MathematicsWorksheet - FE Success Success (Mechanical) MATH Worksheet.pdfb m is the x axis intercept...

x axis

y axis

−3 −2 −1 0 1 2 3 4

A(4,-3)

M(a,a)

B(p,q)

odd functionf(-x) = - f(x)

Mathematics Worksheet

UNSOLVED Mathematics problems (with answers)for practice for FE-Mechanical exam

Prof. Prashant Morewww.FEsuccess.com

Written, typeset and published by

Prof. Prashant More

www.FEsuccess.com

fesuccess@outlook.com

First Edition November 2019.

ISBN No.

electronic format without prior written permission of the publisher.

FEsuccess.com Page 2

I took the Fundamentals of Engineering (FE) Exam in April 2001. I could

manage to pass it in the first attempt mostly because I used to work as a

tutor during my graduate studies at the Student Development Center of

University of Toledo, Ohio. I assisted and tutored almost every subject of

Mechanical Engineering. In October 2003, I helped a Civil Engineering

graduate pass this exam in his first attempt. During last 15 years of my

teaching career I have taught Pneumatics and Hydraulics, Manufacturing

Operations Management, Entrepreneurship and Management, Engineer-

ing Mechanics, Statics and Strength of Materials, Finite Element Analysis,

Mechanical and Electrical Equipment for building, Engineering Graphics

with AutoCAD, Solidworks, Autodesk Inventor, CAD/CAM, Developmen-

tal as well as Applied Technical Math. I had my own tutoring center for

Engineering Mathematics for Mumbai University students. At high school

level, I have taught Computer Math, Geometry and Algebra at Front Royal,

Virginia.

So basically, I had a large collection of hand-written notes and near-

infinite numbers of question papers created which I felt will be helpful if I

compile them in a nice ready-to-publish format.

This book is a humble attempt to encompass the topics required for study

for the FE Exam. This book is primarily written for test takers who have

been out of academic environment for a long time.

I would like to offer my sincere thanks to my previous supervisor, Dr. Alex

Echeverria (PE). Working under his guidance made me realize the true

spirit of teaching. His guidance at early stage of my career and the con-

fidence he had in me made me select teaching as my lifelong passion.

Having the material ready and compiling it to be published is altogether

a different story. I had to learn Latex typesetting language for the same.

For this purpose, I would like to thank Mr. Durga Prasad Rao who was

my colleague at University of Mumbai and Mr. Aneez Kundukulathil, who

worked with me at Muscat, Oman. Mr. Sampat Dhanawade maintained

the necessary documentation updated.

It took me almost eight years to prepare this book as it was done mostly in

my spare time while working full time as an Assistant Professor. However,

I assure that this will not prevent me from listening your suggestions and

incorporating fresh ideas into the content and presentation of the book.

Contents

1 Analytic Geometry 6

1.1 Straight Lines: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Conics: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.3 Circle: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.4 Ellipse: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.5 Parabola: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.6 Hyperbola: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1.7 Distance Formula: . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2 Calculus 78

2.1 Limits: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2.2 Derivatives: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

2.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

2.4 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

2.5 Integration by Partial Fractions: . . . . . . . . . . . . . . . . . . . 112

2.6 Integration Techniques for Trigonometric Functions: . . . . . . . 114

2.7 Definite Integrals: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

2.8 Area Bounded by a Curve: . . . . . . . . . . . . . . . . . . . . . . . 121

2.9 Volume of Revolution of Solids: . . . . . . . . . . . . . . . . . . . . 127

3 Linear Algebra 134

3.1 Types of Matrices: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

3.2 Determinant of a Matrix: . . . . . . . . . . . . . . . . . . . . . . . . 138

3.3 Matrix Operations: . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

4 Vector Analysis 156

4.1 Unit Vector: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

4.2 Direction Ratios and Direction Cosines: . . . . . . . . . . . . . . . 160

4.3 Addition/Subtraction of Two Vectors: . . . . . . . . . . . . . . . . 162

4.4 Dot Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

4.5 Cross Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

4.6 The Del Operator: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

4.7 Gradient: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

4.8 Divergence: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

4.9 Curl: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

5 Differential Equations 196

5.1 General Terminology: . . . . . . . . . . . . . . . . . . . . . . . . . 197

5.2 First Order Differential Equations: . . . . . . . . . . . . . . . . . . 201

5.3 Higher Order Differential Equations: . . . . . . . . . . . . . . . . 213

5.4 Finding Differential Equation from Solution: . . . . . . . . . . . . 220

5.5 Applying Boundary Conditions: . . . . . . . . . . . . . . . . . . . 228

6 Numerical Methods 241

6.1 Newton’s Method of Root Extraction: . . . . . . . . . . . . . . . . 242

6.2 Newton’s Method of Minimization: . . . . . . . . . . . . . . . . . . 247

6.3 Forward Rectangular Rule: . . . . . . . . . . . . . . . . . . . . . . 250

6.4 Trapezoidal Rule: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

6.5 Simpson’s Rule: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

7 Answer Keys- Mathematics 258

1 Analytic Geometry

Introduction

Analytic Geometry studies the representation of shapes by algebraic equa-

tions. These equations indicate related key points and distance between

them. In this module, geometrical shapes obtained by section of a cone

are studied in details.

x axis

y axis

0 1 2 3 40

Figure 1.1: Slope of a line as a ratio of Riseand Run

1.1 Straight Lines:

L E A R N I N G O B J E C T I V E S

After successful completion of this module, you will be able

• Find and interpret equation of a straight line in various

forms.

• Perform slope calculations including parallel and

perpendicular lines.

• Find angle between two coplanar, non-parallel lines.

Slope of a straight line gives you an idea about its inclination with ref-

erence to x-axis. Slope is also referred as gradient.

Slope of a Straight Line:

Slope of a straight line = Rise

= ∆y

m = y2 − y1

x2 −x1(1.1)

Equation of a Straight Line:

Slope Intercept Format:

y = mx +b (1.2)

where m = slope and b = y-intercept

For the above line, y-intercept = 1, and

slope =Ri se

So the equation will be,

y = 0.5x +1

2y = x +2

x −2y +2 = 0

Slope Point Format:

y − y1 = m(x −x1) (1.3)

where x1 and y1 are the coordinates of the point through which the line

passes.

Double Intercept Format:

b= 1 (1.4)

where a = x intercept and b = y intercept

For parallel lines slopes are equal i.e. m1 = m2

For perpendicular lines

m1 ∗m2 = -1

m2 = −1

m1(1.5)

Angle between Two Straight Lines:

Angle between two straight lines is given by:

α= tan−1(

m2 −m1

1+m1m2

)(1.6)

Solved Examples:

S O LV E D E X A M P L E - A A A - 0 1

A N G L E B E T W E E N T W O S T R A I G H T L I N E S

1. The angle between lines 2x- 9y + 16=0 and x + 4y + 5 = 0

is given by:

(A) tan−1(

(B) tan−1(−1

(C) tan−1(−1

)(D) The lines are parallel to each other.

