Written as per the revised G Scheme syllabus prescribed by the Maharashtra State Board of Technical Education (MSBTE) w.e.f. academic year

Written as per the revised G Scheme syllabus prescribed by the Maharashtra State Board of Technical Education (MSBTE) w.e.f. academic year 2012-2013 Basic MATHEMATICS First Year Diploma Semester - I First Edition: June 2015 Salient Features Concise content with complete coverage of revised G-scheme syllabus. Simple & lucid language. Illustrative examples showing detailed solution of problems. MSBTE Questions from Winter-2006 to Summer-2015. MSBTE Question Papers of Summer-2014, Winter-2014, Summer-2015. Three Model Question Papers for practice. List of formulae for quick reference. Printed at: Repro India Ltd., Mumbai No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher. TEID : 920

PREFACE Target s Basic Mathematics is compiled with an aim of shaping engineering minds of students while catering to their needs. It is a complete & thorough book designed as per the new revised G-scheme of MSBTE curriculum effective from June 2012. Each unit from the syllabus is divided into chapters bearing specific objectives in mind. The sub-topic wise classification of this book helps the students in easy comprehension. Each chapter includes the following features: Theory of each mathematical concept is explained with appropriate references. Diagrams and illustrations are provided wherever necessary. Italicized definitions are mentioned for important topics. Illustrative Examples are provided in order to explain the method of solving the problems. The detailed step-by-step solution in these problems helps students to understand and remember each minute step with proper justification of the same. Solved Problems covering every type of MSBTE question gives students the confidence to attempt all the questions in the examination. Exercise (With final answers) covers a variety of questions from simple to complex to help the students gain thorough revision in solving various types of problems. MSBTE Problems covering questions from year 2006 to 2015 are solved exactly as they are expected to be solved by the students in the examination. Formulae section is provided for quick recap and last minute revision of all the formulae at one glance. MSBTE Question Papers of year 2014 and Summer-2015 are added at the end to make students familiar with the MSBTE examination pattern. A set of three Model Question Papers are designed as per MSBTE Pattern for thorough revision and to prepare the students for the final examination. From, Publisher Best of luck to all the aspirants!

SYLLABUS Topic and Contents Hours Marks Topic 1 - Algebra 1.1 Determinant [04] Specific Objectives: Solve simultaneous equations in three variables using Cramer s rule. 04 Definition and expansion of determinant of order 3. Cramer s rule to solve simultaneous equations in three variables. 1.2 Matrices [16] Specific Objectives: Perform all algebraic operations on matrices. Solve simultaneous equations in three variables. Definition of a matrix of order m x n and types of matrices. Algebra of matrices with properties and examples. 10 Transpose of a matrix with properties. Cofactor of an element of a matrix. 32 Adjoint of matrix and inverse of matrix by adjoint method. Solution of simultaneous equations containing two and three unknowns by matrix inversion method. 1.3 Partial Fraction [12] Specific Objectives: Find partial fraction of proper and improper fraction. Definition of fraction, proper, improper fraction and partial fraction. 08 Resolve proper fractions into partial fraction with denominator containing i. non repeated linear factors, ii. repeated linear factors, iii. non repeated quadratic irreducible factors. To resolve improper fraction into partial fraction. Topic 2- Trigonometry 2.1 Trigonometric Ratios of Allied, Compound, Multiple and Sub-Multiple Angles [16] Specific objectives: 40 Solve examples of allied angle, compound angle, multiple and sub-multiple 10 angles. Trigonometric ratios of any angle. Definition of allied angle, compound, multiple and sub-multiple angles. Trigonometric ratios of above angles with proofs. Simple examples

2.2 Factorization and De-factorization Formulae [12] Specific objectives: Derive factorization and de-factorization formulae to solve examples. Formulae for factorization and de-factorization with proof and examples. 2.3 Inverse Trigonometric Ratios [12] Specific objectives: Solve examples of inverse trigonometric ratios. Definition of inverse trigonometric ratios. Principal value of inverse trigonometric ratios. Relation between inverse trigonometric ratios with proof and examples. Topic 3 - Co-ordinate Geometry 3.1 Straight Line [16] Specific objectives: Solve problems with given condition. Angle between two lines with proof. Examples. Condition of parallel and perpendicular lines. Point of intersection of two lines, equation of line passing through point of intersection with given condition. Perpendicular distance between point and line with proof and examples. Distance between two parallel line with proof and examples. Topic 4 - Statistics 4.1 Measures of Dispersion [12] Specific objectives: Find the range, mean deviation, standard deviation and consistency of any data. Measures of dispersion - range, mean deviation from mean and median and standard deviation. Variance and its coefficient. Comparisons of two sets of observations. 08 08 10 16 06 12 TOTAL 64 100

Contents Chapter No. Topic Page No. Unit - I: Algebra 1 1 Determinant 2 2 Matrices 45 3 Partial Fraction 146 4 Unit - II: Trigonometry 212 Trigonometric Ratios of Allied, Compound, Multiple and Sub-Multiple Angles 5 Factorization and De-factorization Formulae 277 6 Inverse Trigonometric Ratios 297 213 Unit - III: Co-ordinate Geometry 327 7 Straight Line 328 Unit - IV: Statistics 373 8 Measures of Dispersion 374 Formulae and Model Question Papers Formulae 407 Model Question Paper I 417 Model Question Paper II 420 Model Question Paper III 423 MSBTE Question Papers Question Paper Summer 2014 426 Question Paper Winter 2014 429 Question Paper Summer 2015 432

