 2) - (9+ 1) = 1 - 17 - 10

1- 2 - ( - 3 - 4) (3+ 2) - 1 - 7 5

- 2 1 - 1 - 2 1 - 1

1 1adj.A= 9 - 17 7 A = adj.A= - 9 - 17 7- 1 A 55 - 10 5 5 - 10 5

Using this in (1)

- 2 1 - 1 13 - 26 3 8 - 15 31 1 1

X= - 9 - 17 7 3 = - 117 - 51 - 56 = - 10 = - 25 5 5

5 - 10 5 - 8 65 - 30 - 40 - 5 1

x = 3, =2,z =1 is the solution

Verification : Consider the first equation

3x- + 2z= 9+ 2+ 2= 13

Characteristic equation, Eigen Values & Eigen Vectors

A- I=LetA& Ibe square matrices of same order and a scalar then 0 is called Characteristic equation and the roots ofthis equation ie values of are called Eigen Values or Characteristic roots. The matrixXsatisfyingAX= Xis called EigenVector.

Example - 1

1 4Find the eigen roots and eigen vectors of the matrix

2 3

1 4 1 01 4- = 0Characteristic equation isLet A=

2 3 0 12 3

1- 4ie = 0 (1 - )(3 - ) - 8= 0

4 3 -

2 23 - 3 - + - 8

= 0 - 4 - 5 = 0


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( - 5 ,)( +1) =0 = - 1 5

Eigen roots are 1 & 5.

To find eigen vector X, corresponding to 1,

AX =1

1 4 x - xie =2 3 -

x+ 4 - x=

2x+3 -

x+ 4 = -x ix= - 4 x=

2x+ 3 = - ie x= - 2 - 2 1

Eigen vector corresponding to eigen value 1 is (2, 1)

To find the eigen vector corresponding to 5.

1 4 x x x+4x 5x1 1 1 2 1ie = 5 =

2 3 x x 2x +3x 5x2 2 1 2 2

x +4x = 5x x x1 2 1 1 2ie x =x ie =1 2x + x = x2 3 5 1 1

1 2 2Eigen vector is (1, 1)

Eigen vector corresponding to eigen root 5 is (1, 1)

Example - 2

6 - 2 2- 2 3 - 1Find the eigen roots and eigen vectors of the matrix 2 - 1 3

6 - 2 2 6- - 2 2Let A= - 2 3 - 1 Characteristic equation is 0- 2 3- - 1 =

2 - 1 3 2 - 1 3-[] [][]ie (6- ) (3- ) - 1+ 2- 2(3 - ) + 2+ 2 2- 2(3 - ) = 02[] [][]ie (6- ) 9+ - 6 - 1 + 2- 6+ 2 + 2+ 2 2- 6 + 2 ) = 02

ie (6- )( - 6 + 8) + 2(2 - 4) + 2(2 - 4) = 02

ie 6 - 36 + 48 - +6 - 8 + 4 - 8+4 - 8= 02 3 2

ie - + 12 - 36 + 32 =0 ie - 12 +36 - 32 = 03 2 3 2

which is the characteristic equation, by inspection 2 is a root

3 2dividing - 12 + 36 - 32 by - 2,we have

0( - 2)( - 10 + 16) =2

= 2 ,, 28ie ( - 2)( - 2 )( - 8 ) = 0


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To find eigen vector or = 2

ConsiderAX= 2X

6 - 2 2 x x x1 1 1

ie - 2 3 - 1 x = 2x where X= x2 2 2

2 - 1 3 x x x3 3 3

x - x + x = x 4x - 2x +2x =06 2 2 21 2 3 1 1 2 3

- x+ x - x = x - 2x +x - x = 02 3 21 2 3 2 1 2 3

x - x + x = x 2x - x +x = 02 3 21 2 3 3 1 2 3

the above three equations represent one equation 2xx+ x= 0.1 2 3

xx x x xLet x =0 ,then 2x = x ie = ie = =1 2 1 2 3

3 1 2 1 2 1 2 0Eigen Vector is (1, 2, 0)

To find the eigen vector for = 8

ConsiderAX= 8X

6 - 2 2 x 8x1 1

ie - 2 3 - 1 x = 8x2 2

2 - 1 3 x 8x3 3ie 6x - 2x + 2x =8x ie - 2x - 2x +2x =0

1 2 3 1 1 2 3- 2x +3x - x =8x - 2x - 5x - x = 0

1 2 3 2 1 2 32x - x +3x =8x 2x - x - 5x = 0

1 2 3 3 1 2 3(1)ie x+x - x =0

1 2 3(2)2x +5x +x =0

1 2 3(3)2x - x - 5x =0

1 2 3adding (1) & (2) we get 3x +6x =0

1 2

x xie x+2x = 0 ie x =- 2x = =K(Say)1 2

1 2 1 2- 2 1

then ,x = - 2K x = K1 2

subsitituting in (1)- 2K+K- x = 0 x = -K

3 3

x =- 2 &K, x = K x = -K1 2 3

Eigen vect oris (- 2, 1, - 1) or (2, - 1, 1)

Eigen roots are 2,2,8

& 1Ei gen vect ors are (1, 2 , 0 ) & (2, - 1, )

Properties of Eigen values

(1) The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal.

(2) The product of the eigen values of a matrix is equal to the value of its determinant.

