Engineering Mathematics By Sivaramakrishna Das Pdf
Engineering Mathematics
P. Sivaramakrishna Das, C. Vijayakumari
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About Pearson Pearson is the world's learning company, with presence across 70 countries worldwide. Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world. We believe learning opens up opportunities, creates fulfilling careers and hence better lives. We hence collaborate with the best of minds to deliver you classleading products, spread across the Higher Education and K12 spectrum. Superior learning experience and improved outcomes are at the heart of everything we do. This product is the result of one such effort. Your feedback plays a critical role in the evolution of our products and you can reachus@pearson.com. We look forward to it. A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 1 3/2/2017 6:17:51 PM This page is intentionally left blank Engineering Mathematics P. Sivaramakrishna Das Professor of Mathematics and Head of the P.G. Department of Mathematics (Retired) Ramakrishna Mission Vivekananda College Mylapore, Chennai Presently Professor of Mathematics and Head of the Department of Science and Humanities K.C.G College of Technology (a unit of Hindustan Group of Institutions Karapakkam, Chennai) C. Vijayakumari Professor of Mathematics (Retired) Queen Mary's College (Autonomous) Mylapore, Chennai A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 3 3/2/2017 6:17:52 PM Copyright © 2017 Pearson India Education Services Pvt. Ltd Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128, formerly known as TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia. No part of this eBook may be used or reproduced in any manner whatsoever without the publisher's prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISB; N 978-93-325-1912-1 eISBN 978-93-325-8776-2 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India. Registered office: 4th Floor, Software Block, Elnet Software City, TS-140, Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India. Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email: companysecretary.india@pearson.com A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 4 3/2/2017 6:17:52 PM Dedicated to Our Beloved Parents A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 5 3/2/2017 6:17:52 PM This page is intentionally left blank Brief Contents Prefacexxxi About the Authors xxxiii A. ALGEBRA 1. Matrices 1.1 2. Sequences and Series 2.1 B. CALCULUS 3. Differential Calculus 3.1 4. Applications of Differential Calculus 4.1 5. Differential Calculus of Several Variables 5.1 6. Integral Calculus 6.1 7. Improper Integrals 7.1 8. Multiple Integrals 8.1 9. Vector Calculus 9.1 C. DIFFERENTIAL EQUATIONS 10. Ordinary First Order Differential Equations 10.1 11. Ordinary Second and Higher Order Differential Equations 11.1 12. Applications of Ordinary Differential Equations 12.1 13. Series Solution of Ordinary Differential Equations and Special Functions 13.1 14. Partial Differential Equations 14.1 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 7 3/2/2017 6:17:52 PM viii n Brief Contents D. COMPLEX ANALYSIS 15. Analytic Functions 15.1 16. Complex Integration 16.1 E. SERIES AND TRANSFORMS 17. Fourier Series 17.1 18. Fourier Transforms 18.1 19. Laplace Transforms 19.1 F. APPLICATIONS 20. Applications of Partial Differential Equations Index A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 8 20.1 I.1 3/2/2017 6:17:52 PM Contents Prefacexxxi About the Authors xxxiii 1. Matrices 1.1 1.0 Introduction 1.1 1.1 Basic Concepts 1.1 1.1.1 Basic Operations on Matrices 1.4 1.1.2 Properties of Addition, Scalar Multiplication and Multiplication 1.5 1.2 Complex Matrices 1.7 Worked Examples 1.10 Exercise 1.1 1.13 Answers to Exercise 1.1 1.14 1.3 Rank of a Matrix 1.14 Worked Examples 1.16 Exercise 1.21.23 Answers to Exercise 1.21.24 1.4 Solution of System of Linear Equations 1.24 1.4.1 Non-homogeneous System of Equations 1.24 1.4.2 Homogeneous System of Equations 1.25 1.4.3 Type 1: Solution of Non-homogeneous System of Equations 1.26 Worked Examples1.26 1.4.4 Type 2: Solution of Non-homogeneous Linear Equations Involving Arbitrary Constants 1.34 Worked Examples1.34 1.4.5 Type 3: Solution of the System of Homogeneous Equations 1.38 Worked Examples1.38 1.4.6 Type 4: Solution of Homogeneous System of Equation Containing Arbitrary Constants 1.41 Worked Examples1.41 Exercise 1.31.44 Answers to Exercise 1.31.45 1.5 Matrix Inverse by Gauss–Jordan method 1.46 Worked Examples1.47 Exercise 1.41.53 Answers to Exercise 1.41.53 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 9 3/2/2017 6:17:52 PM x n Contents 1.6 Eigen Values and Eigen Vectors 1.54 1.6.0 Introduction 1.54 1.6.1 Vector 1.54 Worked Examples 1.55 1.6.2 Eigen Values and Eigen Vectors 1.56 1.6.3 Properties of Eigen Vectors 1.57 Worked Examples 1.58 1.6.4 Properties of Eigen Values 1.67 Worked Examples 1.70 Exercise 1.5 1.72 Answers to Exercise 1.5 1.73 1.6.5 Cayley-Hamilton Theorem 1.73 Worked Examples 1.75 Exercise 1.6 1.82 Answers to Exercise 1.6 1.83 1.7 Similarity Transformation and Orthogonal Transformation 1.83 1.7.1 Similar Matrices 1.83 1.7.2 Diagonalisation of a Square Matrix 1.84 1.7.3 Computation of the Powers of a Square Matrix 1.85 1.7.4 Orthogonal matrix 1.86 1.7.5 Properties of orthogonal matrix 1.86 1.7.6 Symmetric Matrix 1.87 1.7.7 Properties of Symmetric Matrices 1.88 1.7.8 Diagonalisation by Orthogonal Transformation or Orthogonal Reduction 1.89 Worked Examples 1.90 1.8 Real Quadratic Form. Reduction to Canonical Form 1.96 Worked Examples 1.99 Exercise 1.7 1.111 Answers to Exercise 1.7 1.112 Short Answer Questions1.113 Objective Type Questions1.114 Answers1.116 2. Sequences and Series 2.0 Introduction 2.1 Sequence 2.1.1 Infinite Sequence 2.1.2 Finite sequence 2.1.3 Limit of a sequence 2.1.4 Convergent sequence 2.1.5 Oscillating sequence 2.1.6 Bounded sequence 2.1.7 Monotonic Sequence Worked Examples Exercise 2.1 Answers to Exercise 2.1 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 10 2.1 2.1 2.1 2.1 2.2 2.2 2.2 2.2 2.2 2.3 2.3 2.9 2.9 3/2/2017 6:17:52 PM Contents n xi 2.2 Series 2.9 2.2.1 Convergent Series 2.9 2.2.2 Divergent series2.10 2.2.3 Oscillatory series2.10 2.2.4 General properties of series2.10 2.3 Series of Positive Terms 2.10 2.3.1 Necessary Condition for Convergence of a Series 2.10 2.3.2 Test for convergence of positive term series2.11 2.3.3 Comparison tests2.11 Worked Examples2.13 Exercise 2.22.17 Answers to Exercise 2.22.18 2.3.4 De' Alembert's Ratio Test 2.18 Worked Examples2.21 Exercise 2.32.25 Answers to Exercise 2.32.26 2.3.5 Cauchy's Root Test 2.27 Worked Examples2.28 2.3.6 Cauchy's Integral Test 2.30 Worked Examples2.32 Exercise 2.42.36 Answers to Exercise 2.42.36 2.3.7 Raabe's Test 2.36 Worked Examples2.37 Exercise 2.52.41 Answers to Exercise 2.52.42 2.3.8 Logarithmic Test 2.42 Worked Examples2.44 2.4 Alternating Series 2.46 2.4.1 Leibnitz's Test 2.46 Worked Examples2.47 2.5 Series of Positive and Negative Terms 2.50 2.5.1 Absolute Convergence and Conditional Convergence 2.50 2.5.2 Tests for absolute convergence2.50 Worked Examples2.51 Exercise 2.62.55 Answers to Exercise 2.62.55 2.6 Convergence of Binomial Series 2.56 2.7 Convergence of the Exponential Series 2.57 2.8 Convergence of the Logarithmic Series 2.57 2.9 Power Series 2.58 2.9.1 Hadmard's Formula 2.59 2.9.2 Properties of power series2.60 Worked Examples2.60 Exercise 2.72.66 Answers to Exercise 2.72.67 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 11 3/2/2017 6:17:52 PM xii n Contents Short Answer Questions Objective Type Questions Answers 2.67 2.69 2.70 3. Differential Calculus 3.1 3.0 Introduction 3.1 3.1 Successive Differentiation 3.2 Worked Examples 3.3 Exercise 3.1 3.6 3.1.1 The nth Derivative of Standard Functions 3.7 Worked Examples 3.11 Exercise 3.2 3.16 Answers to Exercise 3.2 3.17 Worked Examples 3.18 Exercise 3.33.24 3.2 Applications of Derivative 3.25 3.2.1 Geometrical Interpretation of Derivative 3.25 3.2.2 Equation of the Tangent and the Normal to the Curve y = f(x)3.25 Worked Examples3.26 Exercise 3.43.33 Answers to Exercise 3.43.34 3.2.3 Length of the Tangent, the Sub-Tangent, the Normal and the Sub-normal 3.34 Worked Examples3.36 Exercise 3.53.38 Answers to Exercise 3.53.38 3.2.4 Angle between the Two Curves 3.38 Worked Examples3.39 Exercise 3.63.42 Answers to Exercise 3.63.43 3.3 Mean-value Theorems of Derivatives 3.43 3.3.1 Rolle's Theorem 3.43 Worked Examples3.44 3.3.2 Lagrange's Mean Value Theorem 3.47 Worked Examples3.49 3.3.3 Cauchy's Mean Value Theorem 3.53 Worked Examples3.54 Exercise 3.73.56 Answers to Exercise 3.73.58 3.4 Monotonic Functions 3.58 3.4.1 Increasing and Decreasing Functions 3.58 3.4.2 Piece-wise Monotonic Function 3.58 3.4.3 Test for increasing or decreasing functions3.59 Worked Examples3.60 Exercise 3.83.65 Answers to Exercise 3.83.66 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 12 3/2/2017 6:17:52 PM Contents n xiii 3.5 Generalised Mean Value Theorem 3.66 3.5.1 Taylor's Theorem with Lagrange's Form of Remainder 3.66 3.5.2 Taylor's series 3.68 3.5.3 Maclaurin's theorem with Lagrange's form of remainder 3.68 3.5.4 Maclaurin's series 3.68 Worked Examples 3.69 Exercise 3.9 3.74 Answers to Exercise 3.9 3.74 3.5.5 Expansion by Using Maclaurin's Series of Some Standard Functions 3.75 Worked Examples 3.75 3.5.6 Expansion of Certain Functions Using Differential Equations 3.78 Worked Examples 3.78 Exercise 3.10 3.81 Answers to Exercise 3.10 3.82 3.6 Indeterminate Forms 3.82 0 3.6.1 General L'Hopital's Rule for Form 3.84 0 Worked Examples 3.85 Exercise 3.11 3.94 Answers to Exercise 3.11 3.94 3.7 Maxima and Minima of a Function of One Variable 3.94 3.7.1 Geometrical Meaning 3.96 3.7.2 Tests for Maxima and Minima 3.96 Summary 3.97 Worked Examples 3.97 Exercise 3.12 3.103 Answers to Exercise 3.123.104 3.8 Asymptotes 3.104 Worked Examples3.105 3.8.1 A General Method 3.108 3.8.2 Asymptotes Parallel to the Coordinates Axes 3.110 Worked Examples 3.110 3.8.3 Another Method for Finding the Asymptotes 3.113 Worked Examples 3.114 3.8.4 Asymptotes by Inspection 3.115 Worked Examples 3.116 3.8.5 Intersection of a Curve and Its Asymptotes 3.116 Worked Examples 3.116 Exercise 3.13 3.121 Answers to Exercise 3.13 3.122 3.9 Concavity 3.122 Worked Examples3.124 Exercise 3.143.127 Answers to Exercise 3.143.128 3.10 Curve Tracing 3.128 3.10.1 Procedure for Tracing the Curve Given by the Cartesian Equation f(x, y) = 0. 3.128 Worked Examples3.129 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 13 3/2/2017 6:17:52 PM xiv n Contents 3.10.2 Procedure for Tracing of Curve Given by Parametric Equations x = f(t), y = g(t)3.137 Worked Examples3.137 3.10.3 Procedure for Tracing of Curve Given by Equation in Polar Coordinates f(r, u) = 0 3.141 Worked Examples3.142 Exercise 3.153.146 Answers to Exercise 3.153.146 Short Answer Questions3.148 Objective Type Questions3.149 Answers3.152 4. Applications of Differential Calculus 4.1 Curvature in Cartesian Coordinates 4.1.0 Introduction 4.1.1 Measure of Curvature 4.1.2 Radius of Curvature for Cartesian Equation of a Given Curve 4.1.3 Radius of Curvature for Parametric Equations Worked Examples 4.1.4 Centre of Curvature and Circle of Curvature 4.1.5 Coordinates of the Centre of Curvature Worked Examples Exercise 4.1 Answers to Exercise 4.1 4.1.6 Radius of Curvature in Polar Coordinates Worked Examples 4.1.7 Radius of Curvature at the Origin Worked Examples 4.1.8 Pedal Equation or p - r Equation of a Curve Worked Examples 4.1.9 Radius of Curvature Using the p - r Equation of a Curve Worked Examples Exercise 4.2 Answers to Exercise 4.2 4.2 Evolute 4.2.1 Properties of Evolute 4.2.2 Procedure to Find the Evolute Worked Examples Exercise 4.3 Answers to Exercise 4.3 4.3 Envelope 4.3.1 Method of Finding Envelope of Single Parameter Family of Curves Worked Examples 4.3.2 Envelope of Two Parameter Family of Curves Worked Examples A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 14 4.1 4.1 4.1 4.1 4.2 4.4 4.4 4.11 4.12 4.13 4.15 4.16 4.17 4.19 4.22 4.23 4.25 4.26 4.28 4.29 4.30 4.31 4.31 4.31 4.34 4.34 4.41 4.41 4.42 4.42 4.43 4.45 4.45 3/2/2017 6:17:53 PM Contents n xv 4.3.3 Evolute as the Envelope of Normals 4.48 Worked Examples4.49 Exercise 4.44.52 Answers to Exercise 4.44.53 Short Answer Questions4.54 Objective Type Questions4.54 Answers4.56 5. Differential Calculus of Several Variables 5.1 5.0 Introduction 5.1 5.1 Limit and Continuity 5.1 Worked Examples 5.4 Exercise 5.1 5.6 Answers to Exercise 