Target Publications Pvt. Ltd. Basic Physics (F.Y.Dip.Sem.-1) Unit I: Algebra MSBTE UNIT I Algebra Chapter-1 Determinant 1.1 Introduction 1.2 Determinant of Second Order 1.3 Determinant of Third Order 1.3 (a) Definition and elements of determinant of third order 1.3 (b) Minors and Cofactors 1.3 (c) Value of a determinant of third order Chapter-2 Matrices 2.1 Introduction 2.2 Definition of a Matrix 2.3 Types of Matrices 2.3.(a) Rectangular Matrices 2.3.(b) Square Matrices 2.3.(c) Zero Matrices / Null Matrices 2.4 Algebra of Matrices/Operation on Matrices 2.4.(a) Scalar Multiplication of Matrices 2.4.(b) Equality of Matrices 2.4.(c) Addition of Matrices 1.4 Properties of Determinants 1.5 Cramer s Rule 1.5 (a) Cramer s Rule for equations involving two unknown variables 1.5 (b) Cramer s Rule for equations involving three unknown variables 2.4.(d) Subtraction of Matrices 2.4.(e) Multiplication of Matrices 2.5 Transpose of Matrices 2.6 Determinant of a Square Matrix 2.7 Minors and Cofactors 2.7.(a) Minor of an element of a Matrix 2.7.(b) Cofactor of an element of a Matrix 2.8 Adjoint of a Matrix 2.9 Inverse of a Matrix 2.10 Matrix Form of System of Equations Chapter-3 Partial Fraction 3.1 Introduction 3.2 Polynomials 3.3 Proper and Improper Fractions 3.4 Factorization of the Denominator of a proper fraction and determining its type 3.5 Resolve a given proper fraction into partial fractions 3.5.(a) Resolve a proper fraction with Distinct linear factors as denominator, into partial fractions 3.5.(b) 3.5.(c) Resolve a proper fraction with Repeated linear factors as denominator, into partial fractions Resolve a proper fraction with Distinct Irreducible Quadratic factor as denominator, into Partial fractions 3.6 Resolve a given Improper fraction into partial fractions 1

01 Determinant Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE 1.1 Introduction Determinants are arrangement of numbers (real, imaginary, complex) in equal number of rows (horizontal arrangement) and columns (vertical arrangement) enclosed between a pair of vertical segments.they are denoted by or D or A. Determinants are scalar quantities. 1.2 Determinant of Second Order A determinant of order 2 is an arrangement of four numbers in two rows and two columns, enclosed between two vertical segments. Following is an example of a determinant of order 2. D = 1 2 3 4 Row 1 Row 2 D = 1 2 3 4 Thus; 1 and 2 are elements of Row 1, 3 and 4 are elements of Row 2, 1 and 3 are elements of Column 1, 2 and 4 are elements of Column 2. Column 1 Column 2 The set of elements from top left corner to the bottom right corner in a determinant is called the Main Diagonal or Leading Diagonal or Principal Diagonal of that determinant. 1 2 D = Here, 1 and 4 are elements of Principal Diagonal. 3 4 Principal Diagonal The set of elements from bottom left corner to the top right corner in a determinant is called the Reverse or Secondary Diagonal of that determinant. 1 2 D = Here, 3 and 2 are elements of Secondary Diagonal. 2 3 4 Secondary Diagonal To calculate the value of the determinant, consider the determinant of numbers a, b, c, d as given below, D = a c b d = a d c b = ad cb.

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE i.e. Value of determinant D = Product of elements in the Principal Diagonal Product of elements in the Secondary Diagonal. 1 2 D = = (1)(4) (3)(2) = 4 6 = 2 3 4 Illustrative Example 2 5 1. Evaluate: 3 4 2 5 Let D = 3 4 Now, D = Product of elements in the Principal Diagonal Product of elements in the Secondary Diagonal. D = [2(4)] [3(5)] = 8 15 D = 23 Solved Problems 2 3 1. Evaluate: 4 8 Let D = 2 3 = (2)(8) (4)(3) = 16 + 12 = 28 4 8 x 3 2. Find x, if = 0. 1 x 2 x 3 Given, = 0 1 x 2 x(x + 2) (1)(3) = 0 x 2 + 2x 3 = 0 x 2 + 3x x 3 = 0 x(x + 3) 1(x + 3) = 0 (x + 3)(x 1) = 0 x = 3 or x = 1 2 7 3. Solve : x 1 3 12 9 2 2 7 Given, x 1 3 12 9 2 (2)(12) (3)(7) = (x)(2) (9)(1) 24 21 = 2x 9 3 + 9 = 2x 2x = 12 x = 6 3

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE 4. 1 3 2x x1 Find x if,. 2 0 3x x 2 Given, 1 3 2 x x 1 2 0 3x x 2 (1)(0) (2)(3) = (2x)(x + 2) (3x)(x 1) 0 + 6 = 2x 2 + 4x 3x 2 + 3x 6 = x 2 + 7x x 2 7x + 6 = 0 x 2 6x x + 6 = 0 x(x 6) 1(x 6) = 0 (x 6) (x 1) = 0 x 6 = 0 or x 1 = 0 x = 6 or x = 1 Exercise 1.1 1. Evaluate the following determinants: 3 5 12 16 i. ii. 7 2 4 3 2. Find x, if x 5 2 i. x 2 7 = 0 ii. x 2 3 = 3 3x 2x iii. 2 3 = x 3 4 5 2x 5 iv. 2 3x x x +x x1 = 1 2 5x 1 1.3 Determinant of Third Order 1.3 (a) Definition and elements of determinant of third order: A determinant of order 3 is an arrangement of nine numbers in three rows and three columns, enclosed between two vertical segments. Following is an example of a determinant of order 3. D = 1 9 2 8 3 7 4 6 5 Row 1 Row 2 Row 3 D = 1 9 2 8 3 7 4 6 5 4 Thus; 1, 9 and 2 are elements of Row 1, 8, 3 and 7 are elements of Row 2, 4, 6 and 5 are elements of Row 3, 1, 8 and 4 are elements of Column 1, 9, 3 and 6 are elements of Column 2, 2, 7 and 5 are elements of Column 3. Column 1 Column 3 Column 2