1(3) If is an eigen value of Athen is the eigen value of A .1


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Cayley - Hamilton Theorem

Every square matrix satisfies its characteristic equation.

a- ba b=0Characteristic equation isLet A=

c d-c d

which on simplification becomes a quadric equation in in the form 0 +a +a = where a,aare constants.21 2 1 2

A +aA+aI=2Cayley Hamilton Theorem states that 0 where Iis a unit matrix of order 2 & 0 is a null matrix of order 2.1 2a b c

1 1 1If A= a b c

2 2 2a b c

3 3 3

a - b c1 1 1

then characteristic equation is 0a b - c =2 2 2

a b c -3 3 3

which on simplification becomes 0 which is a cubic equation.3 2- +a +a +a =1 2 3

Then as per Cayley Hamilton Theorem 0- A +aA +aA+aI= where Iis a unit matrix of order 3 & 0 is a null3 21 2 3

matrix of order 3.

In general ifAis a square matrix of ordernthen characteristic equation will be of the form

- -(- 1) +a +a + ........ +aI= 0n n n1 nz2 nand by Cayley Hamilton Theorem

n n n-1 n-z(- 1) A +aA +aA + ........ +aI= 02 n

whereIis a unit matrix of ordern& 0 is a null matrix of ordern.

Note :- If we put = 0 in the characteristic equation then a = An

If a = 0, matrix Ais singular & the matrix Ais non-singular & hence inverse exists and we can find thea 0nn

inverse ofAusing Cayley Hamilton Theorem.

Example - 1

a bLet A=

c da- bCharacteristic equation is = 0

c d-

ie +a +a = 0where a,aare constants.21 2 1 2

By Cayley Hamilton Theorem

2A +aA+aI= 01 2

multiply both sides byA1

2 -1 -1 -1then AA +aAA +aA = 01 2-1ie A+aI+aA = 0

1 2aA = - (A+aI)-1

2 11

A = - (A+aI)- 11a

2


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Example - 2

The characteristic equation of a matrixAof order 2 is - 5 + 10 = 0 find A.2

Solution : put = 0 in C.E. then the constant 10 is .A

Example - 3

2 - 1Find the inverse of using Cayley Hamilton Theorem.

- 3 4

2 - 1 2 - - 1Solution : Let A= C.E.is = 0

- 3 4 - 3 4-

ie 8- 4 - 2 + - 3=02ie ( 2- )( 4 - ) - 3= 0

ie - 6 + 5=02

by Cayley Hamilton Theorem

2A - 6A+ 5I= 0

multiply both sides byA1

A- 6I+5A =0- 1

-15A = -A+ 6I

2 - 1 1 0 - 2+6 1+ 0 4 1= - + 6 = =

- 3 4 0 1 3+0 - 4+ 6 3 2

4 11A =- 1

5 3 2

Diagonalisation of Matrices

IfAis a square matrix of ordernwhere all the eigen values are linearly independent then a matrix Pcan be found such thatP APis a Diagonal Matrix.1

LetAbe a square matrix of order 3 and let , , be the eigen values, corresponding to these. LetX,X,Xbe three1 2 3 1 2 3

vectors where

x

z

1 1 1X = x , X = , X = z1 2 2 2 3 2

x z3 3 3

x z 0 01 1 1 1

-1LetP= x z ThenP AP= 0 02 2 2 2

x z 0 03 3 3 3

Example - 1

1 - 2Let A=

- 5 4

1 - - 2C.E. is =0 (1 - )( 4 - ) - 10 =0

- 5 4 -


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ie - 5 - 6 = 0 ( + 1)( - 6) = 02 = - 1,6 are eigen valu es

For , = - 1 let AX= -X

1 - 2 x - x1 1=ie

- 5 4 x - x2 2

x - 2x =-x 1x x1 2 1ie x = x = .... ei gen vect oris1 21 2- x + x =- x5 4 1 1 1

1 2 2For = 6, AX= 6X

1 - 2 x 6x1 1ie =

- 5 4 x 6x2 2

x - 2x =-x1 2 1 5x =- 2x

1 2- 5x +4x = -x1 2 2

x x - 2ie = eigen vect or is1 2

5- 2 5

1 2 5 2- 1Let P= Then P =- 1

1 5 7 - 1 1

5 2 1 - 2 1 21 --1P AP=

7 - 1 1 - 5 4 1 5

5 - 10 - 10 +8 1 2 - 5 - 2 1 21 - 1 -= =

7 - 1 - 5 2+4 1 5 7 - 6 6 1 5

- 5- 2 10 - 10 7 0 1 01 1- -= = =

7 - 6+ 6 12 +30 7 0 42 0 6

1 2 1 - 2-Thus P= diagonaliz ethe matrix

1 5 - 5 4

Example - 2

1 1 3

Let A= 1 5 13 1 1

1- 1 31 5- 1 =Characteristic equation is 03 1 1-

[][][]- - - - - - - + - - =ie (1 ) (5 )(1 ) 1 11 3 3 1 3( 5 ) 02ie (1- )( - 6 + 4) - (- - 2) + 3(- 14+3 ) = 0

ie - 6 +4 - +6 - 4 + +2 - 42 + 9 = 02 3 2

ie - + 7 - 36 =0 ie - 7 +36 =03 2 3 2

by inspection 2 is a root + 2 is a factor. equation becomes2( + 2

)( - 9

+ 18 )= 0


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6 = - 2 ,, 3ie ( +2) ( - 3 ( - 6 ) =0

ie characteristic roots are 2, 3, 6.