Solution:

G R A P H O F A B S O L U T E VA L U E F U N C T I O N

2. A graph of the equation, y =| x−3 | would consist of which

one of the following?

(A) One straight line.

(B) Two parallel straight lines.

(C) One curved line.

(D) Two straight lines.

Solution:

X A N D Y- I N T E R C E P T S O F A S T R A I G H T L I N E

3. Given the slope-intercept form of a line as y = mx + b,

which one of the following is true?

bis the y axis intercept

(B) b is the x axis intercept

(C) y - b is the slope of the line

(D) -b

mis the x axis intercept

Solution:

R E FL E C T I O N O F A P O I N T

4. A ray of light coming from the point (1, 2) is reflected at

a point A on the x-axis and then passes through the point

(5, 3). The coordinates of the point A are:

(A) (13

(B) (5

(C) (- 7, 0)

(D) None of these

Solution:

M I D P O I N T O F I N T E R C E P T B E T W E E N C O O R D I N AT E

A X E S

5. If the co-ordinates of the middle point of the portion of

a line intercepted between coordinate axes (3,2), then the

equation of the line will be:

(A) 2x+3y=12

(B) 3x+2y=12

(C) 4x-3y=6

(D) 5x-2y=10

Solution:

I M A G E O F A P O I N T W I T H R E S P E C T T O A L I N E

6. The image of the point (4, -3) with respect to the line y =

(A) (-4, -3)

(B) (3, 4)

(C) (-4, 3)

(D) (-3, 4)

Solution:

A R E A O F T R I A N G L E F O R M E D B Y L I N E S

7. A line L passes through the points (1, 1) and (2, 0) and

another line L passes through (1

2,0) and perpendicular to

L. Then the area of the triangle formed by the lines L, L

and y- axis, is:

Solution:

D O U B L E - I N T E R C E P T E Q U AT I O N

8. The area of triangle formed by the lines

x = 0,

y = 0 andx

(A) ab

(C) 2ab

Solution:

E Q U AT I O N O F D I A G O N A L S O F A PA R A L L E L O G R A M

9. The diagonals of a parallelogram PQRS are along the lines

x+3y=4 and 6x-2y=7. Then PQRS must be a:

(A) Rectangle

(B) Square

(C) Cyclic quadrilateral

(D) Rhombus

Solution:

A B S O L U T E VA L U E F U N C T I O N A C R O S S QU A D R A N T S

10. The area enclosed within the curve |x|+|y|=1 is:

Solution:

P E R P E N D I C U L A R B I S E C T O R O F A L I N E

11. The line 3x +2y =24 meets y -axis at A and x-axis at B. The

perpendicular bisector of AB meets the line through (0,-1)

parallel to x-axis at C. The area of the triangle ABC is:

(A) 182 sq.units

(B) 91 sq.units

(C) 48 sq.units

(D) None of these

Solution:

D I S TA N C E O F A P O I N T F R O M A L I N E

12. The equation of the line joining the point (3, 5) to the

point of intersection of the lines 4x +y -1 =0 and 7x -3y

-35 =0 is equidistant from the points (0, 0) and (8, 34).

(A) True

(B) False

(C) Nothing can be said

(D) None of these

Solution:

A N G L E B E T W E E N D I A G O N A L S O F A QU A D R I L AT E R A L

13. The sides AB, BC, CD and DA of a quadrilateral are

x +2y = 3,

x = 1,

x −3y = 4,

5x + y +12 = 0

respectively. The angle between diagonals AC and BD is:

(A) 45◦

(B) 60◦

(C) 90◦

(D) 30◦

Solution:

P E R P E N D I C U L A R T O I N T E R I O R B I S E C T O R O F A N

A N G L E

14. Given vertices

A(1,1),

B(4,−2) and

C (5,5)

of a triangle, then the equation of the perpendicular

dropped from C to the interior bisector of the angle A is:

(A) y - 5 =0

(B) x - 5 =0

(C) y + 5 =0

(D) x + 5 =0

Solution:

15. A line 4x+y=1 passes through the point A(2,-7) meets the

line BC whose equation is 3x-4y+1=0 at the point B. The

equation to the line AC so that AB = AC, is:

(A) 52x + 89y + 519 =0

(B) 52x + 89y - 519 =0

(C) 89x + 52y + 519 =0

(D) 89x + 52y - 519 =0

Solution:

16. If the equation of base of an equilateral triangle is 2x-y=1

and the vertex is (-1, 2), then the length of the side of the

triangle is:

(B)2p15

Solution:

17. Let PS be the median of the triangle with vertices:

P (2,2),

Q (6,1) and

R (7,3)

The equation of the line passing through (1,-1) and paral-

lel to PS is:

(A) 2x-9y-7=0

(B) 2x-9y-11=0

(C) 2x+9y-11=0

(D) 2x+9y+7=0

Solution:

18. The equation of perpendicular bisectors of the sides AB

and AC of a triangle ABC are x - y+5=0 and x+2y=0 respec-

tively. If the point A is (1,-2) , then the equation of line

BC is:

(A) 23x + 14y - 40=0

(B) 14x - 23y + 40=0

(C) tan−1(2)

(D) 14x + 23y - 40=0

Solution:

Figure 1.2: Various conic sections, Source:Stackexchange, Latex code by Marmot

1.2 Conics:

• Understand origin of a conic section as a section of cone

by a cutting plane at various angles.

• Recognize the conic sections from their functions in

standard from and from their graphs.

• Convert a function of a conic section to standard form to

determine whether it yields a circle, a parabola, an ellipse,

or a hyperbola.

• Graph each of the conic sections from its function in

standard form.

• Create the function of a conic section, given information

on its important points and distances.

When a cone is cut by a plane, the resulting section of cone has differ-

ent possibilities:

Eccentricity:

• For Ellipse e < 1

• For parabola e = 1

• For hyperbola e > 1

Conic Section Equation:

The general form of the conic section equation is:

Ax2 +B x y +C y2 +Dx +E y +F = 0 (1.7)

where both A and C are non-zero.

• If B2 - 4AC <0, an ellipse is defined.

• If B2 - 4AC >0, a hyperbola is defined.

• If B2 - 4AC = 0, the conic is a parabola.

• If A = C and B = 0, a circle is defined.

• If A = B = C = 0, a straight line is defined.

x2 + y2 +2ax +2by + c = 0 (1.8)

is the normal form of the conic section equation, if that conic section has

a principal axis parallel to a coordinate axis.

h = a;k = b

r =√

a2 +b2 − c

• If a2 + b2 - c is positive, a circle, center (-a, -b).

• If a2 + b2 - c equals zero, a point at (-a, -b).