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE The elements in Principal Diagonal and the Secondary Diagonal are represented below. 1 9 2 1 9 2 D = 8 3 7 D = 8 3 7 4 6 5 4 6 5 Principal Diagonal Secondary Diagonal Here 1, 3 and 5 are elements of Principal Diagonal, and 4, 3 and 2 are elements of Secondary Diagonal. 1.3 (b) Minors and Cofactors: i. Minor of an element of a determinant is the determinant obtained by removing the row and the column containing that element. a11 a12 a13 Let D = a21 a22 a23 Here a ij = Element in i th row and j th column of D. a31 a32 a33 The minor of a ij is the value of the determinant obtained by eliminating the i th row and j th column of D. We denote the minor of a ij by M ij In case of the determinant D, a22 a23 M 11 = minor of a 11 = = a 22 a 33 a 32 a 23 a32 a33 a21 a23 M 12 = minor of a 12 = = a 21 a 33 a 31 a 23 a31 a33 a21 a22 M 13 = minor of a 13 = = a 21 a 32 a 31 a 22 a31 a32 a12 a13 M 21 = minor of a 21 = = a 12 a 33 a 32 a 13 a32 a33 a11 a13 M 22 = minor of a 22 = = a 11 a 33 a 31 a 13 a31 a33 a11 a12 M 23 = minor of a 23 = = a 11 a 32 a 31 a 12 a31 a32 a12 a13 M 31 = minor of a 31 = = a 12 a 23 a 22 a 13 a22 a23 a11 a13 M 32 = minor of a 32 = = a 11 a 23 a 21 a 13 a21 a23 a11 a12 M 33 = minor of a 33 = = a 11 a 22 a 21 a 12 a a 21 22 Now, consider the following example, 1 3 2 D = 3 1 2 1 0 3 5

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE 6 ii. i.e. The minor of each element of the above determinant is given as follows: M 11 = minor of a 11 = 1 2 = (1)(3) (0)(2) = 3 0 = 3 0 3 M 12 = minor of a 12 = 3 2 1 3 = (3)(3) (1)(2) = 9 2 = 7 M 13 = minor of a 13 = 3 1 = (3)(0) (1)(1) = 0 1 = 1 1 0 M 21 = minor of a 21 = 3 2 = (3)(3) (0)(2) = 9 0 = 9 0 3 M 22 = minor of a 22 = 1 2 = (1)(3) (1)(2) = 3 2 = 1 1 3 M 23 = minor of a 23 = 1 3 = (1)(0) (1)(3) = 0 3 = 3 1 0 M 31 = minor of a 31 = 3 2 = (3)(2) (1)(2) = 6 2 = 4 1 2 M 32 = minor of a 32 = 1 2 = (1)(2) (3)(2) = 2 6 = 4 3 2 M 33 = minor of a 33 = 1 3 = (1)(1) (3)(3) = 1 9 = 8 3 1 Cofactor of an element of a determinant is obtained by multiplying the minor of that element with (1) i+j, where i and j are the row number and column number respectively of that element. Cofactor of a ij is the product of the factor (1) i+j and its corresponding minor. Cofactor of a ij = (1) i+j minor of a ij Cofactor = Minor, if (i + j) is an even number. Cofactor = Minor, if (i + j) is an odd number. 1 3 2 Thus, for the determinant D = 3 1 2 we get the cofactor of each element as follows: 1 0 3 Cofactor of a 11 = (1) 1+1 minor of a 11 = (1) 2 M 11 = +M 11 = + (3) = 3 Cofactor of a 12 = (1) 1+2 minor of a 12 = (1) 3 M 12 = M 12 = (7) = 7 Cofactor of a 13 = (1) 1+3 minor of a 13 = (1) 4 M 13 = +M 13 = + ( 1) = 1 Cofactor of a 21 = (1) 2+1 minor of a 21 = (1) 3 M 21 = M 21 = (9) = 9 Cofactor of a 22 = (1) 2+2 minor of a 22 = (1) 4 M 22 = +M 22 = + (1) = 1 Cofactor of a 23 = (1) 2+3 minor of a 23 = (1) 5 M 23 = M 23 = ( 3) = 3 Cofactor of a 31 = (1) 3+1 minor of a 31 = (1) 4 M 31 = +M 31 = + (4) = 4 Cofactor of a 32 = (1) 3+2 minor of a 32 = (1) 5 M 32 = M 32 = (4) = 4 Cofactor of a 33 = (1) 3+3 minor of a 33 = (1) 6 M 33 = +M 33 = + (8) = 8

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE 1.3 (c) Value of a determinant of third order: We consider the cofactors of first row to define the value of the determinant of third order. Thus the value of the determinant is found by following formula, Value of 3 rd order determinant = Element a 11 Cofactor of element a 11 + Element a 12 Cofactor of element a 12 + Element a 13 Cofactor of element a 13. = a 11 (1) 1+1 minor of a 11 + a 12 (1) 1+2 minor of a 12 + a 13 (1) 1+3 minor of a 13. = a 11 (1) minor of a 11 + a 12 (1) minor of a 12 + a 13 (1) minor of a 13. Value of determinant of third order = a 11 minor of a 11 a 12 minor of a 12 + a 13 minor of a 13. 1. Evaluate: Illustrative Example Let D = 1 3 2 3 1 2. 1 0 3 1 3 2 Row 1 3 1 2 1 0 3 D = a 11 minor of a 11 a 12 minor of a 12 + a 13 minor of a 13. 1 3 2 1 3 2 1 3 2 D = (1) 3 1 2 (3) 3 1 2 + (2) 3 1 2 1 0 3 1 0 3 1 0 3 = (1) 1 2 0 3 (3) 3 2 1 3 + (2) 3 1 1 0 = (1)[(1)(3) (0)(2)] (3)[(3)(3) (1)(2)] + (2)[(3)(0) (1)(1)] = (1) (3 0) (3) (9 2) + (2) (0 1) = (1)(3) (3)(7) + (2)(1) = 3 21 2 D = 20 Solved Problems 1. Expand the following determinants: 3 1 2 1 2 3 i. 0 0 1 ii. 12 13 14 3 5 0 33 34 35 iii. 1 1 1 10 11 12 100 101 102 7