5.1 5.6 5.2 Partial Derivatives 5.6 ∂z ∂z 5.7 5.2.1 Geometrical Meaning of , ∂x ∂y 5.2.2 Partial Derivatives of Higher Order 5.8 5.2.3 Homogeneous Functions and Euler's Theorem 5.8 Worked Examples 5.9 5.2.4 Total Derivatives 5.15 Worked Examples 5.17 Exercise 5.25.24 Answers to Exercise 5.25.26 5.3 Jacobians 5.26 5.3.1 Properties of Jacobians 5.27 Worked Examples5.29 5.3.2 Jacobian of Implicit Functions 5.35 Worked Examples5.35 Exercise 5.35.37 Answers to Exercise 5.35.38 5.4 Taylor's Series Expansion for Function of Two Variables 5.38 Worked Examples5.39 Exercise 5.45.44 Answers to Exercise 5.45.44 5.5 Maxima and Minima for Functions of Two Variables 5.45 5.5.1 Necessary Conditions for Maximum or Minimum 5.46 5.5.2 Sufficient Conditions for Extreme Values of f (x, y ) 5.46 5.5.3 Working Rule to Find Maxima and Minima of f (x, y ) 5.46 Worked Examples5.47 5.5.4 Constrained Maxima and Minima 5.51 5.5.5 Lagrange's Method of (undetermined) Multiplier 5.51 5.5.6 Method to Decide Maxima or Minima 5.52 Worked Examples5.56 Exercise 5.55.65 Answers to Exercise 5.55.66 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 15 3/2/2017 6:17:53 PM xvi n Contents 5.6 Errors and Approximations Worked Examples Exercise 5.6 Answers to Exercise 5.6 Short Answer Questions Objective Type Questions Answers 5.67 5.68 5.72 5.73 5.73 5.74 5.76 6. Integral Calculus 6.1 6.0 Introduction 6.1 6.1 Indefinite Integral 6.1 6.1.1 Properties of Indefinite Integral 6.1 6.1.2 Integration by Parts 6.3 6.1.3 Bernoulli's Formula 6.3 6.1.4 Special Integrals 6.3 Worked Examples 6.4 Exercise 6.1 6.9 Answers to Exercise 6.1 6.9 6.2 Definite Integral (Newton–Leibnitz formula) 6.10 6.2.1 Properties of Definite Integral 6.10 Worked Examples6.15 Exercise 6.26.27 Answers to Exercise 6.26.27 b 6.3 Definite Integral ∫ f ( x) dx as Limit of a Sum 6.28 a 6.3.1 Working Rule 6.28 Worked Examples6.29 Exercise 6.36.32 Answers to Exercise 6.36.33 6.4 Reduction Formulae 6.33 n n 6.4.1 The Reduction Formula for (a) ∫ sin x dx and (b) ∫ cos x dx 6.33 ∫ tan x dx and (b) ∫ cot x dx 6.36 The Reduction Formula for (a) ∫ sec x dx and (b) ∫ cosec x dx 6.37 6.4.2 The Reduction Formula for (a) n n n n 6.4.3 Worked Examples6.38 6.4.4 The Reduction Formula for ∫ sin m x cos n x dx , Where m, n are Non-negative Integers6.45 Worked Examples6.47 6.4.5 The Reduction Formula for (a) xm (log x) ndx, (b) xn sin mx dx, (c) xn cos mx dx6.49 ax m ax n 6.4.6 The Reduction Formula for (a) ∫ e sin x dx and (b) ∫ e cos x dx 6.51 ∫ A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 16 ∫ ∫ 3/2/2017 6:17:54 PM Contents n xvii 6.4.7 The Reduction Formula for (a) ∫ cos m x sin nx dx and (b) ∫ cos m x cos nx dx 6.52 Exercise 6.4 6.55 Answers to Exercise 6.4 6.55 6.5 Application of Integral Calculus 6.55 6.5.1 Area of Plane Curves 6.56 6.5.1 (a) Area of Plane Curves in Cartesian Coordinates 6.56 Worked Examples 6.57 Exercise 6.5 6.66 Answers to Exercise 6.5 6.67 6.5.1 (b) Area in Polar Coordinates 6.67 Worked Examples 6.68 Exercise 6.6 6.72 Answers to Exercise 6.6 6.72 6.5.2 Length of the Arc of a Curve 6.72 6.5.2 (a) Length of the Arc in Cartesian Coordinates 6.72 Worked Examples 6.73 Exercise 6.7 6.78 Answers to Exercise 6.7 6.79 6.5.2 (b) Length of the Arc in Polar Coordinates 6.79 Worked Examples 6.80 Exercise 6.8 6.81 Answers to Exercise 6.8 6.81 6.5.3 Volume of Solid of Revolution 6.82 6.5.3(a) Volume in Cartesian Coordinates 6.82 Worked Examples 6.83 Exercise 6.9 6.89 Answers to Exercise 6.9 6.90 6.5.3 (b) Volume in Polar Coordinates 6.91 Worked Examples 6.91 Exercise 6.10 6.93 Answers to Exercise 6.10 6.93 6.5.4 Surface Area of Revolution 6.93 6.5.4(a) Surface Area of Revolution in Cartesian Coordinates 6.93 Worked Examples 6.94 Exercise 6.11 6.99 Answers to Exercise 6.11 6.99 6.5.4 (b) Surface Area in Polar Coordinates 6.100 Worked Examples6.100 Exercise 6.126.102 Answers to Exercise 6.126.103 Short Answer Questions6.103 Objective Type Questions6.103 Answers6.106 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 17 3/2/2017 6:17:55 PM xviii n Contents 7. Improper Integrals 7.1 7.1 Improper Integrals 7.1 7.1.1 Kinds of Improper Integrals and Their Convergence 7.1 Worked Examples 7.4 Exercise 7.17.13 Answers to Exercise 7.17.13 7.1.2 Tests of Convergence of Improper Integrals 7.14 Worked Examples 7.15 Exercise 7.27.27 Answers to Exercise 7.27.27 7.2 Evaluation of Integral by Leibnitz's Rule 7.27 7.2.1 Leibnitz's Rule—Differentiation Under Integral Sign for Variable Limits 7.28 Worked Examples7.28 Exercise 7.3 7.47 Answers to Exercise 7.37.47 7.3 Beta and Gamma functions 7.47 7.3.1 Beta Function 7.47 7.3.2 Symmetric property of beta function7.48 7.3.3 Different forms of beta function7.48 7.4 The Gamma Function 7.49 7.4.1 Properties of Gamma Function 7.50 7.4.2 Relation between Beta and Gamma Functions 7.51 Worked Examples7.55 Exercise 7.47.69 Answers to Exercise 7.47.69 7.5 The Error Function 7.70 7.5.1 Properties of Error Functions 7.70 7.5.2 Series expansion for error function7.71 7.5.3 Complementary error function7.71 Worked Examples7.72 Exercise 7.57.76 Answers to Exercise 7.57.76 Short Answer Questions7.76 Objective Type Questions7.77 Answers7.78 8. Multiple Integrals 8.1 Double Integration 8.1.1 Double Integrals in Cartesian Coordinates 8.1.2 Evaluation of Double Integrals Worked Examples Exercise 8.1 Answers to Exercise 8.1 8.1.3 Change of Order of Integration Worked Examples A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 18 8.1 8.1 8.1 8.2 8.3 8.6 8.7 8.7 8.8 3/2/2017 6:17:55 PM Contents n xix Exercise 8.28.15 Answers to Exercise 8.28.15 8.1.4 Double Integral in Polar Coordinates 8.16 Worked Examples8.16 8.1.5 Change of Variables in Double Integral 8.19 Worked Examples8.19 Exercise 8.38.26 Answers to Exercise 8.38.27 8.1.6 Area as Double Integral 8.27 Worked Examples8.28 Exercise 8.48.31 Answers to Exercise 8.48.31 Worked Examples8.32 Exercise 8.58.37 Answers to Exercise 8.58.37 8.2 Area of a Curved Surface 8.37 8.2.1 Surface Area of a Curved Surface 8.38 8.2.2 Derivation of the Formula for Surface Area 8.38 8.2.3 Parametric Representation of a Surface 8.41 Worked Examples8.41 Exercise 8.68.49 Answers to Exercise 8.68.49 8.3 Triple Integral in Cartesian Coordinates 8.49 Worked Examples8.50 Exercise 8.78.55 Answers to Exercise 8.78.56 8.3.1 Volume as Triple Integral 8.56 Worked Examples8.56 Exercise 8.88.63 Answers to Exercise 8.88.64 Short Answer Questions8.64 Objective Type Questions8.64 Answers8.66 9. Vector Calculus 9.0 Introduction 9.1 Scalar and Vector Point Functions 9.1.1 Geometrical Meaning of Derivative 9.2 Differentiation Formulae 9.3 Level Surfaces 9.4 Gradient of a Scalar Point Function or Gradient of a Scalar Field 9.4.1 Vector Differential Operator 9.4.2 Geometrical Meaning of ∇φ 9.4.3 Directional Derivative 9.4.4 Equation of Tangent Plane and Normal to the Surface A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 19 9.1 9.1 9.1 9.2 9.3 9.4 9.4 9.4 9.4 9.5 9.5 3/2/2017 6:17:55 PM xx n Contents 9.4.5 Angle between Two Surfaces at a Common Point 9.6 9.4.6 Properties of gradients 9.6 Worked Examples 9.8 Exercise 9.1 9.20 Answers to Exercise 9.1 9.21 9.5 Divergence of a Vector Point Function or Divergence of a Vector Field 9.22 9.5.1 Physical Interpretation of Divergence 9.22 9.6 Curl of a Vector Point Function or Curl 9.23 1 of a Vector Field 9.6.1 Physical Meaning of Curl wF = curlv 9.23 2 Worked Examples 9.24 Exercise 9.2 9.30 Answers to Exercise 9.2 9.31 9.7 Vector Identities 9.31 Worked Examples 9.37 9.8 Integration of Vector Functions 9.39 9.8.1 Line Integral 9.40 Worked Examples 9.40 Exercise 9.3 9.46 Answers to Exercise 9.3 9.47 9.9 Green's Theorem in a Plane 9.47 9.9.1 Vector Form of Green's Theorem 9.50 Worked Examples 9.50 9.10 Surface Integrals 9.56 9.10.1 Evaluation of Surface Integral 9.57 9.11 Volume Integral 9.58 Worked Examples 9.58 9.12 Gauss Divergence Theorem 9.62 9.12.1 Results Derived from Gauss Divergence Theorem 9.64 Worked Examples 9.68 9.13 Stoke's Theorem 9.81 Worked Examples 9.83 Exercise 9.4 9.97 Answers to Exercise 9.49.100 Short Answer Questions9.100 Objective Type Questions9.101 Answers9.102 10. Ordinary First Order Differential Equations 10.0 Introduction 10.1 Formation of Differential Equations Worked Examples Exercise 10.1 Answers to Exercise 10.1 10.2 First Order and First Degree Differential Equations 10.2.1 Type I Variable Separable Equations A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 20 10.1 10.1 10.2 10.2 10.5 10.6 10.6 10.6 3/2/2017 6:17:55 PM Contents n xxi Worked Example 10.6 Exercise 10.2 10.9 Answers to Exercise 10.2 10.9 10.2.2 Type II Homogeneous Equation 10.10 Worked Examples10.10 Exercise 10.310.13 Answers to Exercise 10.310.14 10. 2.3 Type III Non-Homogenous Differential Equations of the First Degree 10.14 Worked Examples10.16 Exercise 10.410.21 Answers to Exercise 10.410.21 10.2.4 Type IV Linear Differential Equation 10.22 Worked Examples10.23 Exercise 10.510.27 Answers to Exercise 10.510.27 10.2.5 Type V Bernoulli's Equation 10.28 Worked Examples10.28 Exercise 10.610.31 Answers to Exercise 10.610.31 10.2.6 Type VI Riccati Equation 10.31 Worked Examples10.33 Exercise 10.710.36 Answers to Exercise 10.710.36 10.2.7 Type VII First Order Exact Differential Equations 10.37 Worked Examples10.39 Exercise 10.810.41 Answers to Exercise 10.810.42 10.3 Integrating Factors 10.42 Worked Examples10.43 10.3.1 Rules for Finding the Integrating Factor for Non-Exact Differential Equation Mdx + Ndy = 0 10.45 Worked Examples10.46 Exercise 10.910.56 Answers to Exercise 10.910.56 10.4 Ordinary Differential Equations of the First Order but of Degree Higher than One 10.56 10.4.1 Type 1 Equations Solvable for p 10.57 Worked Examples10.57 Exercise 10.1010.59 Answers to Exercise 10.1010.60 10.4.2 Type 2 Equations Solvable for y10.60 Worked Examples10.61 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 21 3/2/2017 6:17:55 PM xxii n Contents 10.4.3 Type 3 Equations Solvable for x10.64 Worked Examples10.65 Exercise 10.1110.67 Answers to Exercise 10.1110.67 10.4.4 Type 4 Clairaut's Equation 10.67 Worked Examples10.68 Exercise 10.1210.71 Answers to Exercise 10.1210.71 Short Answer Questions10.71 Objective Type Questions10.72 Answers10.74 11. Ordinary Second and Higher Order Differential Equations 11.1 11.0 Introduction 11.1 11.1 Linear Differential Equation with Constant Coefficients 11.1 11.1.1 Complementary Function 11.1 11.1.2 Particular Integral 11.2 Worked Examples 11.3 Exercise 11.111.19 Answers to Exercise 11.111.19 11.2 Linear Differential Equations with Variable Coefficients 11.21 11.2.1 Cauchy's Homogeneous Linear Differential Equations 11.21 Worked Examples11.22 11.2.2 Legendre's Linear Differential Equation 11.29 Worked Examples11.30 Exercise 11.211.32 Answers to Exercise 11.211.33 11.3 Simultaneous Linear Differential Equations with Constant Coefficients 11.34 Worked Examples11.34 Exercise 11.311.43 Answers to Exercise 11.311.44 11.4 Method of Variation of Parameters 11.44 11.4.1 Working rule11.45 Worked Examples11.45 Exercise 11.411.51 Answers to Exercise 11.411.52 11.5 Method of Undetermined Coefficients 11.52 Worked Examples11.54 Exercise 11.511.60 Answers to Exercise 11.511.60 Short Answers Questions11.60 Objective Type Questions11.61 Answers 11.63 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 22 3/2/2017 6:17:55 PM Contents n xxiii 12. Applications of Ordinary Differential Equations 12.1 12.0 Introduction 12.1 12.1 Applications of Ordinary Differential Equations of First Order 12.1 12.1.1 Law of Growth and Decay 12.1 12.1.2 Newton's Law of Cooling of Bodies 12.2 Worked Examples 12.2 Exercise 12.1 12.7 Answers To Exercise 12.1 12.8 12.1.3 Chemical Reaction and Solutions 12.8 Worked Examples 12.9 Exercise 12.212.12 Answers to Exercise 12.212.13 12.1.4 Simple Electric Circuit 12.13 Worked Examples12.14 Exercise 12.312.19 Answers to Exercise 12.312.19 12.1.5 Geometrical Applications 12.20 12.1.5 (a) Orthogonal Trajectories in Casterian Coordinates 12.20 Worked Examples12.21 12.1.5 (b) Orthogonal Trajectories in Polar Coordinates 12.23 Worked Examples12.24 Exercise 12.412.26 Answers to Exercise 12.412.27 12.2 Applications of Second Order Differential Equations 12.27 12.2.1 Bending of Beams 12.27 Worked Examples12.29 12.2.2 Electric Circuits 12.34 Worked Examples12.34 Exercise 12.512.38 Answers to Exercise 12.512.39 12.2.3 Simple Harmonic Motion (S.H.M) 12.40 Worked Examples12.41 Exercise 12.612.43 Answers to Exercise 12.612.44 Objective Type Questions12.44 Answers12.45 13. Series Solution of