• If a2 + b2 - c is negative, locus is imaginary

Solved Examples:

S O LV E D E X A M P L E - A A B - 0 2

I D E N T I F Y I N G C O N I C S E C T I O N - I

19. 6x2 + 12x + 6y2 - 8y =100 is an example of a:

(A) circle

(B) parabola

(C) ellipse

(D) hyperbola

Solution:

I D E N T I F Y I N G C O N I C S E C T I O N - I I

20. 4x2 + 12x + y2 - 8y = 64 is an example of a:

(A) circle

(B) parabola

(C) ellipse

(D) hyperbola

Solution:

I D E N T I F Y I N G C O N I C S E C T I O N - I I I

21. 6y = 3x2 - 15 is an example of a:

(A) circle

(B) parabola

(C) ellipse

(D) hyperbola

Solution:

−2 −1 0 1 2

Figure 1.3: Circle with center at Origin

−2 −1 0 1 2 3

Figure 1.4: Circle with center at a pointother than Origin

1.3 Circle:

• Find equation of circle in standard and non-standard

• Given the equation of a circle, find its center and radius

or diameter.

Equation of circle with center at Origin:

Equation of a circle with center at (0,0) and radius = r

x2 + y2 = r 2

Equation of circle with center at point other that Origin:

Equation of a circle with center at (h,k) and radius = r

(x −h)2 + (y −k)2 = r 2 (1.9)

or in another form:

x2 + y2 +2g x +2 f y + c = 0

which has center at (-g,-f) and

radius =√

g 2 + f 2 − c

A given equation in x and y can represent a circle, if it follows following

three conditions:

1. The highest power of x is 2. Also, the highest power of y term is 2.

2. Coefficient of x2 is same as coefficient of y2.

3. There is NO product term xy.

Solved Examples:

S O LV E D E X A M P L E - A AC - 0 1

22. Consider an ant crawling along the curve

(x −2)2 + y2 = 4

where x and y are in meters. The ant starts at the point

(4,0) and moves counter-clockwise with a speed of 1.57

meters per second. The time taken by the ant to reach

the point (2,2) is (in seconds):

(D) π/2

Solution:

23. What is the radius of the circle

x2 + y2 −6y = 0?

Solution:

24. What are the coordinates of the center of the curve

x2 + y2 −2x −4y −31 = 0?

(A) (-1, -1)

(B) (-2, -2)

(C) (1, 2)

(D) (2, 1)

Solution:

25. A circle whose equation is:

x2 + y2 +4x +6y −23 = 0

has its center at:

(A) (2, 3)

(B) (3, 2)

(C) (-3, 2)

(D) (-2, -3)

Solution:

26. What is the radius of a circle with the equation:

x2 −6x + y2 −4y −12 = 0

(A) 3.46

Solution:

D I A M E T E R O F A C I R C L E

27. The diameter of a circle described by

9x2 +9y2 = 16

Solution:

D I S TA N C E B E T W E E N C E N T E R O F C I R C L E A N D Y- A X I S

28. How far from the y-axis is the center of the curve

2x2 +2y2 +10x −6y −55 = 0

(A) 2.5

(B) 3.0

(C) 2.75

(D) 3.25

Solution:

D I S TA N C E B E T W E E N C E N T E R O F C I R C L E S

29. What is the distance between the centers of the circles

x2 + y2 +2x +4y −3 = 0 and

x2 + y2 −8x −6y +7 = 0?

(A) 7.07

(B) 7.77

(C) 8.07

(D) 7.87

Solution:

S H O R T E S T D I S TA N C E F R O M A P O I N T T O A C I R C L E

center

30. The shortest distance from A (3, 8) to the circle

x2 + y2 +4x −6y = 12

is equal to?

(A) 2.1

(B) 2.3

(C) 2.5

(D) 2.7

Solution:

P R O P E R T I E S F R O M E Q U AT I O N O F A C I R C L E

31. The equation circle

x2 + y2 −4x +2y −20 = 0

describes:

(A) A circle of radius 5 centered at the origin.

(B) An eclipse centered at (2, -1).

(C) A sphere centered at the origin.

(D) A circle of radius 5 centered at (2, -1).

Solution:

A R E A O F C I R C L E F R O M I T S E Q U AT I O N

32. Find the area (in square units) of the circle whose equa-

tion is

x2 + y2 = 6x −8y

(A) 20 π

(B) 22 π

(C) 25 π

(D) 27 π

Solution:

E Q U AT I O N O F C I R C L E F R O M G I V E N C O N D I T I O N S

33. Determine the equation of the circle whose radius is 5,

center on the line x = 2 and tangent to the line 3x - 4y +

11 = 0.

(A) (x −2)2 +(y −2)2 = 5

(B) (x −2)2 + (y +2)2 = 25

(C) (x −2)2 + (y +2)2= 5

(D) (x −2)2 +(y −2)2 = 25

Solution:

E Q U AT I O N O F A C I R C L E G I V E N I T S C E N T E R A N D

TA N G E N T S

34. Find the equation of the circle with the center at (-4, -5)

and tangent to the line 2x + 7y - 10 = 0.

(A) x2 + y2 + 8x - 10y - 12 = 0

(B) x2 + y2 + 8x - 10y + 12 = 0

(C) x2 + y2 + 8x + 10y - 12 = 0

(D) x2 + y2 8x + 10y + 12 = 0

Solution:

C O N D I T I O N F O R P O I N T C I R C L E

35. Find the value of k for which the equation

x2 + y2 +4x −2y −k = 0

represents a point circle.

(C) -6

(D) -5

Solution:

E Q U AT I O N O F S P H E R E F R O M I T S C E N T E R A N D

R A D I U S

36. The equation of a sphere with center at (-3, 2, 4) and of

radius 6 units is?

(A) x2 + y2 +z2+6x - 4y - 8z = 36

(B) x2 +y2+ z2 +6x - 4y - 8z = 7

(C) x2 +y2 + z2 +6x - 4y + 8z = 6

(D) x2 + y2 + z2 +6x - 4y + 8z = 36

Solution:

−3 −2 −1 0 1 2 3

Figure 1.5: Ellipse with center at a pointother that Origin

−3 −2 −1 0 1 2 3

Figure 1.6: Ellipse with center at Origin

1.4 Ellipse:

• Find equation of ellipse types, regular and rectangular,

eccentricity, directrix, focus, latus rectum, relation

between a and b.

Equation of an ellipse when center is at (h,k) is:

(x −h)2

a2 + (y −k)2

b2 = 1 (1.10)

Equation of an ellipse when center is at (0,0) is:

a2 + y2

b2 = 1 (1.11)

Foci: (±ae,0) = (± c, 0) where e is the eccentricity of the ellipse.

c2 = a2 −b2 (1.12)

Here, a is called the semi major axis, and b is called semi minor axis.

(Assuming a > b) For the ellipse shown in the figure, the semi major axis=

3 and semi minor axis = 1.