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE iv. 2 1 2 2 1 3 5 4 9 3 1 2 i. 0 0 1 = 3 0 1 5 0 3 5 0 v. (1) 0 1 3 0 0 1 5 2 1 3 3 1 4 + (2) 0 0 3 5 vi. 0 a b a 0 c b c 0 = 3(0 5) + 1(0 + 3) 2(0 0) = 15 + 3 0 = 12 ii. 1 2 3 12 13 14 33 34 35 13 14 = 1 34 35 2 12 14 33 35 + 3 12 13 33 34 = 1(455 476) 2(420 462) + 3(408 429) = 21 + 84 63 = 0 iii. 1 1 1 11 12 10 12 10 11 10 11 12 = 1 1 + 1 101 102 100 102 100 101 100 101 102 = 1(1122 1212) 1(1020 1200) + 1(1010 1100) = 90 + 180 90 = 0 iv. v. 2 1 2 2 1 3 5 4 9 0 1 5 2 1 3 3 1 4 = 1 3 2 4 1 2 3 9 5 9 + (2) 2 1 5 4 = 2(9 + 12) 1(18 + 15) 2(8 5) = 6 + 3 6 = 3 = 1 3 0 1 1 2 3 4 3 4 + (5) 2 1 3 1 = 0(4 + 3) 1( 8 9) 5(2 + 3) = 0 + 17 25 = 8 vi. 0 a b a 0 c b c 0 = 0 c 0 a a c + b a 0 c 0 b 0 b c = 0(0 c 2 ) a(0 bc) + b(ac 0) = 0 + abc + abc = 2abc 2. Find x, if 2 1 x 1 i. 1 3 4 = 0 ii. 0 5 3 0 3 x iv. x 1 3 1 = 0 v. 4 1 5 x 1 2 x 2 1 2x 1 3 = 29 iii. 3 x 2 = 5 3 4 5 1 3 1 1 x 2 x 1 2x 2 4x 1 2 4 = 0 vi. 1 4 16 = 0 4 6 9 1 1 1 8

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE 2 1 x 1 i. 1 3 4 = 0 2 0 5 3 3 4 1 5 3 1 4 0 3 + (x + 1) 1 3 0 5 = 0 2(9 20) 1(3 + 0) + (x + 1)(5 0) = 0 22 + 3 + 5x + 5 = 0 5x = 14 x = 14 5 x 1 2 ii. 2x 1 3 = 29 x 1 3 (1) 2 x 3 4 5 3 5 3 4 5 x(5 12) + 1(10x + 9) + 2(8x 3) = 29 7x + 10x + 9 16x 6 = 29 13x + 3 = 29 13x = 26 x = 2 + 2 2 x 1 3 4 = 29 x 2 1 x 2 iii. 3 x 2 = 5 x 3 1 1 3 1 x(x + 6) 2(3 + 2) + 1(9 x) = 5 x 2 + 6x 10 + 9 x = 5 x 2 + 5x 6 = 0 x 2 + 6x x 6 = 0 x (x + 6) 1 (x + 6) = 0 (x + 6)(x 1) = 0 x = 6 or x = 1 2 3 2 1 1 + 1 3 x = 5 1 3 iv. 0 3 x x 1 3 1 = 0 0 3 1 1 5 ( 3) x 1 1 4 5 4 1 5 + x x 1 3 4 1 = 0 0(15 1) + 3[5(x + 1) 4] + x[(x + 1) 12] = 0 0 + 3 (5x + 5 4) + x (x 11) = 0 3 (5x + 1) + x 2 11x = 0 15x + 3 + x 2 11x = 0 x 2 + 4x + 3 = 0 x 2 + 3x + x + 3 = 0 x (x + 3) + 1(x + 3) = 0 (x + 3) (x + 1) = 0 x = 3 or x = 1 9

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE v. 1 x 2 x 1 2 4 4 6 9 = 0 1 2 4 6 9 x 1 4 4 9 + x2 1 2 4 6 = 0 1(18 24) x(9 16) + x 2 (6 8) = 0 6 + 7x 2x 2 = 0 2x 2 7x + 6 = 0 2x 2 4x 3x + 6 = 0 2x(x 2) 3(x 2) = 0 (x 2)(2x 3) = 0 x = 2 or x = 3 2 vi. 1 2x 2 4x 1 4 16 1 1 1 = 0 1 4 16 1 1 1(4 16) 2x(1 16) + 4x 2 (1 4) = 0 12 + 30x 12x 2 = 0 2x 2 5x + 2 = 0 2x 2 4x x + 2 = 0 2x(x 2) 1(x 2) = 0 (x 2)(2x 1) = 0 2x 1 16 1 1 + 4x 2 1 4 1 1 = 0 x = 2 or x = 1 2 Exercise 1.2 1. Expand the following determinants: 2 3 5 i. 7 1 2 ii. 3 4 1 0 x y x y 0 z z 0 iii. 1 3 5 2 6 10 31 11 38 iv. 1 3 2 4 1 2 v. 3 5 2 6 3 2 2 1 2 10 5 2 2. Find x, if 9 1 x i. 13 3 2 + 6 = 0 ii. 14 1 x 1 5 3 7 2 1 2 = 0 9 1 x iii. 8 3 2 5 6 x = 0 18 15 10 iv. x 1 2 3 x 3 = 6 1 3 2 10

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE 1.4 Properties of Determinants i. The value of determinant remains unchanged if its rows and columns are interchanged. a1 a2 a3 Row 1 a1 b1 c1 Thus if D = b1 b2 b then the value of 3 Row 2 a2 b2 c is also D. 2 c c c Row 3 a b c 1 2 3 3 3 3 1 3 2 Row 1 Column 1 Column 3 Column 2 Example: If D = 5 4 6 Row 2 9 8 7 Row 3 Then by interchanging the Rows and Columns of determinant D we get new determinant which is equal to the given determinant, 1 5 9 D = 3 4 8 2 6 7 ii. Column 1 Column 3 Column 2 If any two rows or columns of a determinant are interchanged, then the sign of the determinant is changed. a1 b1 c1 Row 1 Thus if,d 1 = a b c 2 2 2 a3 b3 c3 Row 3 Then by interchanging Row 1 with Row 3, we get a3 b3 c3 Row 1 D 2 = a b c 2 2 2 a b c 1 1 1 Row 3 D 2 = D 1 1 3 2 Row 1 Example: If D 1 = 5 4 6 9 8 7 Row 3 Now by R 1 R 3, i.e. interchanging elements of Row 1 with corresponding elements of Row 3 of determinant D we get new determinant with change in sign, 9 8 7 Row 1 D 2 = 5 4 6 1 3 2 Row 3 D 2 = D 1 1 3 2 9 8 7 5 4 6 = 5 4 6 9 8 7 1 3 2 The above is also true when any two columns are interchanged. 11