To find the eigen vector for = 2 Consider AX = - 2X1 1

x1

where X

= x

1 2x3

1 1 3 x - 2x1 1

ie 1 5 1 x = - 2x2 2

3 1 1 x - 2x3 3

(1)ie x +x + 3x = - 2x ie 3x +x + 3x =01 2 3 1 1 2 3

(2)x +5x +x = - 2x x + 7x +x = 01 2 3 2 1 2 3

(3)3x +x +x = - 2x 3x +x + 3x =01 2 3 3 1 2 3

(1) & (3) are same.

Put x =0 in (1) or (2), then x +x = 02 1 3

1-x x eigen vectorX = 0ie x =- x =1 3 11 3- 1 1

1

LetXbe the eigen vector for = 3.2

ie AX = 3X2 2

1 1 3 31 1 1ie 1 5 1 = 3 where X =

2 2 2 2

3 1 1 3 3 3 3

(1)+ + = ie - 2 + + 3 =0ie 3 31 2 3 1 1 2 3

(2)+ + = + 2 + = 05 3 31 2 3 2 1 2 3

(3)+ + = 3 + - 2 =03 3 31 2 3 3 1 2 3

Let us eliminate from (1) & (2)1

(1) 1is - 2 + +3 = 01 2 3

(2) 2 is 2 + 4 + 2 = 01 2 3

adding 5 +5 = 02 3

(4)+ =0 ie = - =2 32 3 2 3

- 1 1

Let us eliminate from (2) & (3)2

(2)1is +2 + = 01 2 3

(3)2 is 6 + 2 - 4 = 01 2 3

subtractin g - 5 + 5 =02 3

(5)5 =5 = =1 3

1 3 1 31 1


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1X = - 11 2 3From (4) & (5) = = 2

1 - 1 11

Next, letXbe the eigen vector for = 63

ie AX =6X3 3

1 1 3 z 6z z1 1 1

ie 1 5 1 z = 6z where X = z2 2 3 2

3 1 1 z 6z z3 3 3

(1)z +z + z = z ie - 5z +z +3z = 0ie 3 61 2 3 1 1 2 3

(2)z + z +z = z z - z +z =05 61 2 3 2 1 2 3

(3)z +z +z = z 3z +z - 5z = 03 61 2 3 3 1 2 3

adding (1)& (2), - 4z + 4z =01 3

z z(4)1 3ie z =z =

1 3 1 1

Let us eliminatezfrom (2) & (3)3

(2) 5 is 5z - 5z + 5z = 01 2 3(3)1is 3z +z + - 5 = 0

1 2 3adding 8z - 4z = 0

1 2

z z1 2 (5)ie 2z = z =

1 2 1 2

1z z z

X = 2From (4)& (5), = =1 2 33

1 2 11

- 1 1 1 - 2 0 0- 1Let P= 0 - 1 2 ThenP AP= 0 3 0

1 1 1 0 0 6

- 1 1 1 1 1 3P= 0 - 1 2 diagonalize 1 5 1

1 1 1 3 1 1

Exercise

1 2 31. Evaluate - 1 2 3.

2 3 1

a-b b-c c- a2. Evaluateb-c c-a a-b.

c-a a-b b-c


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1 2 33 . Evaluate 2 3 4.

3 4 4

2 0 44. If 6 x 5= 0,then findx.

- 1 - 3 1

1 -q5 . Evaluate - 1 r .

q - r 1

2 06. Find the adjoint of .

2 3

1 2 37. If A= 2 3 4 findthe co- factorof 1.

3 4 2

2 1 5' '8. If A= find AA &AA.

0 3 7

sec tan9. Find the inverse of .

tan sec

2 3 4 1 2 3-1 0 . If A= andB= find5A- 3Band 6B- 7A.

- 1 0 5 0 4 2

4 2 5 2 0 4-11 . If 3A+B= and 2B+ A= findAand B.

3 7 6 - 1 2 3

1 x x x 4 5+ -12 . If + = find xand .

5 x 7 - 12 4x

5 6 7 () ''1 3 . If A= verify th

at A =

A

.

8 9 10

2 4- 3 4 5

' '14. IfA= andB= - 4 3 findA+BandA- B.6 - 2 1

3 - 2

3 4215 . 0If A= provethat A - 10A+I= .

5 7

6 516. Findthe inverse of .

3 2

1 217. Findthe characteri stic equation of .

0 4


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2 318. Findthe eigen valu es of .

0 4

1 9 . S olve 2x+z=- 1, 2 + x= 5, z- = - 2byCramer's Rule.

- + =5x 4z 520. Solve

2x+ 3 + 5z= 27x- 2 + 6z= 5

by matrix method.

2 1-21. Findthe characteri stic roots of .

0 1

- 1 0 1

22. F indthe characteristic r oots of 1 2 1 .

2 2 3

1 223. Verify Cayley - Hamilton Theorem for the matrix .

3 4

2 024. Verify Cayley - Hamilton Theorem for the matrix .

1 - 1

1 225. Findthe eigen vect orsfor the matrix .

2 1

6 - 2 226. Findthe eigen valuesandeigen vectors for thematrix - 2 3 - 1.