Eccentricity:

e =√

1− b2

a2 = c

a(1.13)

Also, for an ellipse,

e < 1 (1.14)

Solved Examples:

S O LV E D E X A M P L E - A A D - 0 1

G E N E R A L E Q U AT I O N O F A C O N I C S E C T I O N

37. The general equation of a conic section is given by the

following equation:

Ax2 +B x y +C y2 +Dx +E y +F = 0

A curve maybe identified as an ellipse by which of the fol-

lowing conditions?

(A) B 2- 4AC <0

(B) B 2 - 4AC = 0

(C) B 2 - 4AC >0

(D) B 2 - 4AC = 1

Solution:

E Q U AT I O N O F L O C U S G I V E N A C O N D I T I O N

38. Point P (x, y) moves with a distance from point (0, 1) one-

half of its distance from line y = 4. The equation of its lo-

cus is?

(A) 2x2 - 4y2 = 5

(B) 4x2 + 3y2 = 12

(C) 2x2 + 5y3 = 3

(D) x2 + 2y2 = 4

Solution:

L E N G T H O F L AT U S R E C T U M O F A N E L L I P S E

39. What is the length of the latus rectum of

4x2 +9y2 +8x −32 = 0?

(A) 2.5

(B) 2.7

(C) 2.3

(D) 2.9

Solution:

D I S TA N C E B E T W E E N F O C I O F A N E L L I P S E

40. The lengths of the major and minor axes of an ellipse are

10 m and 8 m, respectively. Find the distance between

the foci.

Solution:

C E N T E R O F A N E L L I P S E

41. The equation

25x2 +16y2 −150x +128y +81 = 0

has its center at?

(A) (3, -4)

(B) (3, 4)

(C) (4, -3)

(D) (3, 5)

Solution:

M A J O R A X I S O F A N E L L I P S E

42. Find the major axis of the ellipse

x2 +4y2 −2x −8y +1 = 0

(B) 10

Solution:

L E N G T H O F L AT U S R E C T U M O F A N E L L I P S E

43. An ellipse with an eccentricity of 0.65 and has one of its

foci 2 units from the center. The length of the latus rec-

tum is nearest to?

(A) 3.5 units

(B) 3.8 units

(C) 4.2 units

(D) 3.2 units

Solution:

E C C E N T R I C I T Y O F A N E L L I P S E

44. The earth’s orbit is an ellipse with the sun at one of the

foci. If the farthest distance of the sun from the earth is

105.5 million km and the nearest distance of the sun from

the earth is 78.25 million km, find the eccentricity of the

ellipse.

(A) 0.15

(B) 0.25

(C) 0.35

(D) 0.45

Solution:

E Q U AT I O N O F D I R E C T R I X O F A N E L L I P S E

45. An ellipse with center at the origin has a length of major

axis 20 units. If the distance from center of ellipse to its

focus is 5, what is the equation of its directrix?

(A) x = 18

(B) x = 20

(C) x = 15

(D) x = 16

Solution:

−3 −2 −1 1 2 3

f (x) = x2

Figure 1.7: Parabola: Plot of y = x2

−3 −2 −1 1 2 3

f (x) =−x2

Figure 1.8: Parabola: Plot of y = −x2

1.5 Parabola:

• Find equation of parabola- Types of parabolas,

eccentricity directrix, focus, latus rectum, asymptote.

The standard form of a parabola’s equation is generally expressed:

y2 = 4ax

The role of ’a’

If a >0, the parabola opens towards right

if a <0, it opens towards left.

x2 = 4by

The role of ’b’

If b >0, the parabola opens upwards

if b <0, it opens downwards.

Key Properties of a parabola:

• Vertex is the minimum / maximum point of the parabola.

• Focus is a point on the axis of symmetry of the parabola that is a set

distance from the vertex of the parabola.

• Directrix is a line (not a point) that is perpendicular to the axis of sym-

metry of the parabola and does not intersect with the parabola.

• Eccentricity of a parabola is always equal to 1 because any point on

parabola is equidistant from focus and directrix.

Solved Examples:

S O LV E D E X A M P L E - A A E - 0 1

F I N D I N G C U R V E F R O M A N E Q U AT I O N

46. 3x2 + 2x - 5y + 7 = 0. Determine the curve.

(A) Parabola

(B) Ellipse

(C) Circle

(D) Hyperbola

Solution:

F O C U S O F A PA R A B O L A

47. The focus of the parabola y2 = 16x is at

(A) (4, 0)

(B) (0, 4)

(C) (3, 0)

(D) (0, 3)

Solution:

E Q U AT I O N O F V E R T E X O F A PA R A B O L A

48. Where is the vertex of the parabola x2 = 4(y - 2)?

(A) (2, 0)

(B) (0, 2)

(C) (3, 0)

(D) (0, 3)

Solution:

E Q U AT I O N O F D I R E C T R I X O F A PA R A B O L A

49. Find the equation of the directrix of the parabola y2 =

(A) x = 2

(B) x = -2

(C) x = 4

(D) x = -4

Solution:

V E R T E X O F A PA R A B O L A

50. Given the equation of a parabola

3x +2y2 −4y +7 = 0

Locate its vertex.

(A) (5/3, 1)

(B) (5/3, -1)

(C) (-5/3, -1)

(D) (-5/3, 1)

Solution:

F I N D I N G FA C I N G D I R E C T I O N O F A C U R V E

51. In the equation

y =−x2 +x +1

where is the curve facing?

(A) Upward

(B) Facing left

(C) Facing right

(D) Downward

Solution:

F O C U S O F PA R A B O L A

52. Find the location of the focus of the parabola

y2 +4x −4y −8 = 0

(A) (2.5, -2)

(B) (3, 1)

(C) (2, 2)

(D) (-2.5, -2)

Solution:

E Q U AT I O N O F A X I S O F S Y M M E T RY

53. Find the equation of the axis of symmetry of the function

y = 2x2 −7x +5

(A) 7x + 4 = 0

(B) 4x + 7 = 0

(C) 4x - 7 = 0

(D) x - 2 = 0

Solution:

E Q U AT I O N O F PA R A B O L A F R O M D I R E C T R I X A N D

F O C U S

54. A parabola has its focus at (7, -4) and directrix y = 2. Find

its equation.

(A) x2 + 12y - 14x + 61 = 0

(B) x2 - 14y + 12x + 61 = 0

(C) x2 - 12x + 14y + 61 = 0

(D) none of the above

Solution:

E Q U AT I O N O F D I R E C T R I X O F A PA R A B O L A

55. Given a parabola

(y −2)2 = 8(x −1)

What is the equation of its directrix?

(A) x = -3

(B) x = -1

(C) y = -3

(D) y = 3

Solution:

−4 −2 2 4

Figure 1.9: Hyperbola. The dotted lines areasymptotes.

a2− y2

cF2 F1V

Center

Figure 1.10: Hyperbola. The foci pointsare shown with their relationship with theasymptotes. F1 and F2 are the Foci points,` is the Semi-Latus Rectum, V is the Vertex

1.6 Hyperbola:

• Find equation of hyperbola- regular and rectangular,

eccentricity, directix, focus, latus rectum, asymptote,

relation between a and b.