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE iii. If any two rows or columns of a determinant are identical, then its value is zero. Thus for the following determinant a1 b1 c1 a b c = 0...[ R 1 and R 2 are identical] 1 1 1 a b c 3 3 3 1 3 2 Example: For D = 1 3 2 = 0.[ R 1 and R 2 are identical] 9 8 7 The above is also true when any two columns are identical. iv. 12 If all the elements of any row or column are multiplied by a constant k, then the value of new determinant so obtained is k times the value of the original determinant. Thus we have: ka1 kb1 kc1 a1 b1 c1 a2 b2 c2 = k a2 b2 c2 a b c a b c 3 3 3 Example: If D = 3 3 3 1 3 2 5 4 6 9 8 7 Then by R 1 5R 1, i.e. multiplying all the elements of row 1 by 5, the value of the determinant also gets multiplied by 5. 51 53 52 5 15 10 5D = 5 4 6 = 5 4 6 9 8 7 9 8 7 The same can be applied for a column. By C 1 5C 1, i.e. multiplying all the elements of column 1 by 5, the value of the determinant also gets multiplied by 5. 51 3 2 5 3 2 5D = 55 4 6 = 25 4 6 59 8 7 45 8 7 v. If each element of any row or column is expressed as a sum of two terms, then the determinant can be expressed as the sum of two determinants. Thus we have, a1 x b1 y c1z a1 b1 c1 x y z a2 b2 c2 = a2 b2 c2 + a2 b2 c2 a b c a b c a b c 3 3 3 Example: If D = 5 4 6 1 3 2 9 8 7 3 3 3 = 3 3 3 23 31 42 1 3 2 9 8 7

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE vi. Then by the above property, we have 5 4 6 2 3 4 3 1 2 D = 1 3 2 = 1 3 2 + 1 3 2 9 8 7 9 8 7 9 8 7 The above is also true when elements of any column are expressed as sum of two terms. If a constant multiple of all elements of any row (or column) is added to the corresponding elements of any other row (or column), then the value of new determinant so obtained remains unchanged. Thus we have a1 b1 c1 a1 b1 c1 a2 b2 c = 2 a2 b2 c2 a b c a ka b kb c kc 3 3 3 3 1 3 1 3 1 5 4 6 Example: If D = 1 3 2, then 9 8 7 By R 1 R 1 + 2R 2, i.e. adding second multiple of each element of row 2 to corresponding elements of row 1, the value of determinant doesn t change, 52 (1) 42 (3) 62 (2) 7 10 10 D = 1 3 2 = 1 3 2 9 8 7 9 8 7 vii. If all the elements in any one row or column are zeros. then value of determinant is zero. Thus we have, 0 b1 c1 0 b2 c = 0 2 0 b c 3 3 0 1 2 Example: If D = 0 6 5 0 9 10 Then the value of the determinant is equal to zero. 0 1 2 D = 0 6 5 = 0 0 9 10 Note: In all the above properties when we perform some operations on rows (or columns), we state these operations in symbolic form as follows: i. R i R j or R ij for i j means interchange of i th and j th row. ii. C i C j or C ij for i j means interchange of i th and j th column. iii. (k)r j or (k)c j means multiplication of j th row (or j th column) by constant k. iv. R i R i + kr j (or C i C i + k C j ) means change in i th row (or i th column) by adding corresponding k multiples of elements of j th row (or j th column) in i th row (or i th column). 13

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE Solved Problems 1. Evaluate using properties: 16 29 35 1 18 72 i. 50 100 110 ii. 2 40 148 82 158 180 2 45 150 5 13 17 10 10 10 iii. 30 68 105 iv. 213 211 210 25 66 84 372 375 377 16 29 35 i. Let D = 50 100 110 82 158 180 Applying R 3 R 3 2R 1, we get 16 29 35 D = 50 100 110 50 100 110 D = 0 [R 2 and R 3 are identical] 1 18 72 ii. Let D = 2 40 148 2 45 150 Applying R 2 R 2 2R 1, we get 1 18 72 D = 0 4 4 2 45 150 Applying R 3 R 3 2R 1, we get 1 18 72 D = 0 4 4 = 1(24 36) 18(0 0) + 72(0 0) 0 9 6 D = 12 5 13 17 iii. Let D = 30 68 105 25 66 84 Applying R 2 R 2 6R 1, we get 5 13 17 D = 0 10 3 25 66 84 14

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE Applying R 3 R 3 5R 1, we get 5 13 17 D = 0 10 3 = 5(10 3) 13(0 0) + 17(0 0) 0 1 1 D = 35 10 10 10 iv. Let D = 213 211 210 372 375 377 Applying C 2 C 2 C 1, we get 10 0 10 D = 213 2 210 372 3 377 Applying C 3 C 3 C 1, we get 10 0 0 D = 213 2 3 = 10(10 + 9) 0 + 0 372 3 5 D = 10 2. Evaluate using properties: 1 1 x i. 1 x 2 x ii. 1 2 x 3 x 1 1 x i. Let D = 1 x 2 x 1 x x 2 3 1 xy xy x y 1 yz yz y z 1 zx zx z x Taking x common from C 3, we get 1 1 1 D = x 1 x x = x(0) [ C 2 and C 3 are identical] 1 2 x 2 x D = 0 1 xy xy x y ii. Let D = 1 yz yz y z 1 zx zx z x Applying C 3 C 3 + (xyz)c 1, we get 1 xy xy x y xyz 1 xy xy x y z D = 1 yz yz yz xyz 1 zx zx zx xyz = 1 yz yz x y z 1 zx zx xyz 15