Ordinary Differential Equations and Special Functions 13.0 Introduction 13.1 Power Series Method 13.1.1 Analytic Function 13.1.2 Regular Point 13.1.3 Singular Point 13.1.4 Regular and Irregular Singular Points Worked Examples Exercise 13.1 Answers to Exercise 13.1 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 23 13.1 13.1 13.1 13.1 13.2 13.2 13.2 13.3 13.9 13.9 3/2/2017 6:17:55 PM xxiv n Contents 13.2 13.3 13.4 Frobenius Method 13.9 Worked Examples13.11 Exercise 13.213.33 Answers to Exercise 13.213.33 Special Functions 13.34 Bessel Functions 13.34 13.4.1 Series Solution of Bessel's Equation 13.34 13.4.2 Bessel's Functions of the First Kind 13.37 Worked Examples13.39 13.4.3 Some Special Series 13.40 13.4.4 Recurrence Formula for Jn (x)13.41 13.4.5 Generating Function for Jn (x) of Integral Order 13.44 Worked Examples13.46 13.4.6 Integral Formula for Bessel's Function Jn (x) 13.49 Worked Examples13.53 13.4.7 Orthogonality of Bessel's Functions 13.56 13.4.8 Fourier–Bessel Expansion of a Function f(x)13.59 Worked Examples13.60 13.4.9 Equations Reducible to Bessel's Equation 13.62 Worked Examples13.62 Exercise 13.313.65 Answers to Exercise 13.313.66 13.5 Legendre Functions 13.66 13.5.1 Series Solution of Legendre's Differential Equation 13.66 13.5.2 Legendre Polynomials 13.71 13.5.3 Rodrigue's Formula 13.71 Worked Examples13.73 13.5.4 Generating Function for Legendre Polynomials 13.74 Worked Examples13.75 13.5.5 Orthogonality of Legendre Polynomials in [-1, 1] 13.77 Worked Examples13.80 13.5.6 Fourier–Legendre Expansion of f(x) in a Series of Legendre Polynomials13.83 Worked Examples13.83 Exercise 13.413.85 Answers to Exercise 13.413.85 14. Partial Differential Equations 14.1 14.0 Introduction 14.1 14.1 Order and Degree of Partial Differential Equations 14.1 14.2 Linear and Non-linear Partial Differential Equations 14.1 14.3 Formation of Partial Differential Equations 14.2 Worked Examples 14.2 Exercise 14.114.15 Answers to Exercise 14.114.15 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 24 3/2/2017 6:17:55 PM Contents n xxv 14.4 Solutions of Partial Differential Equations 14.16 14.4.1 Procedure to find general integral and singular integral for a first order partial differential equation14.17 Worked Examples14.17 Exercise 14.214.20 Answers to Exercise 14.214.20 14.4.2 First Order Non-linear Partial Differential Equation of Standard Types 14.20 Worked Examples14.21 Exercise 14.314.25 Answers to Exercise 14.314.25 Worked Examples14.26 14.4.3 Equations Reducible to Standard Forms 14.33 Worked Examples14.35 Exercise 14.414.38 Answers to Exercise 14.414.38 14.5 Lagrange's Linear Equation 14.39 Worked Examples14.41 Exercise 14.514.48 Answers to Exercise 14.514.48 14.6 Homogeneous Linear Partial Differential Equations of the Second and Higher Order with Constant Coefficients 14.49 14.6.1 Working Procedure to Find Complementary Function 14.50 14.6.2 Working Procedure to Find Particular Integral 14.51 Worked Examples14.53 Exercise 14.614.66 Answers to Exercise 14.614.67 14.7 Non-homogeneous Linear Partial Differential Equations of the Second and Higher Order with Constant Coefficients 14.68 Worked Examples14.69 Exercise 14.714.73 Answers to Exercise 14.714.73 Short Answer Questions14.74 Objective Type Questions14.74 Answers 14.76 15. Analytic Functions 15.0 Preliminaries 15.1 Function of a Complex Variable 15.1.1 Geometrical Representation of Complex Function or Mapping 15.1.2 Extended complex number system 15.1.3 Neighbourhood of a point and region 15.2 Limit of a Function 15.2.1 Continuity of a function 15.2.2 Derivative of f(z) 15.2.3 Differentiation formulae A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 25 15.1 15.1 15.2 15.3 15.4 15.5 15.5 15.6 15.6 15.7 3/2/2017 6:17:55 PM xxvi n Contents 15.3 Analytic Function 15.8 15.3.1 Necessary and Sufficient Condition for f(z)to be Analytic 15.8 15.3.2 C-R equations in polar form15.10 Worked Examples15.11 Exercise 15.115.20 Answers to Exercise 15.115.21 15.4 Harmonic Functions and Properties of Analytic Function 15.21 15.4.1 Construction of an Analytic Function Whose Real or Imaginary Part is Given Milne-Thomson Method 15.23 Worked Examples15.25 Exercise 15.215.32 Answers to Exercise 15.215.33 15.5 Conformal Mapping 15.33 15.5.1 Angle of rotation15.34 15.5.2 Mapping by elementary functions15.36 Worked Examples15.37 Exercise 15.315.72 Answers to Exercise 15.315.74 15.5.3 Bilinear Transformation 15.79 Worked Examples15.82 Exercise 15.415.89 Answers to Exercise 15.415.90 Short Answer Questions15.90 Objective Type Questions15.91 Answers15.92 16. Complex Integration 16.1 16.0 Introduction 16.1 16.1 Contour Integral 16.1 16.1.1 Properties of Contour Integrals 16.1 Worked Examples 16.2 16.1.2 Simply Connected and Multiply Connected Domains 16.4 16.2 Cauchy's Integral Theorem or Cauchy's Fundamental Theorem 16.4 16.2.1 Cauchy-Goursat Integral Theorem 16.5 16.3 Cauchy's Integral Formula 16.6 16.3.1 Cauchy's Integral Formula for Derivatives 16.7 Worked Examples 16.7 Exercise 16.116.12 Answers to Exercise 16.116.13 16.4 Taylor's Series and Laurent's Series 16.14 16.4.1 Taylor's Series 16.14 16.4.2 Laurent's series16.15 Worked Examples16.16 Exercise 16.216.22 Answers to Exercise 16.216.23 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 26 3/2/2017 6:17:56 PM Contents n xxvii 16.5 Classification of Singularities 16.24 16.6 Residue 16.26 16.6.1 Methods of Finding Residue 16.26 16.7 Cauchy's Residue Theorem 16.27 Worked Examples16.28 Exercise 16.316.34 Answers to Exercise 16.316.36 16.8 Application of Residue Theorem to Evaluate Real Integrals 16.36 16.8.1 Type 1 16.36 Worked Examples16.37 16.8.2 Type 2. Improper Integrals of Rational Functions 16.44 Worked Examples16.46 16.8.3 Type 3 16.50 Worked Examples16.50 Exercise 16.416.55 Answers to Exercise 16.416.56 Short Answer Questions16.56 Objective Type Questions16.58 Answers16.60 17 Fourier Series 17.1 17.0 Introduction 17.1 17.1 Fourier series 17.2 17.1.1 Dirichlet's Conditions 17.2 17.1.2 Convergence of Fourier Series 17.3 Worked Examples 17.5 17.2 Even and Odd Functions 17.15 17.2.1 Sine and Cosine Series 17.15 Worked Examples17.16 Exercise 17.117.23 Answers to Exercise 17.117.25 17.3 Half-Range Series 17.26 17.3.1 Half-range Sine Series 17.27 17.3.2 Half-range cosine series17.27 Worked Examples17.28 Exercise 17.217.36 Answers to Exercise 17.217.37 17.4 Change of Interval 17.38 Worked Examples17.39 17.5 Parseval's Identity 17.47 Worked Examples17.47 Exercise 17.317.50 Answers to Exercise 17.317.52 17.6 Complex Form of Fourier Series 17.53 Worked Examples17.55 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 27 3/2/2017 6:17:56 PM xxviii n Contents Exercise 17.417.59 Answers to Exercise 17.417.59 17.7 Harmonic Analysis 17.60 17.7.1 Trapezoidal Rule 17.60 Worked Examples17.62 Exercise 17.517.68 Answers to Exercise 17.517.69 Short Answer Questions17.69 Objective Type Questions17.70 Answers17.72 18. Fourier Transforms 18.1 18.0 Introduction 18.1 18.1 Fourier Integral Theorem 18.1 18.1.1 Fourier Cosine and Sine Integrals 18.2 Worked Examples 18.2 18.1.2 Complex Form of Fourier Integral 18.6 18.2 Fourier Transform Pair 18.7 18.2.1 Properties of Fourier transforms 18.8 Worked Examples18.12 Exercise 18.118.21 Answers to Exercise 18.118.22 18.3 Fourier Sine and Cosine Transforms 18.23 18.3.1 Properties of Fourier Sine and Cosine Transforms 18.24 Worked Examples18.29 Exercise 18.218.39 Answers to Exercise 18.218.39 18.4 Convolution Theorem 18.40 18.4.1 Definition: Convolution of Two Functions 18.40 18.4.2 Theorem 18.1: Convolution theorem or Faltung theorem18.41 18.4.3 Theorem 18.2 : Parseval's identity for Fourier transforms or Energy theorem18.41 Worked Examples18.43 Exercise 18.318.51 Answers to Exercise 18.318.52 Short Answer Questions 18.52 Objective Type Questions18.53 Answers18.54 19. Laplace Transforms 19.0 Introduction 19.1 Condition for Existence of Laplace Transform 19.2 Laplace Transform of Some Elementary Functions 19.3 Some Properties of Laplace Transform Worked Examples A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 28 19.1 19.1 19.1 19.2 19.4 19.5 3/2/2017 6:17:56 PM Contents n xxix Exercise 19.1 19.9 Answers to Exercise 19.119.10 19.4 Differentiation and Integration of Transforms 19.11 Worked Examples19.12 Exercise 19.219.20 Answers to Exercise 19.219.20 19.5 Laplace Transform of Derivatives and Integrals 19.21 Worked Examples19.23 19.5.1 Evaluation of Improper Integrals using Laplace Transform 19.25 Worked Examples19.25 19.6 Laplace Transform of Periodic Functions and Other Special Type of Functions 19.27 Worked Examples19.29 19.6.1 Laplace Transform of Unit Step Function 19.36 19.6.2 Unit impulse function19.37 19.6.3 Dirac-delta function19.37 19.6.4 Laplace transform of delta function19.37 Worked Examples19.38 Exercise 19.319.39 Answers to Exercise 19.319.41 19.7 Inverse Laplace Transforms 19.41 19.7.1 Type 1 – Direct and shifting methods19.43 Worked Examples19.43 19.7.2 Type 2 – Partial Fraction Method 19.44 Worked Examples19.44 19.7.3 Type 3 – 1. Multiplication by s and 2. Division by s19.48 Worked Examples19.48 19.7.4 Type 4 – Inverse Laplace Transform of Logarithmic and Trigonometric Functions 19.50 Worked Examples19.50 Exercise 19.419.53 Answers to Exercise 19.419.54 19.7.5 Type 5 – Method of Convolution 19.55 Worked Examples19.57 Exercise 19.519.60 Answers to Exercise 19.519.61 19.7.6 Type 6: Inverse Laplace Transform as Contour Integral 19.61 Worked Examples19.62 Exercise 19.619.64 Answers to Exercise 19.619.65 19.8 Application of Laplace Transform to the Solution of Ordinary Differential Equations 19.65 19.8.1 First Order Linear Differential Equations with Constant Coefficients 19.65 Worked Examples19.65 19.8.2 Ordinary Second and Higher Order Linear Differential Equations with Constant Coefficients 19.68 Worked Examples19.68 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 29 3/2/2017 6:17:56 PM xxx n Contents 19.8.3 Ordinary Second Order Differential Equations with Variable Coefficients 19.72 Worked Examples19.72 Exercise 19.719.75 Answers to Exercise 19.719.76 19.8.4 Simultaneous Differential Equations 19.77 Worked Examples19.77 19.8.5 Integral–Differential Equation 19.83 Worked Examples19.83 Exercise 19.819.85 Answers to Exercise 19.819.86 Short Answer Questions19.86 Objective Type Questions19.86 Answers19.88 20. Applications of Partial Differential Equations 20.1 20.0 Introduction 20.1 20.1 One Dimensional Wave Equation – Equation of Vibrating String 20.2 20.1.1 Derivation of Wave Equation 20.2 20.1.2 Solution of One-Dimensional Wave Equation by the Method of Separation of Variables (or the Fourier Method) 20.3 Worked Examples 20.5 Exercise 20.120.34 Answers to Exercise 20.120.35 20.1.3 Classification of Partial Differential Equation of Second Order 20.36 Worked Examples20.37 Exercise 20.220.38 Answers to Exercise 20.220.38 20.2 One-Dimensional Equation of Heat Conduction (In a Rod) 20.39 20.2.1 Derivation of Heat Equation 20.39 20.2.2 Solution of Heat Equation by Variable Separable Method 20.40 Worked Examples20.42 Exercise 20.320.62 Answers to Exercise 20.320.63 Worked Examples20.64 Exercise 20.420.68 Answers to Exercise 20.420.69 20.3 Two Dimensional Heat Equation in Steady State 20.69 20.3.1 Solution of Two Dimensional Heat Equation 20.70 Worked Examples20.71 Exercise 20.520.83 Answers to Exercise 20.520.84 Short Answer Questions20.85 Objective Type Questions20.86 Answers20.88 IndexI.1 A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 30 3/2/2017 6:17:56 PM Preface This book Engineering Mathematics is written to cover the topics that are common to the syllabi of various universities in India. Although this book is designed primarily for engineering courses, it is also suitable for Mathematics courses and for various competitive examinations. The aim of the book is to provide a sound understanding of Mathematics. The experiences of both the authors in teaching undergraduate and postgraduate students from diverse backgrounds for over four decades have helped to present the subject as simple as possible with clarity and rigour in a step-by-step approach. This book has many distinguishing features. The topics are well organized to create self-confidence and interest among the readers to study and apply the mathematical tools in engineering and science disciplines. The subject is presented with a lot of standard worked examples and exercises that will help the readers to develop maturity in Mathematics. This book is organized into 20 chapters. At the end of each chapter, short