Equation of a hyperbola when center is at (h,k)

(x −h)2

a2 − (y −k)2

b2 = 1 (1.15)

Equation of a hyperbola when center is at (0,0)

a2 − y2

b2 = 1 (1.16)

• Focus of hyperbola: The two points on the transverse axis. These points

are what controls the entire shape of the hyperbola since the hyper-

bola’s graph is made up of all points, P, such that the distance between

P and the two foci are equal. To determine the foci you can use the for-

mula: a2 +b2 = c2

• Transverse axis: This is the axis on which the two foci are.

• Asymptotes: The two lines that the hyperbolas come closer and closer

to touching. The equation of the asymptotes is always:

y =±a

Solved Examples:

S O LV E D E X A M P L E - A A F - 0 1

E Q U AT I O N O F H Y P E R B O L A F R O M V E R T I C E S

56. Find the equation of the hyperbola having vertices:

(±5,0), foci: (±8,0)

39− y2

14− y2

25+ y2

25− y2

Solution:

E Q U AT I O N O F T H E A S Y M P T O T E O F H Y P E R B O L A

57. What is the equation of the asymptote of the hyperbolax2

9− y2

(A) 2x - 3y = 0

(B) 3x - 2y = 0

(C) 2x - y = 0

(D) 2x + y = 0

Solution:

E Q U AT I O N O F H Y P E R B O L A G I V E N A S Y M P T O T E S

58. Find the equation of the hyperbola whose asymptotes are

y = ± 2x and which passes through (5/2, 3).

(A) 4x2 + y2 + 16 = 0

(B) 4x2 + y2 - 16 = 0

(C) x2 - 4y2 - 16 = 0

(D) 4x2 - y2 = 16

Solution:

I D E N T I F Y I N G C U R V E R E P R E S E N T E D B Y A N E Q U AT I O N

59. 4x2 - y2 = 16 is the equation of a/an?

(A) parabola

(B) hyperbola

(C) circle

(D) ellipse

Solution:

E C C E N T R I C I T Y O F C U R V E

60. Find the eccentricity of the curve:

9x2 −4y2 −36x +8y = 4

(A) 1.80

(B) 1.92

(C) 1.86

(D) 1.76

Solution:

D I S TA N C E F R O M F O C U S O F A H Y P E R B O L A

61. How far from the x-axis is the focus F of the hyperbola

x2 −2y2 +4x +4y +4 = 0?

(A) 4.5

(B) 3.4

(C) 2.7

(D) 2.1

Solution:

1.7 Distance Formula:

• Calculate distance between two points in space.

Distance between two points in the space P1 (x1, y1, z1) and P2 (x2, y2, z2)

is given by:

√(x2 −x1)2 + (y2 − y1)2 + (z2 − z1)2 (1.17)

This can be proved by repeated application of the Pythagorean Theo-

Solved Examples:

S O LV E D E X A M P L E - A AG - 0 1

D I S TA N C E B E T W E E N T W O P O I N T S I N 3 - D

62. The distance between the points (-5,-5,7) and (3,0,5) is:

(A) 4.531

(B) 5.657

(C) 9.643

(D) 93

Solution:

2 Calculus

Introduction

Calculus is the study of how things change. It provides a framework for

modeling systems in which there is change, and a way to deduce the pre-

dictions of such models. It is the branch of mathematics that deals with

the finding and properties of derivatives and integrals of functions. The

two main types are differential calculus and integral calculus.

2.1 Limits:

• Understand concepts of limits- Algebraic, trigonometric,

involving infinity, using L-Hospitals Rule.

A limit is the value that a function or sequence "approaches" as the

input or index approaches some value.

Sometimes, a function is undefined if it takes some exact value of ’x’. In

that case, we can find out where the function was approaching. This can

be done by removing the factors causing undefined nature or sometimes

by using specific mathematical identities.

L Hopital’s rule:

If a limit is in indeterminate form (such as0

∞∞ ) then,

limx→c

g (x)= limx→c f ′(x)

limx→c g ′(x)= limx→c f ′′(x)

limx→c g ′′(x)(2.1)

Solved Examples:

S O LV E D E X A M P L E - A B A - 0 1

1. If a function is continuous at a point, then:

(A) The limit of function may not exist at that point

(B) The function must be derivative at that point.

(C) Limit of the function at that point tends to infinity

(D) The limit must exist at that point and the value of this

limit should be same as value of the function at that

point.

Solution:

2. Let x denote a real number. Find out the INCORRECT

statement.

(A) S = x :x > 3 represents the set of all real numbers

greater than 3

(B) S = x : x2 <0 represents the empty set.

(C) S = x :x ∈ A and x ∈ B represents the union of set A and

set B.

(D) S = x :a < x < b represents the set of all real numbers

between a and b, where a and b are real numbers.

Solution:

3. Consider the function f(x) = | x | in the interval -1 ≤ x ≤ 1.

At the point x = 0, f (x) is:

(A) continuous and differentiable

(B) non-continuous and differentiable

(C) continuous and non-differentiable

(D) neither continuous nor differentiable

Solution:

limx→0

(1−cos x

Solution:

5. What is

limθ→0

equal to?

(A) θ

(B) sinθ

Solution:

limx→0

sin2 x

is equal to:

(D) -1

Solution:

7. Which of the following functions is not differentiable in

the domain [-1,1]?

(A) f (x) = x2

(B) f (x) = x - 1

(C) f (x) = 2

(D) f (x) = maximum (x, - x )

Solution:

8. At x = 0, the function f (x) = x3+ 1 has

(A) a maximum value

(B) a minimum value

(C) a singularity

(D) a point of inflection

Solution:

9. The minimum value of function y = x2 in the interval [1,

5] is:

(C) 25

(D) undefined

Solution:

10. The value of:

limx→8

x13 −2

x −8

Solution:

11. If

f (x) = 2x2 −7x +3

5x2 −12x −9

then l i mx→3 f (x) will be:

(A)−1

Solution:

limx→1

(1−x) tan(πx

(B) π

Solution:

13. Let f: R → R be defined by:

(3x2 +4)cos x

limh→0

f (h)+ f (−h)−8

is equal to :

(D) π

Solution:

x x + h

f(x + h)

Figure 2.1: First Principle of Derivative

2.2 Derivatives:

• Calculate derivative using basic formulae- algebraic,

trigonometric, inverse trigonometric, logarithmic and

exponential.

• Calculate derivative using rules of derivatives such as -

Addition/subtraction, product, quotient and chain rule.

• Apply differentiation to calculate tangent and normal, rate

of change, maxima/minima and errors/approximation.