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE Taking (x + y + z) common from C 3, we get 1 xy xy D = (x + y + z) 1 yz yz = (x + y + z)(0) [ C 2 and C 3 are identical] 1 zx zx D = 0 3. Find the value of x in the following equations: x1 x1 x1 i. x1 x1 x1 = 0 ii. x1 x1 x1 x1 x1 x1 i. x1 x1 x1 = 0 x1 x1 x1 Applying R 1 R 1 R 3, we get 2 0 2 x1 x1 x1 = 0 x1 x1 x1 Applying R 2 R 2 R 3, we get 2 0 2 0 2 2 = 0 x1 x1 x1 2(2x + 2 2x 2) 0 + 2(0 + 2x + 2) = 0 2(4x) + 2(2x + 2) = 0 8x + 4x + 4 = 0 12x = 4 x = 1 3 x 2 x6 x1 x 6 x1 x2 = 0 x 1 x2 x6 ii. 16 x2 x6 x1 x6 x1 x2 = 0 x1 x2 x6 Applying R 2 R 2 R 1, we get x2 x6 x1 4 7 3 = 0 x1 x2 x6 Applying R 3 R 3 R 1, we get x 2 x6 x1 4 7 3 = 0 3 4 7

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE (x + 2)(49 + 12) (x + 6)(28 + 9) + (x 1) ( 16 21) = 0 (x + 2) (37) (x + 6) (37) + (x 1) (37) = 0 37(x + 2 + x + 6 + x 1) = 0 3x + 7 = 0 x = 7 3 Exercise 1.3 1. Evaluate the following equations using properties: 20 27 36 i. 2 3 4 ii. 1 2 3 2. Find the value of x in the following equations: x 1 1 1 i. 1 x 1 1 = 0 ii. 1 1 x 1 x 1 1 iii. 1 x 1 = 0 iv. 1 1 x 41 1 5 79 7 9 29 5 3 x 1 3 5 2 x 2 5 = 0 2 3 x 4 x 1 4 3 2 x 8 6 = 0 3 12 x 9 1.5 Cramer s Rule Cramer s rule is used to solve simultaneous linear equations in two or three variables using determinants. 1.5 (a) Cramer s Rule for equations involving two unknown variables: The solution of a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 in two unknowns x, y is given by x = D x, y = D y provided D 0 D D where, a1 b1 c1 b1 D =, D x = a b c b Coefficients of x 2 2 Coefficients of y 2 2 Coefficients of x is replaced by constant a1 c1, D y = a2 c2 Coefficients of x Coefficients of y Coefficients of y is replaced by constant 1.5 (b) Cramer s Rule for equations involving three unknown variables: The solution of a 1 x + b 1 y + c 1 z = d 1, a 2 x + b 2 y + c 2 z = d 2 and a 3 x + b 3 y + c 3 z = d 3 in three unknowns x, y, z, is given by x = D x, y = D y Dz, z = D D D provided D 0 17

Target Publications Pvt. Ltd. where, a b c D = a b c a b c 1 1 1 2 2 2 3 3 3 Coefficients Coefficients of x Coefficients of z of y d1 b1 c1 D x = d2 b2 c2 d b c Coefficients of x is replaced 3 3 3 by constant Coefficients of y Coefficients of z D y = Coefficients a d c a d c a d c 1 1 1 2 2 2 3 3 3 of x Coefficients of y is replaced by constant Coefficients of z Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE a1 b1 d1 D z = a2 b2 d2 a3 b3 d3 Coefficients Coefficients of x of z is replaced Coefficients by constant of y 1. Solve using Cramer s rule: 2x + 3y = 5 ; 3x 4y = 1 2x + 3y = 5 3x 4y = 1 2 3 D = = (2) (4) (3) (3) = 8 9 D = 17 18 D x = D x = 17 D y = 3 4 Coefficients of x D y = 17 Illustrative Example Coefficients of x 5 3 1 4 Coefficients of x is replaced by constant Coefficients of y 2 5 3 1 Coefficients of y = (5) (4) (1) (3) = 20 + 3 = 2(1) 3(5) = 2 15 Coefficients of y is replaced by constant

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE By Cramer s rule, we have x = D x 17 = = 1 and y = D y D 17 D x = 1, y = 1 = 17 17 = 1 2. Find x, y and z using Cramer s rule if: 5x 3y + 2z = 5, x + 8y z = 14, 6x 7y = 8 Given equations can be written as, 5x 3y + 2z = 5, x + 8y z = 14, 6x 7y = 8 i.e., 6x 7y + 0z = 8 D = Coefficients of x 5 3 2 1 8 1 6 7 0 = (5) 8 1 7 0 Coefficients of z Coefficients of y (3) 1 1 1 8 + (2) 6 0 6 7 = (5)(0 7) (3) (0 + 6) + (2) ( 7 48) = (5)(7) (3)(6) + (2)(55) = 35 + 18 110 D = 127 D x = 5 3 2 14 8 1 8 7 0 Coefficients Coefficients of z of x is replaced by constant Coefficients of y = (5) 8 1 (3) 14 1 + (2) 14 8 7 0 8 0 8 7 = (5)(0 7) (3) (0 8) + (2) ( 98 + 64) = (5)(7) (3)(8) + (2)(34) = 35 24 68 D x = 127 D y = 5 5 2 1 14 1 6 8 0 = (5) 14 1 8 0 Coefficients Coefficients of z of x Coefficients of y is replaced by constant (5) 1 1 1 14 + (2) 6 0 6 8 = (5) (0 8) (5)(0 + 6) + (2) (8 84) = (5)(8) (5)(6) + (2)(92) = 40 30 184 D y = 254 19

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE D z = Coefficients of x 5 3 5 1 8 14 6 7 8 Coefficients of y = (5) 8 14 7 8 Coefficients of z is replaced by constant (3) 1 14 6 8 + (5) 1 8 6 7 = (5)(64 + 98) (3)(8 84) + (5)(7 48) = (5)(34) (3)(92) + (5)(55) = 170 276 275 D z = 381 By Cramer s rule, we have x = D x 127 = = 1 and y = D y D 127 D x = 1, y = 2, z = 3 = 254 127 D z = 2 and z = D = 381 127 Note: 1. For Cramer s rule, always write the equation in linear form i.e., ax + by = c or ax + by + cz = d, using proper substitution. Example: 3 4 a. 2 and 1 3 5 are not linear. x y x y = 3 Substituting 1 x = a and 1 y = b, the equations reduce to 3a 4b = 2 and a + 3b = 5 which are linear equations in variables a and b. Hence, we can use Cramer s rule. The detail of the method is explained in illustrative examples. b. yz + zx + xy = xyz, 3yz + zx + 2xy = 4xyz, 9yz + zx + 4xy = 16xyz. Observe these equations carefully! If we divide each equation throughout by xyz, we get 1 1 1 1 x y z 3 1 2, 4, 9 1 4 16. x y z x y z Substituting 1 x = a, 1 y = b, 1 z = c, equations reduce to 20 a + b + c = 1, 3a + b + 2c = 4, 9a + b + 4c = 16. which are linear equations in variables a, b, c. Hence, we can use Cramer s rule. 2. For equations having missing variables, take zero as their co-efficient. Example: Consider the system of equations x + y = 3, y + z = 3, z + x = 4 These equations in standard form are as follows: x + y + 0z = 3, 0x + y + z = 3, x + 0y + z = 4