answer questions and objective questions are given to enhance the understanding of the topics. Chapter 1 focuses on the applications of matrices to the consistency of simultaneous linear equations and Eigen value problems. Chapter 2 discusses convergence of sequence and series. Chapter 3 deals with differentiation and applications of derivative, Rolle's Theorem, mean value theorems, asymptotes and curve tracing. Chapter 4 deals with the geometrical application of derivative in radius of curvature, centre of curvature, evolute and envelope. Chapter 5 elaborates calculus of several variables. Chapter 6 deals with integral calculus and applications of integral calculus. Chapter 7 discusses improper integrals, and beta and gamma functions. Chapter 8 focuses on multiple integrals. Chapter 9 deals with vector calculus. Chapter 10 discusses solution of various types of first order differential equations. Chapter 11 is concerned with the solution of second order and higher order linear differential equations. Chapter 12 deals with some applications of ordinary differential equations. Chapter 13 conforms to series solution of ordinary differential equations and special functions. Chapter 14 focuses on solution of partial differential equations. A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 31 3/2/2017 6:17:56 PM xxxii n Preface Chapter 15 examines analytic functions. Chapter 16 focuses on complex integration. Chapter 17 deals with Fourier series. Chapter 18 pertains to Fourier transforms. Chapter 19 discusses Laplace transforms. Chapter 20 is concerned with applications of partial differential equations. Mathematics is a subject that can be mastered only through hard work and practice. Follow the maximum, Mathematics without practice is blind and practice without understanding is futile. "Tell me and I will forget Show me and I will remember Involve me and I will understand" —Confucius We hope that this book is student-friendly and that it will be well received by students and teachers. We heartily welcome valuable comments and suggestions from our readers for the improvement of this book, which may be addressed to profpsdas@yahoo.com. ACKNOWLEDGEMENTS P. Sivaramakrishna Das: I express my gratitude to our chairperson, Dr Elizabeth Varghese, and the directors of K.C.G. College of Technology for giving me an opportunity to write this book. I am obliged to my department colleagues for their encouragement. The inspiration to write this book came from my wife, Prof. C. Vijayakumari, who is also the co-author of this book. P. Sivaramakrishna Das and C. Vijayakumari: We are grateful to the members of our family for lending us their support for the successful completion of this book. We are obliged to Sojan Jose, R. Dheepika and C. Purushothaman of Pearson India Education Services Pvt. Ltd, for their diligence in bringing this work out to fruition. A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 32 P. Sivaramakrishna Das C. Vijayakumari 3/2/2017 6:17:56 PM About the Authors Prof. Dr P. Sivaramakrishna Das started his career in 1967 as assistant professor of Mathematics at Ramakrishna Mission Vivekananda College, Chennai, his alma mater and retired as Head of the P.G. Department of Mathematics from the same college after an illustrious career spanning 36 years. Currently, he is professor of Mathematics and Head of the Department of Science and Humanities, K.C.G. College of Technology, Chennai (a unit of Hindustan Group of Institutions). P. Sivaramakrishna Das has done pioneering research work in the field of "Fuzzy Algebra" and possess a Ph.D. in this field. His paper on fuzzy groups and level subgroups was a fundamental paper on fuzzy algebra with over 600 citations and it was the first paper from India. With a teaching experience spanning over 49 years, he is an accomplished teacher of Mathematics at undergraduate and postgraduate levels of Arts and Science and Engineering colleges in Chennai. He has guided several students to obtain their M.Phil. degree from the University of Madras, Chennai. He was the most popular and sought-after teacher of Mathematics in Chennai during 1980s for coaching students for IIT-JEE. He has produced all India 1st rank and several other ranks in IIT-JEE. He was also a visiting professor at a few leading IIT-JEE training centres in Andhra Pradesh. Along with his wife C. Vijayakumari, he has written 10 books covering various topics of Engineering Mathematics catering to the syllabus of Anna University, Chennai, and has also written "Numerical Analysis", an all India book, catering to the syllabi of all major universities in India. Prof. Dr C. Vijayakumari began her career in 1970 as assistant professor of Mathematics at Government Arts College for Women, Thanjavur, and has taught at various Government Arts and Science colleges across Tamil Nadu before retiring as professor of Mathematics from Queen Mary's College (Autonomous), Chennai after an illustrious career of spanning 36 years. As a visiting professor of Mathematics, she has taught the students at two engineering colleges in Chennai. With a teaching experience spanning over 40 years, she is an accomplished teacher of Mathematics and Statistics at both undergraduate and postgraduate levels. She has guided many students to obtain their M.Phil. degree from the University of Madras, Chennai and Bharathiar University, Coimbatore. Along with her husband P. Sivaramakrishna Das, she has co-authored several books on Engineering Mathematics catering to the syllabus of Anna University, Chennai and has also co-authored "Numerical Analysis", an all India book, catering to the syllabi of all major universities in India. A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 33 3/2/2017 6:17:56 PM This page is intentionally left blank Engineering Mathematics A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 35 3/2/2017 6:17:56 PM This page is intentionally left blank 1 Matrices 1.0 INTRODUCTION The concept of matrices and their basic operations were introduced by the British mathematician Arthur Cayley in the year 1858. He wondered whether this part of mathematics will ever be used. However, after 67 years, in 1925, the German physicist Heisenberg used the algebra of matrices in his revolutionary theory of quantum mechanics. Over the years, the theory of matrices have been found as an elegant and powerful tool in almost all branches of Science and Engineering like electrical networks, graph theory, optimisation techniques, system of differential equations, stochastic processes, computer graphics, etc. Because of the digital computers, usage of matrix methods have become greatly fruitful. In this chapter, we review some of the basic concepts of matrices. We shall discuss two important applications of matrices, namely consistency of system of linear equations and the eigen value problems. 1.1 BASIC CONCEPTS Definition 1.1 Matrix A rectangular array of mn numbers (real or complex) arranged in m rows (horizontal lines) and n columns (vertical lines) and enclosed in brackets [ ] is called an m × n matrix. The numbers in the matrix are called entries or elements of the matrix. Usually an m × n matrix is written as ⎡ a11 ⎢ ⎢ a21 ⎢A A= ⎢ ⎢a ⎢ i1 ⎢A ⎢ ⎢⎣ am1 a12 a13 … a1 j a22 a23 … a2 j A A ai 2 ai 3 A am 2 A am 3 … amj A … aij … a1n ⎤ ⎥ … a2 n ⎥ A ⎥ ⎥ … ain ⎥ ⎥ ⎥ ⎥ … amn ⎥⎦ where aij is the element lying in the ith row and jth column, the first suffix refers to row and the second suffix refers to column. The matrix A is briefly written as A = [aij]m × n, i = 1, 2, 3, …, m, j = 1, 2, 3, …, n If all the entries are real, then the matrix A is called a real matrix. Definition 1.2 Square Matrix In a matrix, if the number of rows = number of columns = n, then it is called a square matrix of order n. If A is a square matrix of order n, then A = [aij]n × n, i = 1, 2, 3, …, n; j = 1, 2, 3, …, n. Definition 1.3 Row Matrix A matrix with only one row is called a row matrix. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 1 5/30/2016 4:34:37 PM 1.2 ■ Engineering Mathematics EXAMPLE 1.1 Let A = [a11 a12 a13 … a1n]. It is a row matrix with n columns. So, it is of type 1 × n. EXAMPLE 1.2 Let A = [1, 2, 3, 4]. It is a row matrix with 4 columns. So, it is a row matrix of type 1 × 4. Definition1.4 Column Matrix A matrix with only one column is called a column matrix. EXAMPLE 1.3 ⎡ a11 ⎤ ⎢a ⎥ ⎢ 21 ⎥ A = ⎢a31 ⎥ ⎢ ⎥ ⎢:⎥ ⎢⎣an1 ⎥⎦ Let It is a column matrix with n rows. So, it is of type n × 1. EXAMPLE 1.4 ⎡1⎤ ⎢0⎥ ⎢ ⎥ Let A = ⎢ −2⎥ ⎢ ⎥ ⎢1⎥ ⎢⎣ 3 ⎥⎦ It is a column matrix with 5 rows. So, it is of type 5 × 1. Definition 1.5 Diagonal Matrix A square matrix A = [aij] with all entries aij = 0 when i ≠ j is is called a diagonal matrix. In other words a square matrix in which all the off diagonal elements are zero is called a diagonal matrix. EXAMPLE 1.5 0 … 0⎤ 0 … 0 ⎥⎥ is a diagonal matrix of order n. ⎥ : ⎥ 0 … ann ⎦ ⎡a11 0 ⎢0 a 22 (1) A = ⎢ ⎢: : ⎢ 0 ⎣0 ⎡2 0 0 ⎤ (2) A = ⎢⎢0 3 0 ⎥⎥ is a diagonal matrix of order 3. ⎢⎣0 0 −4 ⎥⎦ ⎡ −1 ⎢0 (3) A = ⎢ ⎢0 ⎢ ⎣0 0 2 0 0 0 0 3 0 0⎤ 0 ⎥⎥ is a diagonal matrix of order 4. 0⎥ ⎥ 0⎦ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 2 5/30/2016 4:34:38 PM Matrices ■ 1.3 Definition 1.6 Scalar Matrix In a diagonal matrix if all the diagonal elements are equal to a non-zero scalar a, then it is called a scalar matrix. EXAMPLE 1.6 ⎡a 0 0 ⎤ A = ⎢⎢ 0 a 0 ⎥⎥ is a scalar matrix. ⎢⎣ 0 0 a ⎥⎦ Definition 1.7 Unit Matrix or Identity Matrix In a diagonal matrix, if all the diagonal elements are equal to 1, then it is called a Unit matrix or identity matrix. EXAMPLE 1.7 ⎡1 0 0⎤ ⎡1 0⎤ ⎢ [1], ⎢ , ⎢0 1 0 ⎥⎥ are identity matrices of orders 1, 2, 3 respectively. They are denoted by I1, I2, I3. ⎥ ⎣0 1⎦ ⎢ ⎣0 0 1 ⎥⎦ In general, In is the identity matrix of order n. Definition 1.8 Zero Matrix or Null Matrix In a matrix (rectangular or square), if all the entries are equal to 0, then it is called a zero matrix or null matrix. EXAMPLE 1.8 ⎡0 0 0 0⎤ ⎡0 0 ⎤ A =⎢ ⎥ , B = ⎢0 0 0 0 ⎥ are zero matrices of types 2 × 2 and 2 × 4. 0 0 ⎦ ⎣ ⎦ ⎣ Definition 1.9 Triangular matrix A square matrix A = [aij] is said to be an upper triangular matrix if all the entries below the main diagonal are zero. That is aij = 0 if i > j A square matrix A = [aij] is said to be a lower triangular matrix if all the entries above the main diagonal are zero. That is aij = 0 if i < j EXAMPLE 1.9 ⎡4 ⎡ −1 2 3 ⎤ ⎢0 (1) The matrices A = ⎢⎢ 0 1 4 ⎥⎥ and B = ⎢ ⎢0 ⎢⎣ 0 0 5 ⎥⎦ ⎢ ⎣0 1 2 0 0 0 2⎤ 3 1 ⎥⎥ are upper triangular matrices. 0 −2⎥ ⎥ 0 5⎦ ⎡1 0 0 ⎤ ⎡ 2 0⎤ ⎢ ⎥ (2) The matrices A = ⎢ ⎥ and B = ⎢ 2 −1 0 ⎥ are lower triangular matrices. ⎣ −1 0 ⎦ ⎢⎣0 2 1 ⎥⎦ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 3 5/30/2016 4:34:39 PM 1.4 ■ Engineering Mathematics 1.1.1 Basic Operations on Matrices Definition 1.10 Equality of Matrices Two matrices A = [aij] and B = [bij] of the same type m × n are said to be equal if aij = bij for all i, j and is written as A = B. Definition 1.11 Addition of Matrices Let A = [aij] and B = [bij] of the same type m × n. Then A + B = [cij], where cij = aij + bij for all i and j and A + B is of type m × n. EXAMPLE 1.10 ⎡ −1 2 3⎤ ⎡1 2 3 ⎤ ⎡ −1 + 1 2 + 2 3 + 3 ⎤ ⎡0 4 6 ⎤ If A = ⎢ ⎥ and B = ⎢1 0 −2⎥ , then A + B = ⎢ 0 + 1 1 + 0 5 − 2⎥ = ⎢1 1 3⎥ 0 1 5 ⎦ ⎣ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ We see that A and B are of type 2 × 3 and A + B is also of type 2 × 3. Definition 1.12 Scalar Multiplication of a Matrix Let A = [aij] be an m × n matrix and k be a scalar, then kA = [kaij]. EXAMPLE 1.11 a ⎡a If A = ⎢ 11 12 a a ⎣ 21 22 a13 ⎤ ⎡ ka , then kA = ⎢ 11 a23 ⎥⎦ ⎣ ka21 ka12 ka22 ka13 ⎤ . ka23 ⎥⎦ −a12 −a13 ⎤ ⎡ −a In particular if k = −1, then − A = ⎢ 11 ⎥. ⎣ −a21 −a22 −a23 ⎦ Multiplication of Matrices If A and B are two matrices such that the number of columns of A is equal to the number of rows of B, then the product AB is defined. Two such matrices are said to be conformable for multiplication. In the product AB, A is known as pre-factor and B is known as post-factor. Definition 1.13 Let A = [aij] be an m × p matrix and B = [bij] be an p × n matrix, then AB is defined and p AB = [cij] is an m × n matrix, where cij = ∑ aik b kj . k =1 That is cij is the sum of the products of the corresponding elements of the ith row of A and the jth column of B. EXAMPLE 1.12 ⎡ 1 1 2⎤ ⎡ 1 2⎤ ⎢ 1 3⎥ and B = ⎢ 3 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 2 2 1 ⎥⎦ ⎢⎣ 2 1 ⎥⎦ Since A is of type 3 × 3 and B is of type 3 × 2, AB is defined and AB is of type 3 × 2. Let A = 0 ⎡ 1 1 2⎤ ⎡ 1 2⎤ ⎡1⋅1 + 1⋅ 3 + 2 ⋅ 2 1⋅ 2 + 1⋅1 + 2 ⋅1 ⎤ ⎡ 8 5 ⎤ A B = ⎢⎢0 1 3⎥⎥ ⎢⎢ 3 1 ⎥⎥ = ⎢⎢ 0 ⋅1 + 1⋅ 3 + 3 ⋅ 2 0 ⋅ 2 + 1⋅1 + 3 ⋅1⎥⎥ = ⎢⎢ 9 4 ⎥⎥ ⎢⎣ 2 2 1 ⎥⎦ ⎢⎣ 2 1 ⎥⎦ ⎢⎣ 2 ⋅1 + 2 ⋅ 3 + 1⋅ 2 2 ⋅ 2 + 2 ⋅1 + 1⋅1⎥⎦ ⎢⎣10 7 ⎥⎦ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 4 5/30/2016 4:34:41 PM Matrices ■ 1.5 Note If A and B are square matrices of order n, then both AB and BA are defined, but not necessarily equal. That is, AB ≠ BA, in general. So, matrix multiplication is not commutative. 