2 - 1 3


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BCA 21 / IMCA 21 / Mathematics SVT 21

ALGEBRAIC STRUCTURES

Abreviations usedN : represent set of natural numbers.Zor I : represent set of +ve and ve integers including zero.

Z : represent non-negative integers ie. +ve integers including zero.+

Q : represent set of rational numbers.

R : represent set of real numbers.

C : represent set of complex numbers.

Z= {0, 1, 2, 3, .............. n 1} ie. Z represent set of integers modulon.n n

Q : represent set of +ve rational numbers.+

z- {o} : represent set of integers except 0.

Q- {o} : represent set of rational numbers except zero.

R- {o} : set of real numbers except zero.A set in general is denoted by S.

: for all

: belongs to

Binary Operation

IfSis a non-empty set then a mapping (function) fromSSto Sis defined as Binary Operation (in short B.O.) and denotedby *(read as star). ie. : SS S(Star maps ScrossStoS)

Another Definition

IfSis non-empty set then * (star) is said to be a Binary operation if a,b S,a*b S.

Examples

(1) on N+ and (ie addition & multiplication) are B.O.2 +3 =5 N 2 3 = 6 N

(2) on Z, +, & are B.O.

4 + 5 Z, 3 - 4 = 1 Z, 4 ,- 3= 1 Z 5 6 = 30 Z

a(3) On Q& R+, & are B.O. butis not a B.O. on Q& R for 0, but on Q- {o} & R- {o} is a B.O.Q Q&R

0

(4) on C, + andare B.O.

(x +iy ) +(x +iy ) = (x +x ) +i( + ) C1 1 2 2 1 2 1 2

(x +iy ) +(x +iy ) = (xx - ) +i(x +x ) C1 1 2 2 1 2 1 2 1 2 2 1


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48 KSOU Algebraic Structures

Definitions

(1 ) A non-empty set Swith one or more binary operations is called an 'Algebraic Structure'.(N, +), (Z, +, ), (Q, +,) are all algebraic structures.

(2) Closure Law : A setSis said to be closed under a B.O. *if a, b S,a*b S.

(3) Associative Law : A B.O. *is said to be associative on Sif a,b,c S(a*b) *c= a*(b*c)

(4) Commutative Law : A B.O. *is said to be commutative on Sif a,b S,a*b = b *a.

(5) Identity Law : An elemente Ssatisfyinga*e = a = e *a. a Sis called an identity element for the B.O. *on S.

Examples

(i) + and (addition and multiplication) are associative and commutative on N,Z,Q& R.

(ii) B.O. (subtraction) is not associative & commutative.

(iii) 1 is an identity for B.O. on Nbut + has no identity on N. Where asOis an identity onZ,Qand Rfor the B.O. +.

1 0(iv) If Sis a set of 2 2 matrices and B.O. is matrix multiplication then is an identity element.I=

0 1

Group

A non-empty set Gtogether with a B.O. *ie (G,*) is said to form a group if the following axioms are satisfied.

G. Closure Law : a,b G,a*b G1

G. Associative Law : a,b,c G, (a*b) *c= a*(b*c)2

G. Identity Law : There exists an elemente Gsuch that a G,a*e = a = e *a.3

G. Inverse : a G, there exists an elementbsuch thata*b = e = b *a. Thisbis called inverse ofaand usually denoted4

asa1

ie.a*a = e = a *a.1 1

In addition to the above four axions if a,b G,a*b = b*a. Then (G, *) is called an'abelian group' or'commutativegroup'.

Note (1 ) If for ( G, *) onlyGis satisfied it is called a 'groupoid'.1

(2 ) If for ( G, *)G& Gare satisfied it is called a 'semi-group'.1 2

(3 ) If for ( G, *)G,G&Gare satisfied it is called 'Monoid'.1 2 3

Examples

(i) ( N, +) is a groupoid and semigroup.

(ii) ( N, ) is a groupoid, semigroup and Monoid (identity for is 1)

(iii) ( Z, +) is a group (identity is Oanda = a) ie. a+ (a) = 0 = a+a.1

Note :- Every group is a monoid but the converse is not true, (Z, +) is a group and also a monoid but (N,) is a monoid butnot a group.


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Properties of Groups

1. Cancellation laws are valid in a group

ie if is a group then( G, *) a,b,c G,

(i () a*b=a*c b=c left cancellati onlaw)

(ii () b*a=c*a b=c right cancellation law)

Proof :-a*b=a*c, as ,a G a G- 1

- -a *(a*b) =a *(a*c)1 1

- 1 -1ie (a *a)*b= (a *a)*c

ie e*b=e*c where eisthe identity.

b=c

Similarly by considering

-1 -1(b*a)*a =(c*a)*awe get b=c

2. In a group G, the equations a*x=b and *a=bhaveunique solutions, a,b G.

Proof :-

a

*x

=

b

Operating on both sides bya1

-1 -1a *(a*x) =a *b

- 1 -1ie )(a *a *x=a *b

- 1ie e*x=a *b

x=a *b-1

To prove that the solution is unique, letx&xbe two solutions of a*x=b.1 2

a*x =b a*x =bie &1 2

a*x =a*x1 2

Operating on both sides bya1-1 - 1Weget a *(a*x) =a *(a*x )

1 2

- 1 -1ie (a *a) *x = (a *a)*x1 2

ie e*x =e*x1 2

x =x1 2

solution is unique.