First Principle of Derivative

f ′(x) = lim∆x→0

f (x +∆x)− f (x)

∆x(2.2)

Some Basic Formulae:

d xc = 0 (2.3)

d x(u ± v ±w ± ....) = du

d x± d v

d x± d w

d x± ..... (2.4)

d x(cu) = c

d x(2.5)

d x(uv) = u

d x+ v

d x(2.6)

v dud x −u

d xv2 (2.7)

Chain Rule :

d x= d y

d x(2.8)

List of Derivative Formulae:

Algebraic Group

d xxn = nxn−1 (2.9)

Trigonometric Group

d xsin x = cos x (2.10)

d xcos x =−sin x (2.11)

d xtan x = sec2 x (2.12)

d xcot x =−csc2 x (2.13)

d xsec x = sec x tan x (2.14)

d xcsc x =−csc x cot x (2.15)

Logarithmic and Exponential Group

d xex = ex (2.16)

d xax = ax log a (2.17)

d xlog x = 1

x(2.18)

Inverse Trigonometric Group

d xsin−1 x = 1p

1−x2(2.19)

d xcos−1 x = −1p

1−x2(2.20)

d xtan−1 x = 1

1+x2 (2.21)

d xcot−1 x = −1

1+x2 (2.22)

d xsec−1 x = 1

x2 −1(2.23)

d xcsc−1 x = −1

x2 −1(2.24)

Solved Problems

S O LV E D E X A M P L E - A B B - 0 1

14. Equation of the line normal to function f (x) = (x −8)23 +1

at P(0, 5) is:

(A) y = 3x - 5

(B) y = 3x + 5

(C) 3y = x + 15

(D) 3y = x - 15

Solution:

15. If

x = a(θ+ sinθ)

y = a(1−cosθ)

thend y

d xwill be equal to:

(A) sin

(B) cos

(C) tan

(D) cot

Solution:

16. For the following parametric function, what is thed 2 y

d t 2 at

x = 1− t 2

y = t − t 3

(A) ∞

(C) -1

Solution:

17. The minimum point of the function

f (x) =(

is at:

(A) x = 1

(B) x = - 1

(C) x = 0

(D) x =1p3

Solution:

18. The right circular cone of largest volume that can be en-

closed by a sphere of 1m radius has a height of:

Solution:

19. If

In = d n

d xn (xn log x),

In −nIn−1 =?

(B) n -1

(C) n!

(D) (n -1)!

Solution:

2.3 Partial Derivatives

• Define a partial derivative.

• Compute partial derivatives by viewing certain variables

as constant.

• Understand notation and computation for higher-order

partial derivatives.

When working with functions of more than one variable, the question

in calculus becomes: how can we evaluate the rate of change? The answer

is called a partial derivative. Given a function f(x, y) or f(x, y, z), the partial

derivative of f with respect to x, ∂ f = fx , is found by treating all variables

other than x as constants.

Technically, were finding

limh→0

f (x +h, y)− f (x, y)

The partial derivatives ∂ f = fy and ∂ f = fz have analogous definitions.

Solved Examples:

S O LV E D E X A M P L E - A B C - 0 1

20. Let f = y x . What is:∂2 f

∂x∂y

at x = 2, y = 1?

(B) ln2

Solution:

21. The function f (x, y) = 2x2 +2x y − y3 has:

(A) only one stationary point at (0, 0)

(B) two stationary points at (0, 0) and

6,−1

)(C) two stationary points at (0, 0) and (1, - 1)

(D) no stationary point

Solution:

22. If u =x + y

x − y, then

∂x+ ∂u

∂y= ??

x − y

x − y2

Solution:

23. If u = log(x2 + y2), then∂2u

∂x2 + ∂2u

∂y2 =??

x2 + y2

(C)x2 − y2

(x2 + y2)2

(D)y2 −x2

(x2 + y2)2

Solution:

2.4 Indefinite Integrals

• Calculate integration of a function using basic formulae -

algebraic, trigonometric, inverse trigonometric,

logarithmic and exponential.

• Calculate integration by substitution, by partial fractions,

by trigonometric substitution, by parts

Integration Formulae:

Algebraic Group

∫xnd x = xn+1

n +1+C (2.25)

Trigonometric Group

∫cos x d x = sin x +C (2.26)

∫sin x d x =−cos x +C (2.27)

∫sec2 x d x = tan x +C (2.28)

∫csc2 x d x =−cot x +C (2.29)

∫sec x tan x d x = sec x +C (2.30)

∫csc x cot x d x =−csc x +C (2.31)

Logarithmic and Exponential Group

∫ex d x = ex +C (2.32)

∫ax d x = ax

log a+C (2.33)

xd x = log x +C (2.34)

Inverse Trigonometric Group

1−x2d x = sin−1 x +C (2.35)

∫ −1p1−x2

d x = cos−1 x +C (2.36)

1+x2 d x = tan−1 x +C or −cot−1 x +C (2.37)

x2 −1d x = sec−1 x +C or −csc−1 x +C (2.38)

Integration by Substitution method

There are occasions when it is possible to perform an apparently difficult

piece of integration by first making a substitution.

I =∫

x sin(x2 +1)d x

Substitute

(x2 +1) = u

2xd x = du

xd x = 1

Then the integral becomes

∫sinudu

2cos(x2 +1)+C

Integration by Parts:

I =∫

x cos x d x

= uv −∫

= x sin x −∫

sin x d x

= x sin x +cos x +C

Solved Examples:

S O LV E D E X A M P L E - A B D - 0 1

24. ∫sec x

sec x + tan xd x

(A) tan x + sec x +C

(B) sec x − tan x +C

(C) tan x − sec x +C

(D) tan x −cot x +C

Solution:

S O LV E D E X A M P L E - A B D - 0 2

25. Evaluate ∫ p1+ sin5x d x

2−cos

Solution:

2.5 Integration by Partial Fractions:

• Distinguish between proper and improper fractions.

• Express an algebraic fraction as the sum of its partial

fractions.

For partial fractions, the given fraction must be proper, i.e the highest

degree of denominator polynomial must be higher than the highest degree

of numerator polynomial. if not, first division must be carried out to make

it a proper polynomial.

Rules on how to set Partial Fractions (with examples):

(x +2)(x +3)= A

(x +2)+ B

(x +3)

(x +2)2 = A

(x +2)+ B

(x +2)2

(x2 +2)(x +3)= Ax +B

(x2 +2)+ C

(x +3)

(x2 +2)(x +3)2 = Ax +B

(x2 +2)+ C

(x +3)+ D

(x +3)2

Solved Examples

S O LV E D E X A M P L E - A B E - 0 1

26. Evaluate: ∫1

x2 −4d x

4ln |x −2|/|x +2|+C

2ln |x −2|/|x +2|+C

4ln |x +2|/|x −2|+C

2ln |x +2|/|x −2|+C

Solution:

2.6 Integration Techniques for Trigonometric Functions:

• Use trigonometric identities to simplify and solve

integration problems.