1.1.2 Properties of addition, scalar multiplication and multiplication 1. If A, B, C are matrices of the same type, then (i) A + B = B + A (ii) A + (B + C) = (A + B) + C (iii) A + 0 = A (iv) A + (−A) = 0 (v) a (A + B) = a A + a B (vi) (a + b)A = a A + b A (vii) a (bA) = (a b)A for any scalars a, b. 2. If A, B, C are conformable for multiplication, then (i) a (AB) = (a A)B = A(a B) (ii) A(BC) = (AB)C (iii) (A + B)C = AC + BC, where A and B are of type m × p and C is of type p × n. (iv) If A is a square matrix, then A2 = A × A, A3 = A2 × A, …, An = An − 1 × A Definition 1.14 Transpose of a Matrix Let A = [aij] be an m × n matrix. The transpose of A is obtained by interchanging the rows and columns of A and it is denoted by AT. ∴ A T = [a ji ] is a n × m matrix. Properties: (i) (AT)T = A (iii) (AB)T = BT AT (ii) (A + B)T = AT + BT (iv) (aA)T = aAT Definition 1.15 Symmetric Matrix A square matrix A = [aij] of order n is said to be symmetric if AT = A. This means [aji] = [aij] ⇒ aji = aij for i, j = 1, 2, …n Thus, in a symmetric matrix elements equidistant from the main diagonal are the same. EXAMPLE 1.13 ⎡1 ⎡a h g ⎤ ⎢ −2 ⎢ ⎥ A = ⎢ h b f ⎥ and B = ⎢⎢ 3 ⎢ ⎢⎣ g f c ⎥⎦ ⎢⎣ 4 −2 3 0 5 5 2 7 8 4 ⎤ ⎥ 7⎥ are symmetric matrices of orders 3 and 4. 8 ⎥ ⎥ 4 ⎥⎦ Definition 1.16 Skew-Symmetric Matrix A square matrix A = [aij] of order n is said to be skew-symmetric if AT = −A. This means [aji] = −[aij] ⇒ aji = − aij for all i, j = 1, 2, …, n In particular, put j = i, then aii = − aii ⇒ 2aii = 0 ⇒ aii = 0 for all i = 1, 2, …, n So, in a skew-symmetric matrix, the diagonal elements are all zero and elements equidistant from the main diagonal are equal in magnitude, but opposite in sign. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 5 5/30/2016 4:34:41 PM 1.6 ■ Engineering Mathematics EXAMPLE 1.14 2 −3⎤ ⎡0 ⎡ 0 1⎤ ⎢ 4 ⎥⎥ are skew-symmetric matrices of orders 2 and 3. A =⎢ ⎥ and B = ⎢ −2 0 1 0 − ⎦ ⎣ ⎢⎣ 3 −4 0 ⎥⎦ Definition 1.17 Non-Singular Matrix A Square matrix A is said to be non-singular if A ≠ 0 ( A means determinant of A). If A = 0, then A is singular. Definition 1.18 Minor and Cofactor of an Element Let A = [aij] be a square matrix of order n. If we delete the row and column of the element aij, we get a square submatrix of order (n − 1). The determinant of this submatrix is called the minor of the element aij and is denoted by Mij. The cofactor of aij in A is A ij = ( −1)i + j M ij EXAMPLE 1.15 ⎡1 6 2⎤ A = ⎢⎢0 −2 4 ⎥⎥ ⎢⎣ 3 1 2 ⎥⎦ The cofactor of a11 = 1 is A 11 = ( −1)1+1 −2 4 = −4 −4 1 2 The cofactor of a12 = 6 is A 12 = ( −1)1+ 2 0 4 = − (−12) = 12 3 2 = −8 1 2 = − (4 −0) = −4 0 4 Similarly, we can determine the cofactors of other elements. The cofactor of a32 = 1 is A 32 = ( −1)3+ 2 Definition 1.19 Adjoint of a Matrix Let A = [aij] be a square matrix. The adjoint of A is defined as the transpose of the matrix of cofactors of the elements of A and it is denoted by adj A. ⎡ A 11 ⎢A Thus, adj A = ⎢ 21 ⎢ : ⎢ ⎣ A n1 A 12 A 21 : A n2 … A 1n ⎤T … A 2n ⎥ ⎥ : ⎥ … A nn ⎥⎦ Properties: If A and B are square matrices of order n, then (ii) (adj A) A = A (adj A) = A In. (i) adj AT = (adj A)T (iii) adj(AB) = (adj A) (adj B) Using property (ii), we define inverse. Definition 1.20 Inverse of a Matrix If A is a non-singular matrix, then the inverse of A is defined as M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 6 adj A and it is denoted by A−1. A 5/30/2016 4:34:44 PM Matrices ■ A −1 = ∴ 1.7 adj A A EXAMPLE 1.16 ⎡1 6 2⎤ Find the inverse of A = ⎢0 −2 4 ⎥ . ⎢ ⎥ ⎢⎣ 3 1 2 ⎥⎦ Solution. ⎡1 6 2⎤ A = ⎢⎢0 −2 4 ⎥⎥ Given ⎣⎢ 3 1 2 ⎥⎦ 1 6 2 ∴ A = 0 −2 4 = 1(−4 −4) −6(0 − 12) + 2(0 + 6) = −8 + 72 + 12 = 76 ≠ 0 3 1 2 adj A . Since A ≠ 0, A is non-singular and hence A−1 exists and A −1 = A We shall find the cofactors of the elements of A −2 4 0 4 A 11 = ( −1)1+1 = ( −4 − 4) = −8, A 12 = ( −1)1+ 2 = −(0 − 12) = 12 1 2 3 2 A 13 = (− −1)1+ 3 0 −2 = (0 + 6) = 6, 3 1 A 21 = ( −1) 2 +1 6 2 = −(12 − 2) = −10 1 2 A 22 = ( −1) 2 + 2 1 2 = ( 2 − 6) = −4, 3 2 A 23 = ( −1) 2 + 3 1 6 = −(1 − 18) = 17 3 1 A 31 = ( −1)3+1 6 2 = ( 24 + 4) = 28, −2 4 A 32 = ( −1)3+ 2 1 2 = −( 4 − 0) = −4 0 4 A 33 = ( −1)3+ 3 1 6 = ( −2 − 0) = −2 0 −2 ⎡ −8 12 6 ⎤ ⎡ −8 −10 28 ⎤ ⎢ ⎥ adj A = ⎢ −10 −4 17 ⎥ = ⎢⎢12 −4 −4 ⎥⎥ ⎢⎣ 28 −4 −2⎥⎦ ⎢⎣ 6 17 −2 ⎥⎦ T ∴ ∴ 1.2 A −1 ⎡ −8 −10 28 ⎤ 1 ⎢ 12 −4 −4 ⎥⎥ = 76 ⎢ ⎢⎣ 6 17 −2 ⎥⎦ COMPLEX MATRICES A matrix with at least one element as complex number is called a complex matrix. Let A = [aij] be a complex matrix. The conjugate matrix of A is denoted by A and A = ⎡⎣aij ⎤⎦ . M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 7 5/30/2016 4:34:46 PM 1.8 ■ Engineering Mathematics EXAMPLE 1.17 ⎡ 2i A =⎢ ⎣3 − 2i −i 0 2⎤ is a complex matrix. 3⎥⎦ ⎡ 2i The conjugate of A is A = ⎢ ⎢⎣3 − 2i −i 0 2⎤ ⎡ −2i i 2⎤ ⎥=⎢ ⎥ 3⎥⎦ ⎣3 + 2i 0 3⎦ [{ conjugate of a + ib = a − ib] We denote ( A ) by A*. ∴ A* is the transpose of the conjugate of A. In the above example T ⎡ −2i 3 + 2i ⎤ A = ⎢⎢ i 0 ⎥⎥ ⎢⎣ 2 3 ⎥⎦ * Note We have ∴ If ⎡⎣ A T ⎤⎦ = ⎡⎣ A T ⎤⎦ ∴ A ∗ = ⎡⎣ A T ⎤⎦ A = [a ji ], then A T = [a ji ], ⎡⎣ A T ⎤⎦ =[a ji ] ∴ A ∗ = [a ji ] Definition 1.21 Hermitian Matrix A complex square matrix A is said to be a Hermitian matrix if A* = A and Skew-Hermitian matrix if A* = −A. A Hermitian matix is also denoted by AH. If A = [aij], then A * = [a ji ] ∴ A* = A ⇒ a ji = aij for all i and j Put j = i, then aii = aii ⇒ aii are real numbers. So, the diagonal elements of a Hermitian matrix are real numbers. The elements equidistant from the main diagonal are conjugates. A* = −A ⇒ a ji = −aij for all i and j Put j = i, then aii = −aii If aii = a + ib, then aii = a − ib ∴ a − ib = −(a + ib) ⇒ 2a = 0 ⇒ a = 0 ∴ aii = ib, which is purely imaginary if b ≠ 0 and 0 if b = 0. ∴ the diagonal elements of a Skew-Hermitian matrix are all purely imaginary or 0 and the elements equidistant from the main diagonal are conjugates with opposite sign. Properties: If A and B are complex matrices, then 1. ( A) = A, 2. A + B = A + B 3. aA = a A 4. A B = A B 7. (aA).* = aA * 5. (A*).* = A 8. (AB).* = B*A* 6. (A + B).* = A* + B* Definition 1.22 Unitary Matrix A complex square matrix is said to be unitary if AA* = A*A = I From the definition it is obvious that A* is the inverse of A. ∴ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 8 A* = A−1 5/30/2016 4:34:50 PM Matrices ■ 1.9 EXAMPLE 1.18 ⎡ 21 Show that A 5 ⎢ ⎣3 1 i Solution. Given 32 i ⎤ is a Hermitian matrix. 2 ⎥⎦ ⎡ −1 3 − i ⎤ A=⎢ 2 ⎥⎦ ⎣3 + i T ∴ T ⎡ −1 3 − i ⎤ ⎡ −1 3 − i ⎤ ⎡ −1 3 + i ⎤ T A * = (A ) = ⎢ = =A ⎥ ⎢3 − i ⎥ = ⎢3 + i 2 2 ⎥⎦ 2 ⎥⎦ ⎣ ⎦ ⎣ ⎢⎣3 + i ∴ A is Hermitian matrix. EXAMPLE 1.19 32i 312 i ⎤ ⎡ 1 ⎢ Show that B 5 ⎢ 31i 0 223i ⎥⎥ is a Hermitian matrix. ⎢⎣322 i 213i 22 ⎥⎦ Solution. Since the diagonal elements are real and elements equidistant from the main diagonal are conjugates, B is a Hermitian matrix. EXAMPLE 1.20 11 i ⎤ ⎡ 2i Show that A 5 ⎢ is a Skew-Hermitian matrix. 0 ⎥⎦ ⎣2(1 2 i ) Solution. 1+ i⎤ ⎡ 2i A =⎢ Given ⎥ ⎣ −(1 − i ) 0 ⎦ Since the diagonal elements are purely imaginary or zero and (1 + i) and − (1 − i) are conjugates with opposite sign, A is Skew-Hermitian matrix. EXAMPLE 1.21 11 i 2 25i ⎤ ⎡ 2i ⎢ Show that B 5 ⎢ 2(1 2 i ) 0 2 1 3i ⎥⎥ is skew-Hermitian. ⎢⎣2( 2 15i ) 2( 2 2 3i ) 3i ⎥⎦ Solution. 1+ i 2 − 5i ⎤ ⎡ 2i Given B = ⎢ −(1 − i ) 0 2 + 3i ⎥⎥ ⎢ ⎢⎣ −( 2 + 5i ) −( 2 − 3i ) 3i ⎥⎦ In B, the diagonal elements are purely imaginary or zero and the elements equidistant from the main diagonal are conjugates with opposite sign. So, B is skew-Hermitian matrix. Note If A is a real matrix, then the definition of unitary ⇒ AAT = ATA = I. In this case A is called an orthogonal matrix. So, if A is an orthogonal matrix, then AT = A−1. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 9 5/30/2016 4:34:52 PM 1.10 ■ Engineering Mathematics WORKED EXAMPLES EXAMPLE 1 ⎡ 2 1i If A 5 ⎢ ⎣ 25 Solution. Given 3 21 13i ⎤ , then show that AA* is a Hermitian matrix. i 4 22 i ⎥⎦ ⎡ 2 + i 3 −1 + 3i ⎤ A =⎢ ⎥ ⎣ −5 i 4 − 2i ⎦ ⎡ 2−i T −1 − 3i ⎤ ∴ A* = [ A ] = ⎢⎢ 3 4 + 2i ⎥⎦ ⎢⎣ −1 − 3i We have to prove AA* is a Hermitian matrix. That is to prove (AA*)* = AA* T Now ⎡2 − i =⎢ ⎣ −5 3 −i ⎡ 2−i ⎡ 2 + i 3 −1 + 3i ⎤ ⎢ AA* = ⎢ ⎥⎢ 3 ⎣ −5 i 4 − 2i ⎦ ⎢ ⎣ −1 − 3i −5 ⎤ −i ⎥⎥ 4 + 2i ⎥⎦ −5 ⎤ −i ⎥⎥ 4 + 2i ⎥⎦ ⎡( 2 + i )( 2 − i ) + 3 ⋅ 3 + ( −1 + 3i )( −1 − 3i ) ( 2 + i )( −5) + 3( −i ) + ( −1 + 3i ) ( 4 + 2i ) ⎤ =⎢ ( −5)( −5) + i ( −i ) + ( 4 − 2i )( 4 + 2i ) ⎥⎦ ⎣ −5( 2 − i ) + i ⋅ 3 + ( 4 − 2i )( −1 − 3i ) ⎡ 22 + 1 + 9 + 1 + 32 −10 − 5i − 3i − 4 + 10i + 6i 2 ⎤ =⎢ ⎥ 2 25 − i 2 + 4 2 + 22 ⎣ −10 + 5i + 3i − 4 − 10i + 6i ⎦ 24 −20 + 2i ⎤ −14 + 2i − 6 ⎤ ⎡ 24 ⎡ =⎢ =⎢ ⎥ 46 46 ⎥⎦ ⎣ −14 − 2i − 6 ⎦ ⎣ −20 − 2i ∴ ⎡ 24 ( A A *)* = ⎢ ⎢⎣ −20 − 2i [{ i 2 = −1] T −20 + 2i ⎤ ⎥ 46 ⎥⎦ −20 − 2i ⎤ ⎡ 24 ⎡ 24 =⎢ =⎢ ⎥ 46 ⎦ ⎣ −20 + 2i ⎣ −20 − 2i ⇒ (AA*)* = AA* Hence, AA* is a Hermitian matrix. T −20 + 2i ⎤ = AA* 46 ⎥⎦ EXAMPLE 2 Show that every square complex matrix can be expressed uniquely as P + iQ, where P and Q are Hermitian matrices. Solution. Let A be any square complex matrix. We shall rewrite A as A= M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 10 1 ⎡1 ⎤ [A + A *] + i ⎢ ( A − A *) ⎥ 2 2 i ⎣ ⎦ 5/30/2016 4:34:54 PM Matrices ■ 1 1 ( A + A *), Q = ( A − A *), 2 2i We shall now prove P and Q are Hermitian. Put P = 1.11 then A = P + iQ. * Now, ⎡1 ⎤ ⎡1 ⎤ 1 P* = ⎢ ( A + A *) ⎥ = ⎢ ( A * + ( A *) *⎥ = ( A * + A ) = P ⎣2 ⎦ ⎣2 ⎦ 2 ∴ P is Hermitian. * and 1 1 1 ⎡1 ⎤ Q* = ⎢ ( A − A*) ⎥ = ( A * −( A*) *] = − [ A * − A] = ( A − A*) = Q [ 2i 2i 2i ⎣ 2i ⎦ ∴ Q is Hermitian. We shall now prove the uniqueness of the expression A = P + iQ. If possible, let A = R + iS where R and S are Hermitian matrices. ∴ R* = R and S* = S Now, A* = (R + iS)* = R* + (iS)* = R* − iS* = R − iS (1) + (2) ⇒ A + A* = 2R ⇒ R = (1) [by property] (2) 1 ( A + A *) = P 2 1 ( A − A *) = Q 2i ∴ the expression A = P + iQ is unique. (1) − (2) ⇒ A − A* = 2iS ⇒ S = EXAMPLE 3 If A is any square complex matrix, prove that (1) A 1 A* is Hermitian and (ii) A 5 B 1 C, where B is Hermitian and C is Skew-Hermitian. Solution. Given A is a square complex matrix. (i) Let P = A + A* ∴ P* = (A + A*)* = A* + (A*)* = A* + A = A + A* = P ∴ P is Hermitian Hence, A + A* is Hermitian. To prove (ii): Since A is square complex matrix, we can write A as 1 1 A = ( A + A*) + ( A − A*) = B + C 2 2 1 1 where B = ( A + A *) is Hermitian by part (i) and C = ( A − A *) 2 2 [by property] * 1 1 1 ⎡1 ⎤ ⇒ C * = ⎢ ( A − A*) ⎥ = [( A * −( A*) *] = [ A * − A] = − [ A − A*] = −C 2 2 2 ⎣2 ⎦ ∴ C is Skew- Hermitian. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 11 5/30/2016 4:34:57 PM 1.12 ■ Engineering Mathematics EXAMPLE 4 1 12 i ⎤ ⎡ 0 21 If A 5 ⎢ ⎥ , then show that (I 2 A) (I 1 A) is a unitary matrix. 2 1 1 2 i 0 ⎣ ⎦ Solution. Given ∴ ∴ 1 + 2i ⎤ ⎡ 0 1 + 2i ⎤ ⎡1 0 ⎤ ⎡ 0 A =⎢ =⎢ . Let I = ⎢ ⎥ ⎥. ⎥ 0 ⎦ ⎣ −(1 − 2i ) 0 ⎦ ⎣ −1 + 2i ⎣0 1 ⎦ 1 + 2i ⎤ 1 + 2i ⎤ ⎡1 0 ⎤ ⎡ 0 ⎡ 1 I + A= ⎢ =⎢ ⎥ + ⎢ −(1 − 2i ) ⎥ 0 1 − − 0 ( 1 2 i ) 1 ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 1 1 + 2i I +A = = 1 + (1 − 2i )(1 + 2i ) = 1 + 1 + 4 = 6 ≠ 0 −(1 − 2i ) 1 Inverse of I + A exists and (I + A ) −1 = adj (I + A ) I +A 1 − 2i ⎤ ⎡ 1 ⎡ 1 =⎢ adj(I + A ) = ⎢ ⎥ 1 ⎦ ⎣ −(1 + 2i ) ⎣1 − 2i T ∴ ∴ (I + A ) −1 = 1⎡ 1 6 ⎢⎣1 − 2i −(1 + 2i ) ⎤ 1 ⎥⎦ −(1 + 2i ) ⎤ 1 ⎥⎦ 1 + 2i ⎤ ⎡ 1 ⎡1 0 ⎤ ⎡ 0 = I −A = ⎢ ⎥ − ⎢ −1 + 2i 0 1 0 ⎥⎦ ⎢⎣1 − 2i ⎣ ⎦ ⎣ ∴ (I − A )(I + A ) −1 = 1⎡ 1 6 ⎢⎣1 − 2i −(1 + 2i ) ⎤ 1 ⎥⎦ ⎡ 1 ⎢1 − 2i ⎣ −(1 + 2i ) ⎤ 1 ⎥⎦ −(1 + 2i ) ⎤ 1 ⎥⎦ = 1 ⎡1 − (1 + 2i )(1 − 2i ) −(1 + 2i ) − (1 + 2i ) ⎤ 6 ⎢⎣ (1 − 2i ) + (1 − 2i ) −(1 − 2i )(1 + 2i ) + 1⎥⎦ = −2(1 + 2i ) ⎤ 1 ⎡1 − (1 + 4) −2(1 + 2i ) ⎤ 1 ⎡ −4 = ⎢ = B , say ⎢ ⎥ −4 ⎥⎦ 6 ⎣ 2(1 − 2i ) −(1 + 4) + 1⎦ 6 ⎣ 2(1 − 2i ) T Now, B* = − 2(1 + 2i ) ⎤ 1 ⎡ −4 ⎢ ⎥ 6 ⎢⎣ 2(1 − 2i ) − 4 ⎥⎦ 2(1 + 2i ) ⎤ −2(1 − 2i ) ⎤ 1 ⎡ −4 1 ⎡ −4 = ⎢ = ⎢ ⎥ −4 ⎥⎦ −4 ⎦ 6 ⎣ 2(1 + 2i ) 6 ⎣ −2(1 − 2i ) T To prove Now, B = (I − A) (I + A)−1 is unitary, verify BB* = I BB * = = −2(1 + 2i ) ⎤ ⎡ −4 2(1 + 2i ) ⎤ 1 ⎡ −4 ⎢ ⎥ ⎢ −4 ⎦ ⎣ −2(1 − 2i ) −4 ⎥⎦ 36 ⎣ 2(1 − 2i ) 1 ⎡16 + 4(1 + 2i )(1 − 2i ) −8(1 + 2i ) + 8(1 + 2i ) ⎤ 36 ⎢⎣ −8(1 − 2i ) + 8(1 − 2i ) 4(1 + 2i))(1 − 2i ) + 16 ⎥⎦ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 12 5/30/2016 4:35:00 PM Matrices ■ = 1.13 0 ⎤ 1 ⎡36 0 ⎤ ⎡1 0 ⎤ 1 ⎡16 + 4(1 + 4) ⎢ ⎥ = 36 ⎢ 0 36 ⎥ = ⎢0 1 ⎥ = I . 