3. In a group the identity element and inverse of an element are unique.

Proof :- To prove identity is unique. If possible lete& eare two identities then1 2

(1)a G,a*e =a=e *a1 1

(2)& a*e =a=e *a2 2

From LHS of (1), (using (2))a*e =a=e *a1 2

(using LHS of (2))ie a*e =e *a=a*e1 2 2


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by left cancellation, Thus the identity is unique.e =e .1 2

To prove that inverse of an element is unique.- -Let if possible let b& care inverses of a ie b =a andc =a1 1a G,

Now, a*b=e & a*c=e

whereeis the identity of the groupa*b=a*c

by left cancellation lawb = c. Thus inverse of an element is unique.

() -14. In a group G, a =a a G-1

Proof :- Asa is the inverse of a 1

- 1 - 1We have a*a =e=a *a() -1

it can be easily seen from above relation that inverse ofa is aie a =a.-11

Note :- Ifb& care elements of G, such that then each is the inverse of the other.b*c=e=c*b

5. In a group (G, *)

- 1 -1 -1a,b G, (a*b) =b *a

Proof :- ) - 1 -1Consider (a*b) *(b *a

-1 - 1 - 1 -1=a*(b*b )* a =a* e*a =a*a =e

(a*b) =b *a-1 - 1 -1

6. A group of order 3 is abelian.

Proof :- Order of group means the number of elements in a group. If a groupGhasnelements. The order ofGis n, whichis denoted asO(G) =n.

Ifnis finite it is called finite group and nis Infinite then it is called Infinite group.

be a group with a binary operation *.Let G= {e,a,b}

eis the identity by definition of identity it commutes with every element Qa*e=a=e*a

So we have to prove that * = *a b b a

Let a*b=a (a *a)*b=a *a-1 -1 e*b=e b=ewhich is not possible.

Let a*b=b-1 -1(a*b)*b =b*b a*e=e a=e

which is not possible.

cannote be equal toaorb*a b a*b=e

Similarly we can prove that

b*a=ea*b=b*a (G, *) is abelian.


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BTC111

Subgroups

A non-empty subsetHof a groupGis said to form a subgroup with respect to the same binary operation * if is a(H, *)group.

Eg. (1) ( z, +) is a subgroup of ( Q, +)(2) is a subgroup of with B.O.H= { 40, 2 , } G={ 50, 1, 2, 3,4, } + mod 6 ie

6

(3) is a subgroup of with respect to the B.O. multiplication.H= { 1- 1, } G={1,- 1,i, - i}

Theorem

A non-empty subsetHof a groupGis a subgroup of Gif and only if . a,b H,ab H- 1

Proof :- case (i)LetHbe a subgroup of GthenHis a group-1 -1a,b G,b G(inverseaxion) &ab G(closure law)

condition is satisfied.

case (ii) LetHbe a non-empty subset of Gwith the property . - 1a,b H,ab H

We have to prove thatHis a subgroup.

Let ,a H,a,a H aa =e H&ea H ea H ie a H-1 -1 -1 - 1

Since all elements ofHare elements of G, associative property is satisfied.

() - 1Letb H, ie closure property is satisfied.b H& fora H,ab ie ab H-1 -1

HenceHis a group and hence a subgroup.

Permutation group

Let S={a ,a ,a ,.......... a }1 2 3 n

Then a one-one and onto mapping or function from Sonto itself is called a Permutation.

a a a .......... aPermutation is denoted as 1 2 3 n

(a ) (a ) (a ) .......... (a )1 2 3 n

There will ben! ie permutations the set of permutations is denoted byS.nn

There is a composite mapping for& denoted as this can be taken as binary operation. Then theLet ,g S . og,n

setS with binary operation 'O' (ie composite mapping) will form a group. For convenience is denoted as f.ogn

1 2 3 4 1 2 3 4 the B.O. composite function is given byLet = & g= S4

3 1 4 2 2 3 4 1

1 2 3 4 1 2 3 4 1 2 3 4og= o =

3 1 4 2 2 3 4 1 ? ? ? ?to fill up the second row, following is the procedure.

in g: g(1) =2, g( 2) =3, g(3 ) = 4 , g(4 ) = 1

& in : (1) = 3, ( 2) =1, ( 3) =4, ( 4) =2

Now og(1) = [g(1)] = ( 2) = 1

og(2 ) = [g(2)]= (3) =4

og( 3) = [g( 3) ] = ( 4) =2

og( 4) = [g(4)]= (1) =3


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1 2 3 4og=

1 4 2 3

for convenience is written as fog

1 2 3 4 1 2 3 4 1 2 3 4ie og= o =

3 1 4 2 2 3 4 1 1 4 2 3

1 2 3 4 1 2 3 4 1 2 3 4& gf = =

2 3 4 1 3 1 4 2 1 4 2 3the composite function is also called product function.