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE Solved Problems Cramer s rule involving two variables: 1. Solve the following equations using Cramer s Rule: i. 3x + 4y 7 = 0, y 2x = 3 ii. 4x 3y 2 = 0, 3x + 4y + 6 = 0. i. Given equations are 3x + 4y 7 = 0 i.e., 3x + 4y = 7 and y 2x = 3 i.e., 2x + y = 3 3 4 D = 2 1 = 3(1) ( 2) 4 = 3 + 8 D = 11 D x = 7 4 = 7(1) (3) 4 = 7 12 3 1 D x = 5 D y = 3 7 = 3(3) (2) 7 = 9 + 14 2 3 D y = 23 By Cramer s Rule, we have x = D x and y = D y D D x = 5 11 and y = 23 11 ii. Given equations are 4x 3y 2 = 0 i.e., 4x 3y = 2 and 3x + 4y + 6 = 0 i.e., 3x + 4y = 6 D = 4 3 = 4(4) 3(3) = 16 + 9 3 4 D = 25 2 3 D x = = 2(4) (6)(3) = 8 18 6 4 D x = 10 D y = 4 2 = 4(6) 3(2) = 24 6 3 6 D y = 30 By Cramer s Rule, we have x = D x = 10 and y = D y = 30 D 25 D 25 x = 2 5 and y = 6 5 21

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE 2. Find x and y using Cramer s Rule, if 1 2 i. x y = 6, 3 1 x y = 8 ii. 1 2 2 x 3 = 10, 3 5 y x y 2 3 = 3. i. Let 1 x = a and 1 y = b given equations become a 2b = 6 3a + b = 8 D = 1 2 = 1(1) 3(2) = 1 + 6 3 1 D = 7 D a = 6 2 = 6(1) 8(2) = 6 + 16 8 1 D a = 22 D b = 1 6 = 1(8) 3(6) = 8 18 3 8 D b = 10 By Cramer s Rule, we have D D a = a and b = b D D a = 22 7 1 x = 22 7 x = 7 22 10 and b = 7 1 and y = 10 7 and y = 7 10 ii. Let 1 2 x = a and 1 3 y = b given equations become a + 2b = 10 3a 5b = 3 D = 1 2 = 1(5) 3(2) = 5 6 3 5 D = 11 D a = 10 2 = 10 (5) (3)2 = 50 + 6 3 5 D a = 44 D b = 1 10 = 1(3) 3(10) = 3 30 3 3 D b = 33 22

Target Publications Pvt. Ltd. By Cramer s Rule, we have D a a = D = 44 11 and b = D b D = 33 11 a = 4 and b = 3 1 2 = 4 and 1 = 3 x 3 y 2 x = 2 2 and 3 y = 3 1 Equating powers from both sides, we get x = 2 and y = 1 x = 2 and y = 1 Cramer s rule involving three variables: 3. Solve the following equations using Cramer s Rule: i. 2x y + 3z = 9, x + y + z = 6, x y + z = 2 ii. x + y z 2 = 0, x 2y + z = 3, 2x y 3z + 1 = 0 iii. x 2y = 3, x + 3z = 5, 4y z = 1 iv. x + 2y + 3z = 6, 2x + 4y + z = 7, 3x + 2y + 9z = 14 v. 2x y + 3z =4, x + y + z = 2, 3x + y z = 2. vi. x + 2y z = 3, 3x y + 2z = 1, 2x 2y + 3z = 2 vii. x 3y + z = 2, 3x + y + z = 1, 5x + y + 3z = 3 i. Given equations are 2x y + 3z = 9 x + y + z = 6 x y + z = 2 2 1 3 D = 1 1 1 = 2(1 + 1) + 1(1 1) + 3(1 1) = 4 + 0 6 1 1 1 D = 2 9 1 3 D x = 6 1 1 = 9(1 + 1) + 1(6 2) + 3(6 2) = 18 + 4 24 2 1 1 D x = 2 2 9 3 D y = 1 6 1 = 2(6 2) 9(1 1) + 3(2 6) = 8 0 12 1 2 1 D y = 4 2 1 9 D z = 1 1 6 = 2(2 + 6) + 1(2 6) + 9( 1 1) = 16 4 18 1 1 2 D z = 6 Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE 23

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE By Cramer s Rule, we have x = D x -2 = D - 2, y = Dy -4 = D - 2, z = D z D = -6-2 x = 1, y = 2, z = 3 ii. Given equations are x + y z 2 = 0 i.e., x + y z = 2 x 2y + z = 3 2x y 3z + 1 = 0 i.e., 2x y 3z = 1 1 1 1 D = 1 2 1 = 1(6 + 1) 1(3 2) 1(1 + 4) = 7 + 5 3 2 1 3 D = 9 2 1 1 D x = 3 2 1 = 2(6 + 1) 1( 9 + 1) 1( 3 2) = 14 + 8 + 5 1 1 3 D x = 27 1 2 1 D y = 1 3 1 = 1( 9 + 1) 2( 3 2) 1(1 6) = 8 + 10 + 7 2 1 3 D y = 9 1 1 2 D z = 1 2 3 = 1(2 + 3) 1(1 6) + 2(1 + 4) = 5 + 7 + 6 2 1 1 D z = 18 By Cramer s Rule, we have x = D x = 27 D 9, y = Dy = 9 D 9, z = D z D = 18 9 x = 3, y = 1, z = 2 iii. Given equations are x 2y = 3 i.e., x 2y + 0z = 3 x + 3z = 5 i.e., x + 0y + 3z = 5 4y z = 1 i.e., 0x + 4y z = 1 1 2 0 D = 1 0 3 = 1(0 12) + 2(1 0) + 0(4 0) = 12 + 2 0 4 1 D = 10 3 2 0 D x = 5 0 3 = 3(0 12) + 2(5 3) + 0(20 0) = 36 16 1 4 1 D x = 52 24