0 4 ( 1 + 4 ) + 6 36 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ∴ B is unitary. Hence, (I − A)(I + A)−1 is unitary. Note Another method: To prove B is unitary, verify B* = B−1 EXERCISE 1.1 ⎡3 ⎡7 0⎤ 1. If A + B = ⎢ ⎥ , A − B = ⎢0 2 5 ⎣ ⎦ ⎣ 2. Find x, y, z and w if 6 ⎤ ⎡ 4 ⎡x y ⎤ ⎡ x 3⎢ ⎥ = ⎢ −1 2w ⎥ + ⎢ z + w z w ⎣ ⎦ ⎣ ⎦ ⎣ 0⎤ , find A and B 3⎥⎦ x + y⎤ 3 ⎥⎦ 3. If matrix A has x rows and x +5 columns and B has y rows and 11 − y columns such that both AB and BA exist, then find x and y. ⎡ 2 3 4⎤ ⎡ 1 3 0⎤ 4. If A = ⎢⎢ 1 2 3⎥⎥ and B = ⎢ −1 2 1 ⎥ , then find AB and BA and test their equality. ⎢ ⎥ ⎢⎣ −1 1 2⎥⎦ ⎢⎣ 0 0 2⎥⎦ ⎡ ⎢ 0 5. If A = ⎢ ⎢ tan a ⎢⎣ 2 a⎤ − tan ⎥ 2 ⎥ , show that I + A = [I − A ] ⎡cos a − sin a ⎤ ⎢ sin a cos a ⎥ ⎣ ⎦ 0 ⎥⎥ ⎦ ⎡ cos a sin a ⎤ T 6. If A = ⎢ ⎥ , then verify that AA = I2. ⎣ − sin a cos a ⎦ 7. If A is a square matrix, then show that A can be expressed as A = P + Q, where P is symmetric and Q is skew-symmetric. ⎡ A + AT A − AT ⎤ ,Q = ⎢ Hint: Take P = ⎥ 2 2 ⎦ ⎣ ⎡2 0 1⎤ ⎢ ⎥ 8. If A = ⎢ 2 1 3⎥ and f(x) = x2 − 5x + 6, then find f(A). ⎢⎣1 −1 0 ⎥⎦ ⎡ 1 −1 1 ⎤ ⎢ 3 0 ⎥⎥, then prove that A(adj A) = 9. If A = 2 ⎢ ⎢⎣18 2 10 ⎥⎦ ⎡0 0 0⎤ ⎢0 0 0⎥ . ⎢ ⎥ ⎢⎣0 0 0 ⎥⎦ ⎡1 0 0 ⎤ 10. Find the inverse of A = ⎢3 3 0 ⎥ in terms of adj A. ⎢ ⎥ ⎢⎣5 2 −1⎥⎦ 1 ⎡1 + i −1 + i ⎤ is unitary. 11. Show that A = ⎢ 2 ⎣1 + i 1 − i ⎥⎦ 12. If A and B are orthogonal matrices of the same order, prove that AB is orthogonal. [Hint: AAT = I, BBT = I. Compute AB(AB)T = A(BBT)AT = AAT = I]. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 13 5/30/2016 4:35:03 PM 1.14 ■ Engineering Mathematics 13. If A and B are Hermitian matrices of the same order, prove that (i) A + B is Hermitian (ii) AB + BA is Hermitian (iii) iA is Skew-Hermitian (iv) AB − BA is Skew-Hermitian 14. Find the inverse of the following matrices. ⎡ 2 −1 4 ⎤ 3 3⎤ ⎡4 (i) ⎢ −3 0 1 ⎥ (ii) ⎢ −1 0 −1⎥ ⎥ ⎢ ⎢ ⎥ ⎢⎣ −1 1 2⎥⎦ ⎢⎣ −4 −4 −3⎥⎦ ⎡ 3 −3 4 ⎤ 15. If A = ⎢⎢ 2 −3 4 ⎥⎥ , then show that A3 = A−1. ⎢⎣0 −1 1 ⎥⎦ ANSWERS TO EXERCISE 1.1 ⎡2 0⎤ ⎡5 0 ⎤ 1. A = ⎢ ⎥ , B = ⎢1 1 ⎥ 1 4 ⎦ ⎣ ⎦ ⎣ 2. x = 2, y = 4, z = 1, w = 3 4. AB ≠ BA 8. 14. 1.3 ⎡ 1 −1 −3 ⎤ f ( A) = ⎢⎢ −1 −1 −10 ⎥⎥ ⎢⎣ −5 4 4 ⎥⎦ 3 3⎤ ⎡4 ⎡ 1 −6 1 ⎤ (i) A−1 = ⎢⎢ 5 8 −14 ⎥⎥ (ii) A−1 = ⎢⎢ −1 0 −1⎥⎥ ⎢⎣ −4 −4 3 ⎥⎦ ⎢⎣ −3 −1 −3 ⎥⎦ 3. x = 3, y = 8 ⎡1 ⎢ 10. A−1 = ⎢ −1 ⎢ ⎢ ⎢3 ⎣ 0 1 3 2 3 0⎤ ⎥ 0⎥ ⎥ ⎥ −1⎥ ⎦ RANK OF A MATRIX Let A = [aij] be an m × n matrix. A matrix obtained by omitting some rows and columns of A is called a submatrix of A. The determinant of a square submatrix of order r is called a minor of order r of A. EXAMPLE 1.22 ⎡ 2 3 4 −1⎤ Consider A = ⎢⎢0 3 4 0 ⎥⎥ ⎢⎣ 3 −2 −1 2 ⎥⎦ 2 3 4 ⎡2 3 4 ⎤ ⎢ ⎥ Omitting the fourth column, we get the submatrix A 1 = ⎢0 3 4 ⎥ and A 1 = 0 3 4 is a minor of order 3. ⎢⎣ 3 −2 −1⎥⎦ 3 −2 −1 ⎡3 −1⎤ Omitting the first and third columns and the third row, we get the submatrix A 2 = ⎢ ⎥ and ⎣3 0 ⎦ 3 −1 is a minor of order 2. Since A 2 = 3 ≠ 0, it is called a non-vanishing minor of order 2. A2 = 3 0 But 3 4 = 0 , so it is called a vanishing minor of order 2. 3 4 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 14 5/30/2016 4:35:06 PM Matrices ■ 1.15 Definition 1.23 Rank of a Matrix Let A be an m × n matrix. A is said to be of rank r if (1) at least one minor of A of order r is not zero and (ii) every minor of A of order (r + 1) (and higher order) is zero. The rank of A is denoted by r(A) or r(A). Note (1) The definition says rank of A is the order of the largest non-vanishing minor of A. (2) The Rank of Zero matrix is zero. (3) All non-zero matrices have rank ≥ 1. (4) The rank of an m × n matrix is less than or equal to the min {m, n}. (5) r(A) = r(AT) (6) If In is the unit matrix of order n, then I n = 1 ≠ 0 and so, r(In) = n. To find the rank of a matrix A, we have to identify the largest non-vanishing minor. This process involves a lot of computations and so it is tedious for matrices of large type. To reduce the computations, we apply elementary transformations and transform the given matrix to a convenient form, namely Echelon form or normal form. Elementary transformations 1. Interchange of any two rows (or columns) 2. Multiplication of elements of any row (or column) by a non-zero number k. 3. Addition to the elements of a row (column), the corresponding elements of another row (column) multiplied by a fixed number. Note When an elementary transformation is applied to a row, it is called a row transformation and when it is applied to a column, it is called a column transformation. Notation: The following symbols will be used to denote the elementary row operations. (i) Ri ↔ Rj means ith row and jth row are interchanged. (ii) Ri → kRi means the elements of ith row is multiplied by k (≠0) (iii) Ri → Ri + kRj means the jth row is multiplied by k and added to the ith row. Similarly we indicate the column transformations by Ci ↔ Cj, Ci → kCi, Ci → Ci + kCj Definition 1.24 Equivalent Matrices Two matrices A and B of the same type are said to be equivalent if one matrix can be obtained from the other by a sequence of elementary row (column) transformations. Then we write A ~ B. Results: 1. The Rank of a matrix is unaffected by elementary transformations. 2. Equivalent matrices have the same rank. Definition 1.25 Echelon Matrix A matrix is called a row-echelon matrix if (1) all zero rows (i.e., rows with zero elements only), if any, are on the bottom of the matrix and (ii) each leading non-zero element is to the right of the leading non-zero element in the preceding row. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 15 5/30/2016 4:35:07 PM 1.16 ■ Engineering Mathematics EXAMPLE 1.23 ⎡2 ⎡1 2 3⎤ ⎢0 A = ⎢⎢0 1 2⎥⎥ , B = ⎢ ⎢0 ⎢⎣0 0 0 ⎥⎦ ⎢ ⎣0 ⎡1 −1 0 4 5⎤ ⎢0 −1 2 1 3⎥ ⎥ C=⎢ ⎢0 0 0 6 1 ⎥ ⎢ ⎥ ⎣0 0 0 0 0 ⎦ 1 −1 0 −1 0 0 0 0 0⎤ 2⎥⎥ , 0⎥ ⎥ 0⎦ ⎡1 2 3⎤ D = ⎢⎢0 2 1 ⎥⎥ are row echelon matrices. ⎢⎣0 0 2⎥⎦ and Note Triangular matrix is a special case of an echelon matrix. Result: If a matrix A is equivalent to a row echelon matrix B, then r(A) = the number of non-zero rows of B. In the above examples, r(A) = 2, r(B) = 2, r(C) = 3, r(D) = 3. WORKED EXAMPLES ⎡1 ⎢2 Find the rank of the matrix A 5 ⎢ ⎢3 ⎢ ⎣6 Solution. ⎡1 2 3 0 ⎤ ⎡1 2 ⎢ 2 4 3 2⎥ ⎢ ⎥ ∼ ⎢0 0 Given A = ⎢ ⎢ 3 2 1 3⎥ ⎢0 −4 ⎥ ⎢ ⎢ ⎣6 8 7 5⎦ ⎣0 −4 EXAMPLE 1 2 4 2 8 3 −3 −8 −11 3 ⎡1 2 ⎢0 0 −3 ∼⎢ ⎢0 −4 −8 ⎢ ⎣0 0 −3 ⎡1 2 3 ⎢0 −4 −8 ∼⎢ ⎢0 0 −3 ⎢ ⎣0 0 −3 3 3 1 7 0⎤ 2 ⎥⎥ , by reducing to an echelon matrix. 3⎥ ⎥ 5⎦ 0⎤ 2⎥⎥ R 2 → R 2 + ( −2)R1 3⎥ R 3 → R 3 + ( −3)R1 ⎥ 5⎦ R 4 → R 4 + ( −6)R1 0⎤ 2⎥⎥ 3⎥ ⎥ 2⎦ R 4 → R 4 − R 3 0⎤ 3⎥⎥ R 2 ↔ R 3 2⎥ ⎥ 2⎦ ⎡1 2 3 0⎤ ⎢0 −4 −8 3⎥ ⎥ ∼⎢ ⎢0 0 −3 2⎥ ⎥ ⎢ 0 0⎦ R 4 → R 4 − R 3 ⎣0 0 ∴ = B, which is a row echelon matrix. r(A) = the number of non-zero rows in B = 3 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 16 5/30/2016 4:35:09 PM Matrices ■ 1.17 EXAMPLE 2 ⎡1 2 3 21⎤ Determine the rank of the matrix A 5 ⎢⎢ 3 6 9 23 ⎥⎥ , by reducing to an echelon matrix. ⎢⎣ 2 4 6 22 ⎥⎦ Solution. ⎡1 2 3 −1⎤ ⎡1 2 3 −1⎤ Given A = ⎢⎢ 3 6 9 −3⎥⎥ ∼ ⎢0 0 0 0 ⎥ R 2 → R 2 + ( −3)R1 ⎢ ⎥ ⎢⎣ 2 4 6 −2⎥⎦ ⎢⎣0 0 0 0 ⎥⎦ R 3 → R 3 + ( −2)R1 = B, which is a row echelon matrix. r(A) = the number of non-zero rows in B = 1 ∴ EXAMPLE 3 ⎡6 ⎢5 Find the value of k if the rank of the matrix ⎢ ⎢3 ⎢ ⎣2 Solution. Let ⎡6 ⎢5 A =⎢ ⎢3 ⎢ ⎣2 3 5 2 3 1 2 1 1 9⎤ ⎡ 1 6 ⎥⎥ ⎢ ⎢ 3 ⎥ ∼ ⎢5 ⎥ ⎢ k ⎦ ⎢3 ⎢⎣ 2 1 2 2 1 1 1 ⎡ ⎢1 2 ⎢ ⎢0 − 1 ⎢ 2 ∼⎢ 1 ⎢0 − ⎢ 2 ⎢ ⎢0 0 ⎢⎣ 1 ⎡ ⎢1 2 ⎢ ⎢0 − 1 ⎢ 2 ∼⎢ ⎢0 0 ⎢ ⎢ ⎢0 0 ⎢⎣ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 17 5 6 3 2 1 3 2 1 1 5 3 2 1 9⎤ 6 ⎥⎥ is 3. 3⎥ ⎥ k⎦ 3⎤ 2⎥ ⎥ 6⎥ 3⎥ ⎥ k ⎥⎦ 5 6 7 − 6 3 − 6 4 − 6 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ k − 3⎥ ⎥⎦ 5 6 7 − 6 4 6 4 − 6 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ k − 3⎥ ⎥⎦ R1 → 3 2 3 − 2 3 − 2 1 R1 6 R 2 → R 2 + ( −5)R1 R 3 → R 3 + ( −3)R1 R 4 → R 4 + ( −2)R1 3 2 3 − 2 R3 → R3 − R 2 5/30/2016 4:35:10 PM 1.18 ■ Engineering Mathematics 1 ⎡ ⎢1 2 ⎢ ⎢0 − 1 ∼⎢ 2 ⎢ ⎢0 0 ⎢ ⎢0 0 ⎣ 5 6 7 − 6 4 6 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ k − 3⎥⎦ R 4 → R 4 + R 3 3 2 3 − 2 =B Given r(A) = 3. So, the number of non-zero rows of B should be 3. ∴ k−3=0 ⇒ k=3 Definition 1.26 Elementary Matrix A matrix obtained from a unit matrix by performing a single elementary row (column) transformation is called an elementary matrix. Since unit matrices are non-singular square matrices, elementary matrices are also non-singular. EXAMPLE 1.24 ⎡1 0 0 ⎤ ⎡1 0 0⎤ I 3 = ⎢⎢0 1 0 ⎥⎥ z ⎢⎢0 1 1 ⎥⎥ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ C 3 → C 3 + C 2 This is an elementary matrix. ⎡1 0 0 ⎤ Similarly, ⎢⎢0 3 0 ⎥⎥ got by R2 → 3R2 is an elementary matrix. ⎢⎣0 0 1 ⎥⎦ Definition 1.27 Normal form of a Matrix Any non-zero matrix A of rank r can be reduced by a sequence of elementary transformations to the ⎡I r 0 ⎤ form ⎢ ⎥ , where Ir is a unit matrix of order r. ⎣0 0⎦ This form is called a normal form of A. ⎡I ⎤ Other normal forms are Ir, ⎢ r ⎥ , [Ir, 0]. ⎣0⎦ Theorem 1.1 Let A be an m × n matrix of rank r. Then there exist non-singular matrices P and Q of orders m and n ⎡I r 0 ⎤ respectively such that PAQ = ⎢ ⎥ ⎣0 0⎦ Note Each elementary row transformation of A is equivalent to pre multiplying A by the corresponding elementary matrix. Each elementary column transformation is equivalent to post multiplying A by the corresponding elementary matrix. So, there exists elementary matrices P1, P2, …, Pk and Q1, Q2, …, Qt such that ⎡I r 0 ⎤ P1 P2 … Pk A Q1 Q2 … Qt = ⎢ ⎥ ⎣0 0⎦ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 18 5/30/2016 4:35:12 PM Matrices ■ 1.19 ⎡I r 0 ⎤ PAQ = ⎢ ⎥ ⎣0 0⎦ P = P1 P2 … Pk, Q = Q1 Q2 … Qt ⇒ where Working rule to find normal form and P, Q: Let A be a non-zero m × n matrix write A = ImAIn(which is obviously true). Reduce A on the L. H. S to normal form by applying elementary row and column transformations on A. Each elementary row transformation of A will be applied to Im on R. H. S and each elementary column transformation of A will be applied to In on R. H. S. ⎡I r 0 ⎤ = PAQ. After a sequence of suitable applications of elementary transformations, we get ⎢ 0 0 ⎥⎦ ⎣ Then the rank of A is the rank of I = r r WORKED EXAMPLES EXAMPLE 1 ⎡0 1 2 1 ⎤ Reduce the matrix ⎢⎢ 1 2 3 1⎥⎥ to normal form and hence find the rank. ⎢⎣ 3 1 1 3 ⎥⎦ Solution. Let ⎡ 0 1 2 1 ⎤ ⎡ 1 0 2 1 ⎤ C1 ↔ C 2 A = ⎢⎢1 2 3 2⎥⎥ ∼ ⎢ 2 1 3 2⎥ ⎢ ⎥ ⎢⎣ 3 1 1 3⎥⎦ ⎢⎣1 3 1 3⎥⎦ ⎡1 0 0 0⎤ ∼ ⎢⎢ 2 1 −1 0 ⎥⎥ C 3 → C 3 + ( −2)C1 ⎢⎣ 1 3 −1 2⎥⎦ C 4 → C 4 − C1 ⎡1 0 0 0⎤ ∼ ⎢⎢0 1 −1 0 ⎥⎥ R 2 → R 2 + ( −2)R1 ⎢⎣0 3 −1 2⎥⎦ R 3 → R 3 − R1 ⎡1 0 0 0⎤ ∼ ⎢⎢0 1 0 0 ⎥⎥ ⎢⎣0 3 2 2⎥⎦ C 3 → C 3 + C 2 ⎡1 ∼ ⎢⎢0 ⎢⎣0 ⎡1 ∼ ⎢⎢0 ⎢⎣0 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 19 0 0 0⎤ 1 0 0 ⎥⎥ 0 2 2⎥⎦ R 3 → R 3 + ( −3)R 2 0 0 0⎤ 1 0 0 ⎥⎥ 1 0 1 1 ⎥⎦ R 3 → R 3 2 5/30/2016 4:35:14 PM 1.20 ■ Engineering Mathematics ⎡1 0 0 0⎤ ∼ ⎢⎢0 1 0 0 ⎥⎥ ⎢⎣0 0 1 0 ⎥⎦ C 4 → C 4 − C 3 = [I 3 : 0 ] This is the normal form of A and so the r(A) = 3 EXAMPLE 2 ⎡1 21 21⎤ Let A 5 ⎢⎢1 1 1 ⎥⎥ . Find matrices P and Q such that PAQ is in the normal form. Also find rank ⎢⎣3 1 1 ⎥⎦ of A. Solution. Given ⎡1 −1 −1⎤ A = ⎢⎢1 1 1 ⎥⎥ ⎢⎣3 1 1 ⎥⎦ 3× 3 Consider A = I3AI3 ⎡1 −1 −1⎤ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎢1 1 1 ⎥ = ⎢0 1 0 ⎥ A ⎢0 1 0 ⎥ ⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢3 1 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ Our aim is to reduce the LHS matrix to normal form. Also row operations to be applied to pre factor and column operations to be applied to post factor. Apply, C 2 → C 2 + C1 ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 1 1 ⎤ ⎢1 2 2 ⎥ = ⎢0 1 0 ⎥ A ⎢0 1 0 ⎥ C 3 → C 3 + C1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣3 4 4 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ and to post factor R 2 → R 2 − R1 R 3 → R 3 + ( −3)R1 ⎡1 0 0 ⎤ ⎡ 1 0 0 ⎤ ⎡1 1 1⎤ ⎢0 2 2⎥ = ⎢ −1 1 0 ⎥ A ⎢0 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 4 4 ⎥⎦ ⎢⎣ −3 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ and to pre factor R 3 → R 3 + ( −2)R 2 and to pre factor 1 R2 → R2 2 and to pre factor ⎡1 0 0⎤ ⎡ 1 0 0⎤ ⎡1 1 1⎤ ⎢0 2 2⎥ = ⎢ −1 1 0 ⎥ A ⎢ 0 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 0 0 ⎥⎦ ⎢⎣ −1 −2 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎡ 1 ⎡1 0 0⎤ ⎢ ⎢0 1 1⎥ = ⎢ − 1 ⎢ ⎥ ⎢ 2 ⎢⎣0 0 0 ⎥⎦ ⎢ ⎣ −1 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 20 0 0⎤ ⎥ ⎡1 1 1⎤ 1 0 ⎥ A ⎢⎢0 1 0 ⎥⎥ 2 ⎥ ⎢0 0 1 ⎥⎦ −2 1 ⎥⎦ ⎣ 5/30/2016 4:35:16 PM Matrices ■ C3 → C3 − C 2 and to post factor ⎡ 1 ⎡1 0 0⎤ ⎢ ⎢0 1 0⎥ = ⎢ − 1 ⎢ ⎥ ⎢ 2 ⎢⎣0 0 0 ⎥⎦ ⎢ ⎣ −1 ⎡I 2 ⎢0 ⎣ ⇒ 1.21 