1. ∫cos2 xd x =

∫1+cos2x

2d x + 1

∫cos2xd x

2+ sin2x

2×2+C

2. ∫sin2 xd x =

∫1−cos2x

2d x − 1

∫cos2xd x

2− sin2x

2×2+C

3. ∫x sin xd x = x(−cos x)− (1)(−sin x)+C

=−x cos x + sin x +C

4. ∫x cos xd x = x(sin x)− (1)(−cos x)+C

= x sin x +cos x +C

5. ∫sin x cos xd x =

∫2sin x cos x

(−cos2x

= −cos2x

Alternatively, the same integration can be calculated by the method of

substitution. ∫sin x cos xd x

sin x = t

cos xd x = d t

∫td t = t 2

= sin2 x

y = f(x)

even functionf(-x) = f(x)

odd functionf(-x) = - f(x)

2.7 Definite Integrals:

• Find the definite integral of algebraic,

trigonometric/inverse trigonometric, exponential and

logarithmic functions.

• Use properties of definite integrals.

af (x)d x = [F (x)]b

a = F (b)−F (a)

In definite integrals, there is no constant of integration, C.

Properties of Definite Integrals:

1. Change of variable has no effect on definite integration.∫ b

af (x)d x =

af (y)d y

2. If limits of definite integration are reversed, then the answer changes

sign. (because now the area under the curve given by f(x) is calculated

backwards.) ∫ b

af (x)d x =−

bf (x)d x

3. Definite integration can be split into two sub-definite integration as

long as the point of division is in the interval of original limits of def-

inite integration.∫ b

af (x)d x =

af (x)d x +

cf (x)d x.....a < c < b

4. special cases for odd and even functions.

For even function, f(-x) = f(x) i.e the graph is symmetrical about y-axis.

In such case, ∫ a

−af (x)d x = 2

0f (x)d x

For odd functions, f(-x) = -f(x) and graph of f(x) is mirrored about x-axis

AND y-axis. In such case, ∫ a

−af (x)d x = 0

This is because the ’negative’ area balances the ’positive’ area after the

origin.

5. ∫ 2a

0f (x)d x =

0f (x)d x +

0f (2a −x)d x

Solved Examples

S O LV E D E X A M P L E - A B G - 0 1

27. ∫ a

−a(sin6 x + sin7 x)d x

is equal to:

(A) 2∫ a

0 (sin6 x)d x

(B) 2∫ a

0 (sin7 x)d x

(C) 2∫ a

0 (sin6 x + sin7 x)d x

(D) zero

Solution:

28. The value of the integral:∫ ∞

−∞d x

(A) −π

(B)−π2

(D) π

Solution:

29. If f (x) is an even function and a is a positive real number,

then∫ a−a f (x)d x equals:

(C) 2a

(D) 2∫ a

0 f (x)d x

Solution:

30. Let f be a continuous and positive real valued function on

[0, 1]. Then ∫ π

0f (sin x)cos x d x

is equal to:

(C) -1

(D) ∞

Solution:

2.8 Area Bounded by a Curve:

• Apply integration to find area under a curve.

Area bounded by a curve is given by:

A =∫ b

af (x)d x

Solved Examples:

S O LV E D E X A M P L E - A B H - 0 1

31. The area bounded by two curves y = x2 and y2 = x be-

tween the lines x = 0.5 and x = 0.7 is:

(A) 0.345

(B) 0.399

(C) 0.921

(D) 0.082

Solution:

32. What is the area (in square units) bounded by the curve

y2 = x and the line x - 4 = 0?

Solution:

33. Find the area bounded by the parabolas y = 6x -x2and y

=x2 - 2x. Note: The parabolas intersect at points (0,0) and

(4,8).

(A) 44/3 square units

(B) 64/3 square units

(C) 74/3 square units

(D) 54/2 square units

Solution:

34. Find the area bounded by the parabola x2= 4y and y = 4.

(A) 21.33

(B) 33.21

(C) 31.32

(D) 13.23

Solution:

35. The area enclosed between the straight line y = x and the

parabola y = x2 in the x -y plane is:

Solution:

2.9 Volume of Revolution of Solids:

• Apply integration to find volume of a solid.

Volume of solid of revolution about x axis is given by,

V =π∫ b

[f (x)

]2 d x

Or, if the function is given in terms of x = f(y), then Volume of solid of rev-

olution is given by,

V =π∫ d

[f (y)

]2 d y

Solved Examples:

S O LV E D E X A M P L E - A B I - 0 1

36. Determine the volume of the solid obtained by rotating

the region bounded by 3p

x, and the x-axis about the x-

(A)128π

(B)144π

(C)96π

(D)72π

Solution:

37. The volume of an object expressed in spherical co-

ordinates is given by:∫ 2π

∫ π3

0r 2 sinφdr dφdθ

The value of the integral is:

Solution:

38. The area enclosed by the ellipse (x2)/9 + (y2)/4 = 1 is re-

volved about the line x = 3. What is the volume gener-

(A) 355.3

(B) 360.1

(C) 370.3

(D) 365.1

Solution:

39. The parabolic arc y =p

x, 1 ≤ x ≤ 2 is revolved around the

x-axis. The volume of the solid of revolution is:

(C)3π

(D)3π

Solution:

40. Find the volume (in cubic units) generated by rotating a

circle x2 + y2 + 6x + 4y + 12 = 0 about the y-axis.

(A) 39.48

(B) 47.23

(C) 59.22

(D) 62.11

Solution:

References:

1. Paul’s Online Math Notes:

http://tutorial.math.lamar.edu

2. http://www.math.umd.edu/~tjp/

131 09.2 lecture notes.pdf

3 Linear Algebra

3.1 Types of Matrices: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

3.2 Determinant of a Matrix: . . . . . . . . . . . . . . . . . . . . . . . . 138

3.3 Matrix Operations: . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Introduction

A matrix is a n-dimensional array of numbers or expressions arranged in

a set of rows and columns. A basic understanding of elementary matrix

algebra is essential for the analysis of engineering systems.

Matrices can be used to solve simultaneous equations.

3.1 Types of Matrices:

• Identify types of matrices such as Square, Diagonal,

Scalar, Row, Column, Identity.

Row Matrix

A matrix which has only one row.

[2 −6 4

Column Matrix:

A matrix which has only one column.

Square Matrix:

A matrix having same no. of rows and columns.3 6 1

2 0 −2

Diagonal Matrix:

A square matrix where only diagonal elements are present, non-diagonal

elements are zero. 3 0 0

0 0 −2

Scalar Matrix:

Diagonal matrix where all diagonal elements are identical.3 0 0

Identity Matrix:

Scalar Matrix where the diagonal elements are one.1 0 0

Singular Matrix:

A Matrix whose determinant is equal to zero.[2 3

−6 −9

Solved Examples:

S O LV E D E X A M P L E - AC A - 0 1

1. For which value of x will the matrix given below become

singular?