Eg. (1) Let S={ 21, }

1 2 1 2then S = ,

2 1 2 2 1Let B.O. be product permutation

1 2 1 2 1 2then =

1 2 2 1 2 1

1 2 1 2 1 2=

1 2 1 2 1 21 2 1 2 1 2

=2 1 2 1 1 2

1 2closure law is satisfied, associative law can be easily verified. inverse e=

1 2

- 11 2 1 2

=1 2 1 2

- 11 2 1 2

=2 1 2 1

S forms a group.2

Eg. (2) Let S={ 31, 2, }

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3S =3 1 2 3 1 3 2 3 2 1 2 3 1 3 1 2 3 2 1

Let us denote the elements as respectively., , , , & 1 2 3 4 5 6

{}ie S = , , , , , 3 1 2 3 4 5 6

The following is the multiplication table.1 2 3 4 5 6

1 1 2 3 4 5 6

2 2 1 4 3 6 5

3 3 5 1 6 2 4

4 4 6 2 5 1 3

5 5 3 6 1 4 2

6 6 4 5 2 3 1


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BTC113

It can be seen from the table that closure law is satisfied.

For associative law

( ) = = Consider3 4 5 6 5 3( ) = =

3 4 5 3 1 3hence associative law is satisfied.( ) = ( )

3 4 5 3 4

5

identity e=1

-1 - 1 -1 -1 -1 - 1= , = = , = = & = 1 1 2 2 3 3 4 5 5 4 6 6

inverse of all elements exists.

Sforms a group, but it is not an abelian group3

Q = but = .3 4 6 4 3 2 3 4 4 3

Examples

(1) Show that the set R- {o} with B.O. forms a group.

Solution : For any elements a,b R- {o}. a*b R- {o}

2, 3 R- {o}, 2 3 = 6 R- {o}

closure law is satisfied.For any three elementsa,b,c R- {o}

(ab)c=a(bc)

(3 4) 5 = 12 5 = 60.

3(4 5) = 3 20 = 60.

associative law is satisfied.

Identity element is 1,

ie. a R- {o}, a1 =a= 1 a.

Leta R- {o} then there exists

1 } 1 1R- {o suc h that a =1= a

a a a

inverse exists for all elements R

- {o}.

(R- {o}, ) forms an abelian group.

(2) Show that forms an abelian group.(Z , )5 5

Solution : Let us construct table for (Z , )5 5

0 1 2 3 45

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3


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From the above table it can be easily seen that closure law is satisfied.

2 (3 4) =2 2= 45 5 5

( 2 3) 4 =0 4= 45 5 5

ie 2 (3 4 ) = ( 2 3) 45 5 5 5

associative law can be versified.

identity element is 0.

inverse of 0 is 0

inverse of 1 is 4

inverse of 2 is 3

inverse of 3 is 2

inverse of 4 is 1

form a group, further it can be seen from the table that for any two element a,b Z(Z , )5 5 5

a b=b a5 5

is an abelian group.(Z , )5 5

(3) Show that G= {1, 2, 3, 4} with B.O. multiplication mod 5 ie is an abelian group.5

Solution : The following in the relevant table for elements

1 2 3 45

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

From the above table it can be seen that closure law is satisfied.

( 2 3) 4 =1 4 =45 5 5

2 (3 4) =2 2 = 45 5 5

associate law is satisfied.

identity is 1.

inverse of 1 is 1

inverse of 2 is 3

inverse of 3 is 2

inverse of 4 is 4

forms a group and it can be seen from the table that it forms an abelian group. (G, )5

4. If Ris the set of real numbers and *is a binary operation defined on .Rasx* = 1+xy, x, RShow that *is commutative but not associative.

Commutativ e property a*b=b*a, a,b R

x* = 1+xy


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BTC115

*x= 1+ x x* = *x

Associativ e property (x* ) *z= x* ( *z)

LHS )=(1+xy) *z=P*z= 1+ z= 1 + (1 + xyz= 1 +z+xyz

RHS )=x*(1+ z) = x*Q= 1 +xQ= 1 +x(1+ z =1 +x+xyz

LHS RHS *is not aassociativ e

5. Show that set of integers with a*b=a- bwhere a,b Iisnot agroupa*b=a- b I closure axiom is satisfied

(a*b) *c a*(b*c), a*b=a- b I, a,b I.

LHS =(a- b) *c=a- b- c

RHS =a* (b- c) =a- b+c associativ e axiom is not satisfied

is not a group.(I, *)When one of the axiom is not satisfied, it is not a group. Hence, we need not have to check the rest of the axioms.

ab6. In a group Find the identity element, inverse of 4 and solve( 2G,*), a*b= . 4*x=5

To finde:a*e=a=e*aae

a*e= =a e= 2 i.e. i denti ty e lement is 2 .2a*a =e= a *a- 1 - 1

aa 4- 1-1 -1a*a =2 = 2 a =

2 a

44 = = 1. inverse of 4 is 1.-1

4

4x 10 54 *x= 5 = 5 x= =

2 4 2

{}G= cos + isin is real7. Prove that is an abelian group under multiplication{}G= cos + isin is real

Letx, ,zbe any three elements ofG.a aTake x= cos + isin , =cos +isin , z= cos +isin .

awhere , , are real numbers.

i) a a a a axy=(cos +isin )(cos +isin ) = cos( + ) +isin( + ) GQ( + ) is realClosure axiom is satisfied

ii) a a(xy)z=[(cos +isin )(cos +isin )](cos +isin )

a a= {cos( + ) +isin( + )}(cos +isin )

a a= cos{( + ) + }+isin{( + ) + }

a a= cos{ + ( + )} +isin{ +( + )}= x( z) multiplication is associative on R Associative axiom is satisfied.Q

iii) 1 = cos 0+isin 0 Gisthe identity element


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iv) ( cos +isin )( cos - isin ) = 1 cos - isin isthe multiplica tive inverse of cos +isin .

v) axy=(cos +isin )(cos +isin )

a a=cos( + ) +isin( + )

a a=cos( + ) +isin( + ) Qiscommutativ e onR= x

Commutative law is satisfied.