0 0⎤ ⎥ ⎡1 1 0 ⎤ 1 0 ⎥ A ⎢⎢0 1 −1⎥⎥ 2 ⎥ ⎢0 0 1 ⎥⎦ −2 1 ⎥⎦ ⎣ 0⎤ = PA Q 0 ⎥⎦ This shows r(A) = 2, and ⎡ 1 ⎢ 1 P = ⎢− ⎢ 2 ⎢ −1 ⎣ 0 0⎤ ⎡1 1 0 ⎤ ⎥ 1 0 ⎥ , Q = ⎢⎢0 1 −1⎥⎥ 2 ⎥ ⎢⎣0 0 1 ⎥⎦ −2 1 ⎥⎦ EXAMPLE 3 ⎡ 1 21 21 2 ⎤ Let A 5 ⎢⎢ 4 2 2 21⎥⎥ . Find the non-singular matrices P and Q, such that PAQ is in the ⎢⎣ 2 2 0 22 ⎥⎦ normal form. Also find the rank of A. Solution. Given ⎡ 1 −1 −1 2 ⎤ A = ⎢⎢ 4 2 2 −1⎥⎥ ⎢⎣ 2 2 0 −2⎥⎦ 3× 4 Consider A = I3AI4 ⇒ ⎡1 ⎡ 1 −1 −1 2 ⎤ ⎡ 1 0 0 ⎤ ⎢ ⎢ 4 2 2 −1⎥ = ⎢ 0 1 0 ⎥ A ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢0 ⎢⎣ 2 2 0 −2⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢ ⎣0 0 0 0⎤ 1 0 0 ⎥⎥ 0 1 0⎥ ⎥ 0 0 1⎦ Our aim is to reduce the LHS matrix to normal form. Also row operations to be applied to pre factor and column operations to be applied to post factor. C 2 → C 2 + C1 C 3 → C 3 + C1 C 4 → C 4 + ( −2)C1 and to post factor ⎡1 ⎡1 0 0 0 ⎤ ⎡1 0 0 ⎤ ⎢ ⎢ 4 6 6 −9 ⎥ = ⎢0 1 0 ⎥ A ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢0 ⎢⎣ 2 4 2 −6 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢ ⎣0 R 2 → R 2 + ( −4)R1 R 3 → R 3 + ( −2)R1 and to pre factor ⎡1 ⎡1 0 0 0 ⎤ ⎡ 1 0 0 ⎤ ⎢ ⎢0 6 6 −9 ⎥ = ⎢ −4 1 0 ⎥ A ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢0 ⎢⎣0 4 2 −6 ⎥⎦ ⎢⎣ −2 0 1 ⎥⎦ ⎢ ⎣0 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 21 1 −2⎤ 0 0 ⎥⎥ 1 0⎥ ⎥ 0 1⎦ 1 1 0 0 1 1 0 0 1 −2⎤ 0 0 ⎥⎥ 1 0⎥ ⎥ 0 1⎦ 5/30/2016 4:35:18 PM 1.22 ■ Engineering Mathematics 1 R2 → R2 3 1 R3 → R3 2 and to pre factor 3 C4 → C4 + C2 2 and to post factor C3 → C3 − C 2 and to post factor R3 → R3 − R 2 and to pre factor 1 R2 → R2 2 R 3 → ( −1)R 3 and to pre factor ⇒ where ⎡ ⎢ 1 ⎡1 0 0 0 ⎤ ⎢ ⎢0 2 2 −3⎥ = ⎢ − 4 ⎢ ⎥ ⎢ 3 ⎢⎣0 2 1 −3⎥⎦ ⎢ ⎢ −1 ⎣ ⎡ ⎢ 1 ⎡1 0 0 0⎤ ⎢ ⎢0 2 2 0 ⎥ = ⎢ − 4 ⎢ ⎥ ⎢ 3 ⎢⎣0 2 1 0 ⎥⎦ ⎢ ⎢ −1 ⎣ 0 1 3 0 0 1 3 0 ⎡ ⎢ 1 ⎡1 0 0 0⎤ ⎢ ⎢0 2 0 0⎥ = ⎢ − 4 ⎢ ⎥ ⎢ 3 ⎢⎣0 2 −1 0 ⎥⎦ ⎢ ⎢ −1 ⎣ ⎡ ⎢ 1 ⎡1 0 0 0 ⎤ ⎢ ⎢0 2 0 0 ⎥ = ⎢ − 4 ⎢ ⎥ ⎢ 3 ⎢⎣0 0 −1 0 ⎥⎦ ⎢ ⎢ 1 ⎢⎣ 3 ⎡ ⎢ 1 ⎡1 0 0 0⎤ ⎢ ⎢0 1 0 0⎥ = ⎢ − 2 ⎢ ⎥ ⎢ 3 ⎢⎣0 0 1 0 ⎥⎦ ⎢ ⎢− 1 ⎢⎣ 3 0 1 3 0 ⎤ ⎡1 0⎥ ⎥ ⎢0 0⎥ A ⎢ ⎥ ⎢0 ⎢ 1 ⎥⎥ ⎣0 2⎦ 1 1 0 0 ⎡ ⎤ ⎢1 1 0⎥ ⎢ ⎥ ⎢ ⎥ 0 A ⎢0 1 ⎥ ⎢ 1 ⎥⎥ ⎢0 0 ⎢⎣0 0 2⎦ ⎡ ⎤ ⎢1 1 0⎥ ⎢ ⎥ ⎢ ⎥ 0 A ⎢0 1 ⎥ ⎢ 1 ⎥⎥ ⎢0 0 ⎢⎣0 0 2⎦ 1 −2⎤ 0 0 ⎥⎥ 1 0⎥ ⎥ 0 1⎦ 1⎤ 1 − ⎥ 2 ⎥ 3 ⎥ 0 2 ⎥ ⎥ 1 0 ⎥ 0 1 ⎥⎦ 1⎤ 0 − ⎥ 2 ⎥ 3 ⎥ −1 2 ⎥ ⎥ 1 0 ⎥ 0 1 ⎥⎦ 1 6 1 3 1⎤ ⎡ ⎤ ⎢1 1 0 − 2 ⎥ 0⎥ ⎥ ⎢ ⎥ 3 ⎥ ⎢ 0 ⎥ A ⎢ 0 1 −1 ⎥ 2 ⎥ ⎥ ⎢ ⎥ 0 ⎥ 1⎥ ⎢0 0 1 ⎢⎣0 0 0 1 ⎥⎦ 2 ⎥⎦ 1⎤ ⎡ ⎤ ⎢1 1 0 − 2 ⎥ 0 ⎥ ⎥ ⎢ ⎥ ⎢ 0 1 −1 3 ⎥ ⎥ 0 A ⎢ ⎥ 2 ⎥ ⎥ ⎢ ⎥ 0 0 1 0 ⎥ 1⎥ ⎢ − ⎢⎣0 0 0 1 ⎥⎦ 2 ⎥⎦ 0 1 6 1 3 1⎤ ⎡ ⎤ ⎢1 1 0 − 2 ⎥ 0 ⎥ ⎢ ⎥ ⎥ ⎢ 0 1 −1 3 ⎥ ⎥ 0 , Q=⎢ ⎥ 2 ⎥ ⎢ ⎥ ⎥ 0 0 1 0 ⎥ 1⎥ ⎢ − ⎢⎣0 0 0 1 ⎥⎦ 2 ⎥⎦ 0 1 3 1 − 3 0 [I3 : 0] = PAQ, ⎡ ⎢ 1 ⎢ 2 P = ⎢− ⎢ 3 ⎢ 1 ⎢− ⎢⎣ 3 and the rank of A = 3 Remark: To find the rank of a matrix, the simplest method is to reduce to row echelon form. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 22 5/30/2016 4:35:19 PM Matrices ■ 1.23 EXERCISE 1.2 Find the rank of the following matrices reducing to echelon form. ⎡ 2 3 −1 ⎢1 −1 −2 1. ⎢ ⎢3 1 3 ⎢ ⎣6 3 0 −1⎤ −4 ⎥⎥ −2⎥ ⎥ −7⎦ 3 0 −2⎤ ⎡4 4. ⎢ 3 4 −1 3 ⎥ ⎢ ⎥ ⎢⎣ −7 −7 1 5 ⎥⎦ ⎡0 ⎢1 2. ⎢ ⎢3 ⎢ ⎣1 1 −3 −1⎤ 0 1 1 ⎥⎥ 1 0 2⎥ ⎥ 1 −2 0 ⎦ ⎡ 1 2 −1 ⎢4 1 2 5. ⎢ ⎢ 3 −1 1 ⎢ ⎣1 2 0 ⎡1 ⎢2 3. ⎢ ⎢3 ⎢ ⎣6 2 4 2 8 3 3 1 7 0⎤ 2⎥⎥ 3⎥ ⎥ 5⎦ ⎡1 1 2 1⎤ ⎢ −1 −1 −2 1 ⎥ ⎢ ⎥ 6. ⎢ 1 2 1 −1⎥ ⎢ ⎥ ⎢ 1 3 0 −3⎥ ⎢⎣ 1 1 2 3 ⎥⎦ 3⎤ 1 ⎥⎥ 2⎥ ⎥ 1⎦ ⎡ 1 2 −2 3 ⎤ ⎡3 1 −5 −1⎤ ⎢ 2 5 −4 6 ⎥ ⎥ 8. ⎢ 9. ⎢1 −2 1 −5⎥ ⎢ ⎥ ⎢ −1 −3 2 −2⎥ ⎢⎣1 5 −7 2 ⎥⎦ ⎢ ⎥ ⎣ 2 4 −4 6 ⎦ ⎡1 3 4 3⎤ 10. Find the rank of the matrix A = ⎢⎢3 9 12 3⎥⎥ , by reducing to an echelon matrix. ⎢⎣1 3 4 1⎥⎦ ⎡1 ⎢1 7. ⎢ ⎢2 ⎢ ⎣3 1 1 3 −2 0 −3 3 −3 1⎤ 1 ⎥⎥ 2⎥ ⎥ 3⎦ ⎡ −2 −1 −1⎤ 11. Find the rank of the matrix A = ⎢⎢12 8 6 ⎥⎥ . ⎢⎣10 5 6 ⎥⎦ ⎡ 2 1 −3 −6 ⎤ 12. Reduce the matrix ⎢ 3 −3 1 2 ⎥⎥ to normal form and hence find the rank. ⎢ ⎢⎣1 1 1 2 ⎥⎦ ⎡ 4 4 −3 ⎢ 1 1 −1 13. Find the values of k if the rank of ⎢ ⎢k 2 2 ⎢ ⎣9 9 k 1⎤ 0 ⎥⎥ is 3. 2⎥ ⎥ 3⎦ ⎡ 2 1 −1 3⎤ 14. Find the values of a and b if the matrix ⎢1 −1 2 4 ⎥ is of rank 2. ⎢ ⎥ ⎢⎣ 7 −1 a b ⎥⎦ ⎡1 −2 3 1 ⎤ 15. Find the values of a and b if the matrix ⎢ 2 1 −1 2⎥ ⎢ ⎥ ⎢⎣6 −2 a b ⎥⎦ ⎡ 1 −1 2 ⎢4 1 0 16. Reduce to normal form and find the rank of ⎢ ⎢0 3 1 ⎢ ⎣0 1 0 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 23 is of rank 2. −3⎤ 2 ⎥⎥ . 4⎥ ⎥ 2⎦ 5/30/2016 4:35:23 PM 1.24 ■ Engineering Mathematics ⎡0 1 2 1 ⎤ 17. Reduce to normal form and find the rank of ⎢1 2 3 2⎥ . ⎢ ⎥ ⎢⎣ 3 1 1 3⎥⎦ ⎡1 1 2 ⎤ 18. If A = ⎢⎢1 2 3 ⎥⎥ , then find non-singular matrices P and Q such that PAQ is in normal form and find its ⎣⎢0 −1 −1⎦⎥ rank. ANSWERS TO EXERCISE 1.2 1. 6. 11. 16. 3 2. 2 3. 3 3 7. 3 8. 3 3 12. 3 13. k = 2 [I4, 0], rank = 4 17. [I3, 0], rank = 3 ⎡ 1 0 0⎤ ⎡1 −1 −1⎤ ⎢ ⎥ 18. P = ⎢ −1 1 0 ⎥ , Q = ⎢⎢0 1 −1⎥⎥ and rank = 2. ⎢⎣ −1 1 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ 1.4 4. 2 9. 3 14. a = 4, b = 18 5. 3 10. 2 15. a = 4, b = 6 SOLUTION OF SYSTEM OF LINEAR EQUATIONS There are many problems in science and engineering whose solution often depends upon a system of linear equations. The equation a1x1 + a2x2 + … + anxn = b is called a non-homogeneous linear equation in n variables x1, x2, …, xn where b ≠ 0 and at least one ai ≠ 0. If b = 0, then the equation a1x1 + a2x2 + … + anxn = 0 is called a homogenous linear equation in x1, x2, …, xn. 1.4.1 Non-homogeneous System of Equations Consider the system of m linear equations in n variables x1, x2, …, xn a11x1 + a12x2 + … + a1nxn = b1 a21x1 + a22x2 + … + a2nxn = b2 : am1x1 + am2x2 + … + amnxn = bm, where at least one bi ≠ 0 ⎡ b1 ⎤ ⎡ x1 ⎤ ⎡ a11 a12 … a1n ⎤ ⎢a ⎥ ⎢ ⎥ ⎢x ⎥ a22 … a2 n ⎥ b If A = ⎢ 21 , B = ⎢ 2 ⎥, X = ⎢ 2⎥, ⎢ : ⎢:⎥ ⎢:⎥ : : ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ … a a a b ⎣ m1 m 2 ⎣ m⎦ ⎣x n ⎦ mn ⎦ then the system of equations can be written as a single matrix equation AX = B. The matrix A is called the coefficient matrix. A solution of the system is a set of values of x1, x2, …, xn which satisfy the m equations. The system of equations is said to be consistent if it has at least one solution. If the system has no solution, then the system of equations is said to be inconsistent. The condition for the consistency of the system is given by Rouche's theorem. We shall state the theorem without proof. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 24 5/30/2016 4:35:24 PM Matrices ■ 1.25 Theorem 1.2 Rouche's Theorem The system of linear equations AX = B is consistent if and only if the coefficient matrix A and the augmented matrix [A, B] have the same rank. That is., r(A) = r([A, B]) Working rule: Let AX = B represent a system of m equations in n variables. 1. 2. 3. 4. Write down the coefficient matrix A and the augmented matrix [A, B]. Find r(A), r([A, B]) If r(A) ≠ r([A, B]), then the system is inconsistent. That is it has no solution. If r(A) = r([A, B]) = n, the number of variables, then the system is consistent with unique solution. If r(A) = r([A, B]) < n, the number of variables, then the system is consistent with infinite number of solutions. If the rank is r, then in this case the solution set will contain n − r parameters or arbitrary constants. To get the solutions we assign arbitrary values to n − r variables and write down the solutions in terms of them. For example, the system x + y + z = 1, 2x − y + 3z = −1, 2x + 5y + z = 5 is consistent with infinite number of solutions. Here r = 2, n = 3. ∴ the solution set will contain n − r = 3 − 2 = 1 parameter. We assign an arbitrary value to one variable, say y. Put y = k and solve for x and z in terms of k. The solution set is x = 4 + 2k, y = k, z = −3 − 3k, where k is any real number. Note If m = n, then A is a square matrix and the system of equations AX = B has unique solution if A is non-singular. That is., A ≠ 0, then r(A) = number of variables n. The unique solution is X = A−1B. 1.4.2 Homogeneous System of Equations Consider the homogeneous system a11x1 + a12x2 + … + a1nxn = 0 a21x1 + a22x2 + … + a2nxn = 0 : am1x1 + am2x2 + … + amnxn = 0 ⎡ a11 a12 ⎢a a22 If A = ⎢ 21 ⎢ : : ⎢ ⎣am 1 am 2 … a1n ⎤ ⎡ x1 ⎤ ⎢ ⎥ … a2 n ⎥ ⎥ , X = ⎢ x 2 ⎥ , then the matrix equation is AX = 0. ⎢:⎥ : ⎥ ⎢ ⎥ ⎥ … amn ⎦ ⎣x n ⎦ For this system x1 = 0, x2 = 0, …, xn = 0 is always a solution. This is called the trivial solution. If A ≠ 0, the r(A) = n and the only solution is the trivial solution. So, the condition for non-trivial solution is A = 0 (or r(A) < n). In solving equations we use only row operations. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 25 5/30/2016 4:35:26 PM 1.26 ■ Engineering Mathematics 1.4.3 Type 1: Solution of Non-homogeneous System of Equations WORKED EXAMPLES (A) Non-homogeneous system with unique solution EXAMPLE 1 Test for consistency and solve 2x 2 y 1 z 5 7, 3x 1 y 2 5z 5 13, x 1 y 1 z 5 5. Solution. The given equations are x+y+z=5 2x − y + z = 7 3x + y − 5z = 13 We have rearranged the equation for convenience in reducing to row echelon form. The coefficient matrix is and the augmented matrix is ⎡1 1 1 ⎤ A = ⎢⎢ 2 −1 1 ⎥⎥ ⎢⎣ 3 1 −5⎥⎦ ⎡1 1 1 : 5 ⎤ ⎢ [A , B ] = ⎢ 2 −1 1 : 7 ⎥⎥ ∼ ⎢⎣ 3 1 −5 : 13⎥⎦ ⎡1 1 1 : 5 ⎤ ⎢0 −3 −1 : −3⎥ R → R − 2R 2 2 1 ⎢ ⎥ ⎢⎣0 −2 −8 : −2⎥⎦ R 3 → R 3 − 3R1 ⎡1 1 1 : 5 ⎤ ⎢ ⎥ 1 1 :1 ⎥ R2 → − R2 ∼ ⎢0 1 3 3 ⎢ ⎥ ⎢0 −1 −4 : −1⎥ ⎣ ⎦ R → 1R 3 3 2 ⎤ ⎡ ⎢1 1 1 : 5⎥ ⎢ ⎥ 1 ∼ ⎢0 1 : 1⎥ ⎢ ⎥ 3 ⎢ ⎥ 11 ⎢0 0 − : 0⎥ R 3 → R 3 + R 2 ⎢⎣ ⎥⎦ 3 From the last matrix, we find M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 26 ⎡ ⎤ ⎢1 1 1 ⎥ ⎢ ⎥ 1 ⎥ A ∼ ⎢0 1 ⎢ 3 ⎥ ⎢ ⎥ ⎢ 0 0 − 11⎥ ⎢⎣ 3 ⎥⎦ 5/30/2016 4:35:27 PM Matrices ■ 1.27 The number of non-zero rows in the equivalent matrices of A and [A, B] are 3. ∴ r(A) = 3, r([A, B]) = 3 ⇒ r(A) = r([A, B]) = 3, the number of variables. So, the equations are consistent with unique solution. From the reduced matrix [A, B], we find the given equations are equivalent to x + y + z = 5, 1 z =1 3 x+1+0=5 y+ ∴ y = 1 and So, the unique solution is x = 4, y = 1, z = 0. and ⇒ 11 z =0 3 x = 4. − ⇒ z=0 EXAMPLE 2 Test for the consistency and solve x 1 2y 1 z 5 3, 2x 1 3y 1 2z 5 5, 3x 2 5y 1 5z 5 2, 3x 1 9y 2 z 5 4. Solution. The given equations are The coefficient matrix is x + 2y + z = 3 2x + 3y + 2z = 5 3x − 5y + 5z = 2 3x + 9y − z = 4. ⎡1 2 1 ⎤ ⎢2 3 2 ⎥ ⎥ A =⎢ ⎢ 3 −5 5 ⎥ ⎢ ⎥ ⎣ 3 9 −1⎦ The augmented matrix is ⎡1 2 1 ⎢2 3 2 [ A, B] = ⎢ ⎢ 3 −5 5 ⎢ ⎣ 3 9 −1 : 3⎤ ⎡1 2 : 5 ⎥⎥ ⎢⎢0 −1 ∼ : 2⎥ ⎢0 −11 ⎥ ⎢ : 4⎦ ⎣0 3 ⎡1 2 ⎢0 −1 ∼⎢ ⎢0 0 ⎢ ⎣0 0 ⎡1 2 ⎢0 −1 ∼⎢ ⎢0 0 ⎢ ⎣0 0 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 27 1 :3 ⎤ 0 : −1⎥⎥ R2 → R2 − 2 R1 2 : −7⎥ R3 → R3 − 3R1 ⎥ −4 : −5⎦ R4 → R4 − 3R1 1 :3 ⎤ 0 : −1⎥⎥ 2 : 4 ⎥ R3 → R3 − 11R2 ⎥ −4 : −8⎦ R4 → R4 + 3R2 1 :3 ⎤ 0 : −1⎥⎥ 1 :2 ⎥ ⎥ 1 :2 ⎦ 1 R3 2 1 R4 → − R4 4 R3 → 5/30/2016 4:35:28 PM 1.28 ■ Engineering Mathematics ⎡1 2 ⎢0 −1 [A , B ] ∼ ⎢ ⎢0 0 ⎢ ⎣0 0 ⇒ From this last matrix we find ⎡1 2 ⎢0 −1 A ∼⎢ ⎢0 0 ⎢ ⎣0 0 1 :3 ⎤ 0 : −1⎥⎥ 1 :2 ⎥ ⎥ 0 : 0 ⎦ R 4 → R4 − R3 1⎤ 0 ⎥⎥ 1⎥ ⎥ 0⎦ The number of non-zero rows in the equivalent matrices of A and [A B] are 3. ∴ ⇒ r(A) = 3, r([A, B]) = 3 r(A) = r([A, B]) = 3, the number of variables. So, the equations are consistent with unique solution From the reduced matrix [A, B], we find the given equations are equivalent to x + 2y + z = 3, −y = −1 and z = 2 ∴ y = 1, z = 2 and so x + 2 ⋅ 1 + 2 = 3 ⇒ x = −1 So, the unique solution is x = −1, y = 1, z = 2. EXAMPLE 3 Solve x2yz 5 e, xy2z3 5 e, x3y2z 5 e using matrices. Solution. The given equations are x2yz = e xy2z3 = e x3y2z = e (1) (2) (3) Taking logarithm to the base e on both sides of (1), (2) and (3), we get log e x 2 yz = log e e ⇒ log e x 2 + log e y + log e z = 1 ⇒ 2 log e x + log e y + log e z = 1 log e xy z = log e e ⇒ log e x + 2 log e y + 3 log e z = 1 2 3 and log x 3 y 2 z = log e e ⇒ 3 log e x + 2 log e y + log e z = 1 For simplicity, put x1 = logex, y1 = logey, z1 = logez ∴ the equations are The coefficient matrix is M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 28 2x1 + y1 + z1 = 1 x1 + 2y1 + 3z1 = 1 3x1 + 2y1 + z1 = 1 ⎡ 2 1 1⎤ A = ⎢⎢1 2 3⎥⎥ ⎢⎣ 3 2 1⎥⎦ 5/30/2016 4:35:29 PM Matrices ■ 1.29 The augmented matrix is ⎡ 2 1 1 : 1⎤ ⎡1 [A , B ] = ⎢⎢1 2 3 : 1⎥⎥ ∼ ⎢⎢ 2 ⎢⎣ 3 2 1 : 1⎥⎦ ⎢⎣ 3 ⎡1 ∼ ⎢⎢0 ⎢⎣0 2 3 : 1⎤ R1 ↔ R 2 1 1 : 1⎥⎥ 2 1 : 1⎥⎦ 2 3 :1 ⎤ −3 −5 : −1⎥⎥ R 2 → R 2 − 2R1 −4 −8 : −2⎥⎦ R 3 → R 3 − 3R1 ⎤ ⎡ ⎢1 2 3 :1 ⎥ ⎥ ⎢ [A , B ] ∼ ⎢0 −3 −5 : −1 ⎥ ⎢ 4 4 2⎥ ⎢0 0 − : − ⎥ R3 → R3 − R 2 3 3 3⎦ ⎣ ⇒ 3 ⎤ ⎡1 2 ⎢0 −3 −5 ⎥ ⎥ A ∼⎢ From the last matrix, we find ⎢ 4⎥ ⎢0 0 − ⎥ 3⎦ ⎣ The number of non-zero rows in the equivalent matrices of A and [A, B] are 3. ∴ r(A) = 3, r([A, B]) = 3 ⇒ r(A) = r([A, B]) = 3, the number of variables. So, the equations are consistent with unique solution. From the reduced matrix [A, B], we find that the given equations are equivalent to x1 + 2y1 + 3z1 = 1 (4) (5) −3y1 − 5z1 = −1 4 2 − z1 = − ⇒ 3 3 and z1 = 1 2 Substituting in (5), we get −3 y1 − 5 ⋅ 1 = −1 ⇒ 2 Substituting in (4) we get ⇒ 1 ⎛ 1⎞ x1 + 2 ⎜ − ⎟ + 3 ⋅ = 1 ⎝ 2⎠ 2 ∴ 3 y1 = 1 − ⇒ x1 = 1 − 1 2 ⇒ x = e2 ⇒ y=e x1 + 1 2 ⇒ log e x = y1 = − 1 ⇒ 2 z1 = 1 2 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 29 ⇒ y1 = − 1 =1 2 ⇒ x1 = 5 3 =− ⇒ 2 2 log e y = − log e z = 1 2 1 2 ⇒ 1 1 = 2 2 1 − z=e 1 2 = e 1 2 1 2 = 1 e = e 5/30/2016 4:35:31 PM 1.30 ■ Engineering Mathematics So, the unique solution is x = e, 1 y= e , z= e. (B) Non-homogeneous system with infinite number of solutions EXAMPLE 4 By investigating the rank of relevant matrices, show that the following equations possess a one parameter family of solutions: 2x 2 y 2 z 5 2, x 12y 1 z 5 2, 4x 2 7y 2 5z 5 2. Solution. The given equations are The coefficient matrix is x +2y + z = 2 2x − y − z = 2 4x − 7y − 5z = 2 ⎡1 2 1 ⎤ A = ⎢⎢ 2 −1 −1⎥⎥ ⎢⎣ 4 −7 −5⎥⎦ The augmented matrix is ⎡ 1 2 1 : 2⎤ ⎡ 1 2 [A , B ] = ⎢⎢ 2 −1 −1 : 2⎥⎥ ∼ ⎢⎢0 −5 ⎢⎣ 4 −7 −5 : 2⎥⎦ ⎢⎣0 −15 ⎡1 2 ⎢ ∼ ⎢0 −5 ⎢⎣0 0 1 :2 ⎤ −3 : −2⎥⎥ R 2 → R 2 − 2R1 −9 : −6 ⎥⎦ R 3 → R 3 − 4 R1 1 :2 ⎤ −3 : −2⎥⎥ 0 : 0 ⎥⎦ R 3 → R 3 − 3R 2 ⎡1 2 1 ⎤ From the last matrix we find A ∼ ⎢⎢0 −5 −3⎥⎥ ⎢⎣0 0 0 ⎥⎦ The number of non-zero rows of equivalent matrices of A and [A, B] are 2 ∴ ⇒ r(A) = 2, r([A, B]) = 2 r(A) = r([A, B]) = 2 < the number of variables 3. So, the equations are consistent with infinite number of solutions involving one parameter, since n − r = 3 − 2 = 1. From the reduced matrix [A, B], we find that the given equations are equivalent to x + 2y + z = 2 − 5y − 3z = − 2 (1) ⇒ 5y + 3z = 2 (2) Assign arbitrary value to one of the variables. Put z = k in (2) ∴ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 30 5y + 3k = 2 ⇒ y = 2 − 3k 5 5/30/2016 4:35:33 PM Matrices ■ 1.31 Substituting in (1), we get, x+ 2( 2 − 3k ) +k = 2 5 ⇒ x = 2− 2( 2 − 3k ) 10 − 4 + 6k − 5k k + 6 −k = = 5 5 5 1 1 ∴ the solution set is x = ( k + 6), y = ( 2 − 3k ), z = k , where k is any real number. 5 5 EXAMPLE 5 Solve, if the equations are consistent: x 2 y 1 2z 5 1, 3x 1 y 1 z 5 4, x 1 3y 23z 5 2, 5x 2 y 1 5z 5 6. Solution. The given equations are The coefficient matrix is x − y + 2z = 1 3x + y + z = 4 x + 3y − 3z = 2 5x − y + 5z = 6 ⎡1 −1 2 ⎤ ⎢3 1 1 ⎥ ⎥ A =⎢ ⎢1 3 −3⎥ ⎢ ⎥ ⎣ 5 −1 5 ⎦ The augmented matrix is ⎡1 −1 2 ⎢3 1 1 [A , B ] = ⎢ ⎢1 3 −3 ⎢ ⎣5 −1 5 From the last matrix, we find : 1 ⎤ ⎡1 −1 2 : 4 ⎥⎥ ⎢⎢0 4 −5 ∼ : 2⎥ ⎢0 4 −5 ⎥ ⎢ : 6 ⎦ ⎣ 0 4 −5 : 1⎤ : 1⎥⎥ R 2 → R 2 − 3R1 : 1⎥ R 3 → R 3 − R1 ⎥ : 1⎦ R 4 → R 4 − 5R1 ⎡ 1 −1 2 ⎢ 0 4 −5 ∼⎢ ⎢0 0 0 ⎢ ⎣0 0 0 : 1⎤ : 1 ⎥⎥ : 0⎥ R 3 → R 3 − R 2 ⎥ : 0⎦ R 4 → R 4 − R 2 ⎡1 −1 2 ⎤ ⎢0 4 −5⎥ ⎥ A ∼⎢ ⎢0 0 0 ⎥ ⎢ ⎥ ⎣0 0 0 ⎦ The number of non-zero rows of the equivalent matrices of A and [A, B] are 2. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 31 5/30/2016 4:35:34 PM 1.32 ■ Engineering Mathematics ∴ r(A) = 2, r([A, B]) = 2 ⇒ r(A) = r([A, B]) = 2 < 3, the number of variables. So, the equations are consistent with infinite number of solutions involving one parameter, since n − r = 3 − 2 = 1. From reduced matrix [A, B], we find that the given equations are equivalent to x − y + 2z = 1 4y − 5z = 1 Put z = k, in (2) then (1) (2) 4y − 5k = 1 ⇒ y = 1 + 5k 4 Substituting in (1), we get 1 + 5k 1 + 5k 4 + 1 + 5k − 8k 5 − 3k x− + 2k = 1 ⇒ x = 1+ − 2k = = 4 4 4 4 ∴ the solution set is 5 − 3k 1 + 5k x= ,y = , z = k , where k is any real number. 4 4 EXAMPLE 6 Test the consistency of the system of equations and solve, if consistent: x1 1 2x2 2 x3 2 5x4 5 4, x1 1 3x2 2 2x3 2 7x4 5 5, 2x1 2 x2 1 3x3 5 3. Solution. The given equations are x1 + 2x2 − x3 − 5x4 = 4 x1 + 3x2 − 2x3 − 7x4 = 5 2x1 − x2 + 3x3 + 0x4= 3 The coefficient matrix is ⎡1 2 −1 −5⎤ A = ⎢⎢1 3 −2 −7⎥⎥ ⎢⎣ 2 −1 3 0 ⎥⎦ The augmented matrix is ⎡1 2 −1 −5 : 4 ⎤ ⎡1 [A , B ] = ⎢⎢1 3 −2 −7 : 5 ⎥⎥ ∼ ⎢⎢0 ⎢⎣ 2 −1 3 0 : 3⎥⎦ ⎢⎣0 ⎡1 ∼ ⎢⎢0 ⎢⎣0 From the last matrix, we find ⎡1 A ∼ ⎢⎢0 ⎢⎣0 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 32 2 −1 −5 : 4 ⎤ 1 −1 −2 : 1 ⎥⎥ R 2 → R 2 − R1 −5 5 10 : −5⎥⎦ R 3 → R 3 − 2R1 2 −1 −5 : 4 ⎤ 1 −1 −2 : 1 ⎥⎥ 0 0 0 : 0 ⎥⎦ R 3 → R 3 + 5R 2 2 −1 −5⎤ 1 −1 −2⎥⎥ 0 0 0 ⎥⎦ 5/30/2016 4:35:36 PM Matrices ■ 1.33 The number of non-zero rows of the equivalent matrices of A and [A, B] are 2. ∴ r(A) = 2, r([A, B]) = 2 ⇒ r(A) = r([A, B]) = 2 < 4, the number of variables. So, the equations are consistent with infinite number of solutions containing two parameters, since n − r = 4 − 2 = 2. From the reduced matrix of [A, B] we find that the given equations are equivalent to x1 + 2x2 − x3 − 5x4 = 4 (1) and x2 − x3 − 2x4 = 1 (2) Put x3 = k1, x4 = k2, then (2) ⇒ Substituting in (1), we get x2 − k1 − 2k2 = 1 ⇒ x2 = 1 + k1 + 2k2 ⇒ x1 = 2 − k1 + k2 x1 + 2(1 + k1 + 2k2) − k1 − 5k2 = 4 ⇒ x1 + 2 + 2k1 + 4k2 − k1 − 5k2 = 4 ⇒ x1 + k1 − k2 = 2 ∴ the solution set is x1 = 2 − k1 + k2, x2 = 1 + k1 + 2k2, x3 = k1, x4 = k2, where k1, k2 are any real numbers. (C) Non-homogeneous system with no solution EXAMPLE 7 Examine for the consistency of the following equations 2x 1 6y 1 11 5 0, 6x 1 20y 2 6z 1 3 5 0, 6y 2 18z 1 1 5 0. Solution. The given equations are 2x + 6y + 0z = −11 6x + 20y − 6z = −3 0x + 6y − 18z = −1 The coefficient matrix is 0 ⎤ ⎡2 6 ⎢ A = ⎢6 20 −6 ⎥⎥ ⎢⎣0 6 −18⎥⎦ The augmented matrix is 11⎤ 1 ⎡ 1 3 0 : − ⎥ R1 → R1 0 : −11⎤ ⎢ ⎡2 6 2 2 ⎢ ⎥ [A , B ] = ⎢⎢6 20 −6 : −3 ⎥⎥ ∼ ⎢6 20 −6 : −3 ⎥ ⎢⎣0 6 −18 : −1 ⎥⎦ ⎢⎣0 6 −18 : −1 ⎥⎦ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 33 5/30/2016 4:35:37 PM 1.34 ■ Engineering Mathematics 11⎤ ⎡ ⎢1 3 0 : − 2 ⎥ ⎢ ⎥ ∼ ⎢0 2 −6 : 30 ⎥ R 2 → R 2 − 6 R1 ⎢ 0 6 −18 : −1 ⎥ ⎣ ⎦ 11⎤ ⎡ ⎢1 3 0 : − 2 ⎥ ⎢ ⎥ [A , B ] ∼ ⎢ 0 2 −6 : 30 ⎥ ⎢ 0 0 0 : −91 ⎥ R → R − 3R ⎣ ⎦ 3 3 2 ⇒ ⎡1 3 0 ⎤ From the last matrix, we find A ∼ ⎢⎢0 2 −6 ⎥⎥ ⎢⎣0 0 0 ⎥⎦ The number of non-zero rows in the equivalent matrices of A and [A, B] are 2 and 3 respectively. ∴ r(A) = 2, r([A, B]) = 3 ⇒ r(A) ≠ r([A, B]). Hence, the equations are inconsistent and the system has no solution. 1.4.4 Type 2: Solution of Non-homogeneous Linear Equations Involving Arbitrary Constants WORKED EXAMPLES EXAMPLE 1 Show that the system of equations 3x 2 y 1 4z 5 3, x 1 2y 2 3z 5 22, 6x 1 5y 1 lz 5 23 has at least one solution for any real number l. Find the set of solutions when l 5 25. Solution. The given equations are x + 2y − 3z = −2, 3x − y + 4z = 3 6x + 5y + lz = −3 ⎡1 2 −3⎤ The coefficient matrix is A = ⎢⎢ 3 −1 4 ⎥⎥ ⎢⎣6 5 l ⎥⎦ and the augmented matrix is : −2⎤ −3 ⎡1 2 −3 : −2⎤ ⎡1 2 [A , B ] = ⎢⎢ 3 −1 4 : 3 ⎥⎥ ∼ ⎢⎢0 −7 : 9 ⎥⎥ R 2 → R 2 − 3R1 13 ⎢⎣6 5 l : −3⎥⎦ ⎢⎣0 −7 l + 18 : 9 ⎥⎦ R 3 → R 3 − 6 R1 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 34 5/30/2016 4:35:38 PM Matrices ■ 1.35 −3 : −2⎤ ⎡1 2 [A , B ] ∼ ⎢⎢0 −7 13 : 9 ⎥⎥ ⎢⎣0 0 l + 5 : 0 ⎥⎦ R 3 → R 3 − R 2 ⇒ −3 ⎤ ⎡1 2 A ∼ ⎢⎢0 −7 13 ⎥⎥ ⎢⎣0 0 l + 5⎥⎦ Case (i): If l + 5 ≠ 0 ⇒ l ≠ −5, the number of non-zero rows in the equivalent matrices of [A, B] and A are 3. From the last matrix we find ∴ ⇒ r(A) = 3, r([A, B]) = 3 r(A) = r([A, B]) = 3, the number of variables. So, the equations are consistent with unique solution. Case (ii): If l + 5 = 0 ⇒ l = −5, then we get ⎡1 2 −3 : −2⎤ [A , B ] ∼ ⎢⎢0 −7 13 : 9 ⎥⎥ ⎢⎣0 0 0 : 0 ⎥⎦ and ⎡1 2 −3⎤ ⎢ A ∼ ⎢0 −7 13 ⎥⎥ ⎢⎣0 0 0 ⎥⎦ The number of non-zero rows of equivalent matrices of [A, B] and A are 2. ∴ r(A) = 2, r([A, B]) = 2 ⇒ r(A) = r([A, B]) = 2 < 3, the number of variables. So, the equations are consistent with infinite number of solutions involving one parameter since n − r = 3 − 2 = 1. From cases (i) and (ii), we find that the equations are consistent for all values of l. We shall now find the solution when l = −5. The solutions will contain one parameter. In this case from the last matrix [A, B], we find the equations are equivalent to Put z 5 k in (2), then Substituting in (1) we get x + 2y − 3z = −2 (1) − 7y + 13z = 9 (2) −7y + 13k = 9 ⇒ 7 y = 13k − 9 ⇒ y = 1 (13k − 9) 7 1 2 x + 2 ⋅ (13k − 9) − 3k = −2 ⇒ x = −2 − (13k − 9) + 3k 7 7 1 1 = [ −14 − 26 k + 18 + 21k ] = [4 − 5k ] 7 7 ∴ the solutions are 1 1 x = [4 − 5k ], y = [13k − 9], z = k , 7 7 M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 35 where k is any real number. 5/30/2016 4:35:39 PM 1.36 ■ Engineering Mathematics EXAMPLE 2 Find the values of a and b if the equations x 1 y 1 2z 5 2, 2x 2 y 1 3z 5 2, 5x 2 y 1 az 5 b have (i) no solutions, (ii) unique solution and (iii) infinite number of solutions. Solution. The given equations are x + y + 2z = 2 2x − y +3z = 2 5x − y + az = b ⎡1 1 2⎤ The coefficient matrix is A = ⎢⎢ 2 −1 3⎥⎥ ⎢⎣ 5 −1 a ⎥⎦ The augmented matrix is 2 ⎡1 1 2 : 2⎤ ⎡1 1 ⎢ ⎢ ⎥ −1 [A , B ] = ⎢ 2 −1 3 : 2⎥ ∼ ⎢0 −3 ⎢ ⎢⎣ 5 −1 a : b ⎥⎦ ⎣0 −6 a − 10 2 ⎡1 1 ⎢ ∼ ⎢0 −3 −1 ⎢⎣0 0 a − 8 From this matrix, we find : 2⎤ : − 2 ⎥⎥ R 2 → R 2 − 2R1 : b − 10 ⎥⎦ R 3 → R 3 − 5R1 : 2⎤ : − 2 ⎥⎥ : b − 6 ⎥⎦ R 3 → R 3 − 2R 2 2 ⎤ ⎡1 1 A ∼ ⎢⎢0 −3 −1 ⎥⎥ ⎢⎣0 0 a − 8⎥⎦ Case (i): The equations have no solution ⇒ r(A) ≠ r([A, B]) This is possible, if r(A) = 2 and r([A, B]) = 3 ⇒ a − 8 = 0 and b − 6 ≠ 0 ⇒ a = 8 and b ≠ 6. Case (ii): The equations have unique solution ⇒ r(A) = r([A, B]) = 3. ∴ a − 8 ≠ 0 and b is any real number. ∴ a ≠ 8 and b is any real number. Case (iii): The equations have infinite number of solutions ⇒ r(A) = r([A, B]) = 2 < 3, the number of variables. This is possible, if a − 8 = 0 and b − 6 = 0 ⇒ a = 8, b = 6 Thus, no solution ⇒ a = 8, b ≠ 6 Unique solution ⇒ a ≠ 8, b is any real number Infinite number of solutions ⇒ a = 8, b = 6. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 36 5/30/2016 4:35:40 PM Matrices ■ 1.37 EXAMPLE 3 For what values of k, the equations x 1 y 1 z 5 1, 2x 1 y 1 4z 5 k and 4x 1 y 1 10z 5 k2 have (i) a unique solution, (ii) infinite number of solutions, (iii) no solution and solve them completely in each case of consistency. Solution. The given equations are x+y+z=1 2x + y + 4z = k 4x + y + 10z = k2 ⎡1 1 1 ⎤ A = ⎢⎢ 2 1 4 ⎥⎥ ⎢⎣ 4 1 10 ⎥⎦ The coefficient matrix is The augmented matrix is :1 ⎤ ⎡1 1 1 : 1 ⎤ ⎡1 1 1 ⎢ ⎢ ⎥ [A , B ] = ⎢ 2 1 4 : k ⎥ ∼ ⎢ 0 −1 2 : k − 2 ⎥⎥ R 2 → R 2 − 2R1 ⎢⎣ 4 1 10 : k 2 ⎥⎦ ⎢⎣0 −3 6 : k 2 − 4 ⎥⎦ R 3 → R 3 − 4 R1 :1 ⎡1 1 1 ⎤ ⎢ ⎥ ∼ ⎢0 −1 2 :k −2 ⎥ ⎢⎣0 0 0 : k 2 − 4 − 3( k − 2) ⎥⎦ R 3 → R 3 − 3R 2 :1 ⎤ ⎡1 1 1 ⎢ [ A , B ] ∼ ⎢ 0 −1 2 : k − 2 ⎥⎥ ⎢⎣0 0 0 : k 2 − 3k + 2⎥⎦ ⇒ From the last matrix, we find ⎡1 1 1⎤ A ∼ ⎢⎢0 −1 2⎥⎥ ⎢⎣0 0 0 ⎥⎦ Since the number of non-zero rows is 2, r(A) = 2 (i) If k2 − 3k + 2 = 0 ⇒ (k − 2)(k − 1) = 0 ⇒ k = 1, k = 2 :1 ⎤ ⎡1 1 1 ⎢ [A , B ] ∼ ⎢0 −1 2 : k − 2⎥⎥ ⎢⎣0 0 0 : 0 ⎥⎦ ∴ ∴ r[ A , B ] = 2 So, r(A) = r([A,B]) = 2 < 3, the number of variables. ∴ the system of equations is consistent with infinite number of solutions if k = 1 or k = 2. M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 37 5/30/2016 4:35:42 PM 1.38 ■ Engineering Mathematics (ii) If k2 − 3k + 2 ≠ 0 ⇒ k ≠ 1 and k ≠ 2, then r([A, B]) = 3 ∴ r(A) ≠ r([A, B]) So, the system is inconsistent and has no solution if k ≠ 1 and k ≠ 2. Now we shall find the solutions if k = 1 and k = 2. ⎡1 1 1 : 1 ⎤ If k = 1, then [A , B ] = ⎢⎢0 −1 2 : −1⎥⎥ ⎢⎣0 0 0 : 0 ⎥⎦ So, the equivalent equations are Put z = k1, then ∴ x + y + z = 1 and −y + 2z = −1 −y + 2k1 = −1 ⇒ y = 1 + 2k1 x + 1 + 2k1 + k1 = 1 ⇒ x = −3k1 ∴ the solutions are x = 3k1, y = 1 + 2k1, z = k1, where k1 is any real number ⎡1 1 1 : 1 ⎤ If k = 2, then [A , B ] = ⎢⎢0 −1 2 : 0 ⎥⎥ ⎢⎣0 0 0 : 0 ⎥⎦ So, the equivalent equations are x + y + z = 1 and −y + 2z = 0 ⇒ y = 2z. Put z = k2, then y = 2k2 ∴ x + 2k2 + k2 = 1 ∴ the solutions are x = 1 − 3k2, y = 2k2, ⇒ x = 1 − 3k2 z = k2, where k2 is any real number. 1.4.5 Type 3: Solution of the System of Homogeneous Equations WORKED EXAMPLES EXAMPLE 1 Find all the non-trivial solutions of 7x 1 y 2 2z 5 0, x 1 5y 2 4z 5 0, 3x 2 2y 1 z 5 0. Solution. The given equations are 7x + y − 2z = 0 x + 5y − 4z = 0 3x − 2y + z = 0 ⎡7 1 −2 ⎤ The coefficient matrix is A = ⎢⎢1 5 −4 ⎥⎥ ⎢⎣ 3 −2 1 ⎥⎦ Since R. H. S of the equations is zero it is enough, we consider A instead of augmented matrix ⎡7 1 −2 : 0 ⎤ [A , B ] = ⎢⎢1 5 −4 : 0 ⎥⎥ , because r(A) = r([A, B]) always. ⎣⎢ 3 −2 1 : 0 ⎦⎥ M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 38 5/30/2016 4:35:43 PM Matrices ■ ⎡1 5 A ∼ ⎢⎢7 1 ⎢⎣ 3 −2 ⎡1 5 ∼ ⎢⎢0 −34 ⎣⎢0 −17 1.39 −4 ⎤ −2 ⎥⎥ R1 ↔ R 2 1 ⎥⎦ −4 ⎤ 26 ⎥⎥ R 2 → R 2 − 7R1 13 ⎥⎦ R 3 → R 3 − 3R1 −4 ⎤ ⎡1 5 1 ⎢ ∼ ⎢0 17 −13⎥⎥ R 2 → − R 2 2 ⎢⎣0 −17 13 ⎥⎦ ⎡1 5 −4 ⎤ ∼ ⎢⎢0 17 −13⎥⎥ ⎢⎣0 0 0 ⎥⎦ R 3 → R 3 + R 2 The number of non-zero rows is 2. ∴ r(A) = 2 < the number of variables. ∴ the number of solutions is infinite containing n − r = 3 − 2 = 1 parameter. From the last equivale Engineering Mathematics By Sivaramakrishna Das Pdf
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