12 6 0

(D) 12

Solution:

3.2 Determinant of a Matrix:

• Find determinant of a matrix.

• Understand the conditions for zero determinant.

Only square matrices have determinant. (i.e. have the same number of

rows as columns).

For a 2 X 2 matrix, the determinant can be calculated as:

∣∣∣∣∣1 2

∣∣∣∣∣= 1×4−2×3 =−2

For a 3 X 3 matrix, the determinant can be calculated as:

∣∣∣∣∣∣∣a b c

∣∣∣∣∣∣∣• Multiply a by the determinant of the 2 × 2 matrix that is not in a’s row

or column.

• Likewise for b, and for c

• Add them up, making sure that b has a negative sign.

∣∣∣A∣∣∣= a

∣∣∣∣∣e f

∣∣∣∣∣−b

∣∣∣∣∣d f

∣∣∣∣∣+ c

∣∣∣∣∣d e

∣∣∣∣∣Conditions for Zero Determinant:

Sometimes, just by inspection, you may be able to predict that the deter-

minant of a matrix is zero. e.g.

1. If two rows are identical

2. If two columns are identical

3. if two rows (or two columns) are multiples of corresponding elements.

4. if all the elements in any row (or any column) are zero.

In the above example, -6 and -9 are (-3) times their counterparts from the

first row.

Solved Examples:

S O LV E D E X A M P L E - AC B - 0 1

2. If A and B are square matrices of size n x n, then which of

the following statement is not true?

(A) det. (AB) = det (A) det (B)

(B) det (kA) = kn det (A)

(C) det (A + B) = det (A) + det (B)

(D) det (AT ) =1/det (A−1)

Solution:

3.3 Matrix Operations:

• Add, subtract and multiply two matrices.

• Find inverse of a matrix

• Find rank of a matrix and understand its importance.

Scalar multiplication of a Matrix:

If a matrix is multiplied by scalar, then ALL the elements get multiplied by

that scalar. e.g.

2 0 −2

9 18 3

12 15 6

6 0 −6

Transpose of a Matrix:

If rows of a matrix are interchanged as column of the matrix, then the re-

sulting matrix is the transpose of the original one. e.g. the transpose of

: 3 6 1

2 0 −2

will be :

1 2 −2

Multiplication of Two Matrices:

A is m X n matrix(means having m rows and n columns), B is n X s matrix

(means having n rows and s columns), then they can be multiplied and the

result matrix C will be a m X s matrix.

In other words, columns of first matrix should match row of second ma-

trix. Also, the common dimension gets eliminated and the product matrix

does not have common dimension, neither as row nor as column. e.g. A

3 X 4 matrix ( 3 rows and 4 columns) can be multiplied with a 4X6 matrix,

because the ’inner’ dimensions (4 and 4) are matching, which gets ’elimi-

nated’ and the answer will be 3 X 6 matrix.

C = ci j =(

n∑l=1

ai l∗bl j

)FEsuccess.com Page 140

e.g., 3 4 2

1 2 −2

2 −2 0

3 2 −1

3+8+6 0−8+4 6+0−2

6+10+0 0−10+0 12+0+0

1+4−6 0−4−4 2+0+2

17 −4 4

16 −10 12

−1 −8 4

In matrix multiplication, the order in which the matrices are multiplied

is important. (In other words, matrix multiplication is NOT commutative.)

A×B 6= B × A

You may take the same matrices mentioned above and verify that the an-

swer is NOT the same.

Inverse of a Matrix :

For a square matrix A, the inverse is written A−1. When A is multiplied by

A−1 the result is the identity matrix I. Non-square matrices do not have

inverses.

Note: Not all square matrices have inverses. A square matrix which has

an inverse is called invertible or non singular, and a square matrix with-

out an inverse is called non-invertible or singular. A square matrix whose

determinant is zero, does not have inverse.

A A−1 = A−1 A = I

(AB)−1 = B−1 A−1

For a 2 X 2 matrix, the inverse can be easily found out by the following

formula:

if A = [a b

Then A−1 = 1

[d −b

−c a

]A−1 = 1

ad −bc

[d −b

−c a

These matrix operations (transpose, determinant, multiplication and

inverse) can be performed using a scientific calculator, so you must be

comfortable with the MATRIX or MAT mode of your calculator.

Solved Examples

S O LV E D E X A M P L E - AC C - 0 1

3. Multiplication of matrices E and F is G. matrices E and G

cosθ −sinθ 0

sinθ cosθ 0

What is the matrix F ?

cosθ −sinθ 0

sinθ cosθ 0

cosθ cosθ 0

−cosθ sinθ 0

cosθ sinθ 0

−sinθ cosθ 0

sinθ −cosθ 0

cosθ sinθ 0

Solution:

Rank of a Matrix

The rank of a matrix is the maximum number of independent rows (or, the

maximum number of independent columns). A square matrix A n × n is

non-singular only if its rank is equal to n.

4. Consider the system of simultaneous equations:

x +2y + z = 6

2x + y +2z = 6

x + y + z = 5

This system has:

(A) unique solution

(B) infinite number of solutions

(C) no solution

(D) exactly two solutions

Solution:

5. Which of the following is the inverse of the matrix?[−5 7

3 −4

[−4 −7

−3 −5

[5 −7

−3 4

(D) The inverse does not exist.

Solution:

6. Compute the product AB of the two matrices below.3 1

∗[−3 1

(A) −12 1

−5 5

−1 12

Solution:

7. If A =

[4 −1

5 −2

]and B =

−2 1

(AB)T =[

Solution:

8. If A =

[4 −1

5 −2

]and (AB)T =

]then B = ??

(A) [1 2

−2 1

(B) [−1 2

−2 1

(C) [−1 2

−2 1

(D) [1 2

Solution:

9. What is the value of matrix element c(2,3), given C = A +

B + A, where:

0 2 −1

1 3 −1

2 −1 −1

−2 1 1

(D) 15

Solution:

10. The solution to the system of equations[2 5

−4 3

(A) 6,2

(B) -6,2

(C) -6,-2

(D) 6,-2

Solution:

11. If A and B be real symmetric matrices of size n x n, then:

(A) AAT = 1

(B) A = A−1

(C) AB = BA

(D) (AB)T = BA

Solution:

x +2y + z = 4

2x + y +2z = 5

x − y + z = 1

The system of algebraic equations given above has:

(A) a unique solution of x = 1, y = 1 and z = 1.

(B) only the two solutions of (x = 1, y = 1, z = 1) and (x = 2,

y = 1, z = 0)

(C) infinite number of solutions

(D) no feasible solution

Solution:

References:

1. http://home.scarlet.be/math/matr.htm

2. http://linear.ups.edu/

version3/pdf/fcla-draft-solutions.pdf

4 Vector Analysis