So all the axioms are satisfied. Hence, Gis an abelian group under multiplication.

8. Show that the cube roots of unity form an abelian group under multiplication

We know that the cube roots of unity are }1, and .LetG= {1, ,2 2

. 1 2here = 131 1 2

4 3& = = .12

12 2

i) All the entries in the table are the same as the elements of the set. This means the closure law is satisfied.

ii) 1 Consider 1 ( ) = 11=2

2 2(1 ) = = 1

2 21 ( ) = (1 ) . Associative lawis satisfied.

iii) The row heading 1 is the same as the topmost row.

1 is the identity element.

iv) Inverse of 1 is 1, inve rse of is and inverse of is .2 2

Every element has a inverse.

v) The table is symmetrical about the principal diagonal

commutative law holds good.

SoGis an abelian group under multiplication.

1 0 - 1 0 1 0 - 1 09. form an abelian group under matrixShow ,that the four matrices , and

0 1 0 1 0 - 1 0 - 1multiplication.

1 0 1 0 1 0 - 1 0-Take =I, =A, =B, =C; G={1, A, B,C}

0 1 0 1 0 - 1 0 - 1

IA=AI=A, IB=BI=B, IC=CI= C

1 0 1 0 - 1+0 0 + 0 - 1 0-AB= = = =C

0 1 0 - 1 0 + 0 0- 1 0 - 1

Similarly, it can be shown that BA= C, AC =CA= B,BC =CB= A, AA= I etc.


60/136

BTC117

The composition table is . I A B C

I I A B C

A A I C B

B B C I A

C C B A I

i) The entries in the table are the same as the elements of the set G. Closure law is satisfied.

ii) Associative law is satisfied.A(BC) =A(A) = I; (AB)C= (C)C=I

iii) Iis the identity element.

iv) Inverses of I, are respectivelyA,B I, is a group under matrix multiplicationA,B,C. Gis a group.(G, )

v) Since the entries on either side of the leading diagonal are symmetric, is an abelian group.(G, )

10. If every element of a group Ghas its own inverse, show that Gis abelian

(1)-Given a =a, a,b G1

(2)- 1 -1 -1we knowthat, a,b G, (ab) =b a

- 1 -1 - 1from Using these in (2),(1),a =a,b =band(ab) =ab. ab=ba, a,b Gthe commutative law is satisfied, so Gis abelian.

11. In a group . (G, ) if Prove that(ab) =ab , a,b G is abelian and conversely.2 2 2 (G, )

(ab) = (ab)(ab) = (aa)(bb)2

a[b(ab)] =a[a(bb)] using L.C.L.

it is abelian=b(ab) =a(bb) (ba)b= (ab)busing RCL ba abConversely

Ifab = ba pre operating byawe get, a(ab) =a(ba) (aa)b= (ab)a

post operating bybwe get [(aa) b] =[(ab)a]b

(aa)(bb) = (ab)(ab) ab = (ab)2 2 2

{}12. Given Q, the set of non zero rational numbers is a multiplicative group and show that His a subgroupnH= 2 n Z,

0ofQunder multiplication.

{}{ }0H= 2 n Z = ... 2 ,2 , 2 2 .....n - 2 - 1 0 , 1

i) 2 , 2 H, 2 2 = 2 H closure lawis satisfiedm n m n m+n

ii) m n r m n r m+n r m n+r( 22 ie ) 2 )2 = 2 (2 ), m,n,r z (2 )2 = 2 (2

(m+n)+r m+ (n+r) m+n+r m+n+r2 = 2 i.e. 2 = 2 Associative law issatisfied.

iii) 2 is the identity element0

iv) 2 ; there exist 2 such that 2 2 = 2 Inverseof 2 is 2 .m -m m -m 0 m -m

His a group under multiplication and ie His a subgroup of Qunder multiplication.H Q0 0


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Exercise

1. If N= {1, 2, 3, ....}, which of the following are binary operation of N.a

(1)a*b=a+ 2b (2 ) a*b=3a- 4b (5 )a*b=b

2. Which of the following operations on the given set are binary

(1) on I, the set of integers, a*b=3a- 4b

(2) 2 2on R,a*b= a -b

(3) on R,a*b= ab

3. If *is given by show that *is not a binary operation of Z.a b ab* = ,

4. Why the set of rationals does not form a group w.r.t. multiplication ?

5. If x, ,zare any three elements of a group G, find (xyz) .1

() - 16. In a group G, a,b G, a b .-1 -1find

7. If the binary operation *on the set Zis defined by ,a*b=a+b+ 5 find the identity element.

8. In the group of non zero integers mod 5. Find the multiplicative inverse of 4.

1 2 3 1 2 39. If = and g= arepermutatio ns inS , find og.3

1 2 3 2 3 1

10. If S= {1, 2, 3, 4, 5, 6} w.r.t. multiplication (mod 7), solve the equation 3 x= 5 in S.

11. Show that S= {1, 2, 3} under multiplication (mod 4) is not a group.

1 2 3 4 1 2 3 412. If = andg= find gf.

3 4 1 2 2 3 1 4

ab13. The binary operation *is defined by on set of rational numbers, show that *is associative.a*b= ,

7

ab14. If *is defined by on the set of real numbers, show that *is both commutative and associative.a*b= ,

2

15. If *is defined by show that *is associative.2 2a*b= a +b ,

16. On the set of real numbers, Show that *is associative.R,a*b=a+b- 1, a,b R.

17. On the set of real numbers, R,*is defined by ,a*b= 2 examine whethera- 3b+ab *is commutative and associative.

18. In the set of rationals except 1, binary operation *is defined by Find the identity and inverse of 2.a*b=a+b- ab.

ab19. On the set of positive rational numbers Find the identity element and the inverse of 8.Q ,a*b= , a,b Q .+ +

4

20. In a group of integers, an operation *is defined by Find the identity eleme nt.a*b=a+b- 1.


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BTC119

Vectors and Scalars

Vector :A physical quantity which has both direction and magnitude is called a 'Vector'.B

Eg.Velocity, acceleration, force, weight etc.

Scalar : A physical quantity which has only magnitude and no direction is called a 'Scalar'. A

Eg.speed, volume, mass, density, temperature etc.Vectors are represented by directed line segments. LetABbe the line segment. Vector fromAtoBis denoted by andAB

vector fromBtoAis denoted by can also be represented by . The length of is magnitude of the vector denotedaBA. AB ABas or or simply a. For Ais the initial point and Bthe terminal point.AB a AB,

Scalar multiplication of a vector

Let be a scalar andaa vector then arepresents a vector whose magnitude is times the magnitude ofaand the directionis same as that of aif is positive but opposite to that of aif is negative. If it represents a null vector denoted by=0'0 ie a' null vector is a vector of magnitude zero but its direction is arbitrary.

A vector whose magnitude is 1 is called a unit vector and a unit vector in the direction ofais written as

a aor ora(readasacap)

aa

Like and unlike vectors

Vectors having same direction are called 'like vectors' and those having the opposite direction are called 'unlike vectors'.

Co-initial vectors : Vectors having the same initial point are called co-initial vectors.

Coplanar vectors : Vectors in the same plane are called 'coplanar vectors'.

Parallel vectors : Vectors having same direction but different initialC

points are called 'parallel vectors'.

Triangle Law for addition of vectors

Let & represents two vectors then representsBC ACAB b

AB+BC ie a+bA B

aParallelogram Law

Let OA= a&OB=bComplete the parallelogramOACB.

then OA+AC =OC by triangl e lawB C

but AC= OB (parallel vectors)

OC =OA+OB =a+b b

Note : - AB=OB- OA

& BA=OA- OBO Aa


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Properties

(i) Vector addition is commutative ie . a+b=b+a( ) ( )

(ii) Vector addition is associative ie a+ b+c = a+b+c.

(iii) Set of vectors V, with binary operation vector addition will form a'Group' . The identity being 0 (null vector) andinverse of .ais - a

Position vectorsY

(i) Let Pbe a point in a plane where Ois the origin and OX&OYare co-ordinate axes. is called position vectors ofP.OP

rDrawPQ toOX, thenOQ=x&QP= P( , )

Let represents unit vectors in the direction ofi& OX &OY.

Then OQ=xi, QP= XOQ iOP=OQ+QP =xi+

2 2OP = x +

Note :- A plane vector is an ordered pair of real numbers and the distance betweenO& Pis the magnitude of .OP

(ii) Let Pbe a point in three dimensional space whereOX,OY&OZare co-ordinate axes. Let (x, ,z) be the co-ordinates ofrP. Draw PQ to the plane XOY& QA&QBparallel toOY& OXrespectives to meet OXatA& OYatB.

be unit vectors in the direction of OX,OY&OZ ZLet i, &k

ThenOA=xi,OB= ,QP =zk

OQ=OA+OB(by parallelog ramlaw)P( , ,z)

k=xi+

O BOP =OQ +QP (by triang le law) Yi

= xi+ +zk

Q2 2 2OP = x + +z

This position vector OPof the point Pis usuallyX

denoted by r ie OP=r

xi+ +zkunit vecto rin the direction of OPisgiven by

x + +z2 2 2


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Zbe any two points in 3-spaceLet P(x, ,z ) &P(x , ,z )

1 1 1 2 2 2to ,find PQ fromthe triangle OPQ

Q(x , z )2 2 , 2

OP+PQ =OQ

P(x, z )PQ =OQ - OP=x i+ +zk- xi- - zk 1

1

, 1

2 2 2 1 1 1= (x - x) i+ ( - ) +(z - z )k

O Y2 1 2 1 2 1and unit vector in the direction of

(x - x) i+( - ) + (z - z )kPQ is 2 1 2 1 2 1

(x - x ) +( - ) +(z - z )2 2 22 1 2 1 2 1

XScalar product of two vectors

are two non-zero vectors, then is defined as 'scalar product'If a=ai+a +ak and b=bi+b +bk ab +ab +ab1 2 3 1 2 3 1 1 2 2 3 3

of two vectors also known as 'dot product'.a&b, denoted asab

Vector product of two vectors

are two non-zero vectors, then

Study Material for B.tech

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