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE D y = 1 3 0 1 5 3 = 1( 5 3) 3(1 0) + 0(1 0) = 8 3 0 1 1 D y = 11 1 2 3 D z = 1 0 5 = 1(0 20) + 2(1 0) + 3(4 0) = 20 2 12 0 4 1 D z = 34 By Cramer s Rule, we have x = D x - = 52 D - 10, y = Dy - = 11 D - 10, z = D z D = - 34-10 x = 26 5, y = 11 10, z = 17 5 iv. Given equations are x + 2y + 3z = 6 2x + 4y + z = 7 3x + 2y + 9z = 14 1 2 3 D = 2 4 1 = 1(36 2) 2(18 3) + 3(4 12) = 34 30 24 3 2 9 D = 20 6 2 3 D x = 7 4 1 = 6(36 2) 2(63 14) + 3(14 56) = 204 98 126 14 2 9 D x = 20 1 6 3 D y = 2 7 1 = 1(63 14) 6(18 3) + 3(28 21) = 49 90 + 21 3 14 9 D y = 20 1 2 6 D z = 2 4 7 = 1(56 14) 2(28 21) + 6(4 12) = 42 14 48 3 2 14 D z = 20 By Cramer s Rule, we have x = D x -20 = D - 20, y = D - y = 20 D - 20, z = D z D = - 20-20 x = 1, y = 1, z = 1 25

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE v. Given equations are 2x y + 3z = 4 x + y + z = 2 3x + y z = 2 2 1 3 D = 1 1 1 = 2(1 1) + 1(1 3) + 3(1 3) = 4 4 6 3 1 1 D = 14 4 1 3 D x = 2 1 1 = 4(1 1) + 1(2 2) + 3(2 2) = 8 4 2 1 1 D x = 12 2 4 3 D y = 1 2 1 = 2(2 2) 4( 1 3) + 3(2 6) = 8 + 16 12 3 2 1 D y = 4 2 1 4 D z = 1 1 2 = 2(2 2) + 1(2 6) + 4(1 3) = 0 4 8 3 1 2 D z = 12 By Cramer s Rule, we have x = D x - = 12 D - 14 y -4 = D - 14 D z D = - 12-14 x = 6 7, y = 2 7, z = 6 7 vi. Given equations are x + 2y z = 3 3x y + 2z = 1 2x 2y + 3z = 2 1 2 1 D = 3 1 2 = 1(3 + 4) 2(9 4) 1(6 + 2) = 1 10 + 4 2 2 3 D = 5 3 2 1 D x = 1 1 2 = 3(3 + 4) 2(3 4) 1(2 + 2) = 3 + 2 2 2 3 D x = 5 1 3 1 D y = 3 1 2 = 1(3 4) 3(9 4) 1(6 2) = 1 15 4 2 2 3 D y = 20 26

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE D z = 1 2 3 3 1 1 = 1(2 + 2) 2(6 2) + 3(6 + 2) = 0 8 12 2 2 2 D z = 20 By Cramer s Rule, we have x = D x = 5 D, y = Dy = 20 5 D, z = D z 5 D = 20 5 x = 1, y = 4, z = 4 vii. Given equations are x 3y + z = 2 3x + y + z = 1 5x + y + 3z = 3 1 3 1 D = 3 1 1 = 1(3 1) + 3(9 5) +1(3 5) = 2 + 12 2 5 1 3 D = 12 2 3 1 D x = 1 1 1 = 2(3 1) + 3(3 3) + 1(1 3) = 4 + 0 2 3 1 3 D x = 2 1 2 1 D y = 3 1 1 = 1(3 3) 2(9 5) + 1(9 5) = 0 8 + 4 5 3 3 D y = 4 1 3 2 D z = 3 1 1 = 1(3 1) + 3(9 5) + 2(3 5) = 2 + 12 4 5 1 3 D z = 10 By Cramer s Rule, we have x = D x = 2 D 12, y = D D x = 1 6, y = 1 5, z = 3 6 y = 4 D, z = z 12 D = 10 12 4. Find x, y, z using Cramer s Rule, if i. 1 1 1 1 2 1 2 1 3 = 2, = 3, = 1 x y z x y z x y z ii. 2 x 1 + 2 y 1 z = 1, 3 2 + 1 x 1 y 1 z = 1, 1 x 1 3 y 1 z = 2 27

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE i. Let 1 x = a, 1 y = b and 1 z = c D = given equations become a + b + c = 2 a 2b + c = 3 2a b + 3c = 1 1 1 1 1 2 1 = 1( 6 + 1) 1(3 2) + 1(1 + 4) = 5 1 + 3 2 1 3 D = 3 2 1 1 D a = 3 2 1 = 2( 6 + 1) 1(9 + 1) + 1(3 2) = 10 10 5 1 1 3 D a = 25 1 2 1 D b = 1 3 1 = 1(9 + 1) 2(3 2) + 1(1 6) = 10 2 7 2 1 3 D b = 1 1 1 2 D c = 1 2 3 = 1(2 + 3) 1(1 6) + 2(1 + 4) = 5 + 7 + 6 2 1 1 D c = 18 By Cramer s Rule, we have Da a = D = 25 Db, b = 3 D = 1 3, c = Dc D = 18 3 a = 25 3, b = 1 3, c = 6 1 x = 25 3, 1 y = 1 3, 1 z = 6 x = 3 1, y = 3, z = 25 6 ii. 28 Let 1 = a, x 1 1 = b and 1 y 1 z = c given equations become 2a b + 2c = 1 3a + 2b c = 1 a 3b 3c = 2

Target Publications Pvt. Ltd. Basic Physics Chapter (F.Y.Dip.Sem.-1) 01: Determinant MSBTE D = 2 1 2 3 2 1= 2( 6 3) + 1( 9 + 1) + 2( 9 2) = 18 8 22 1 3 3 D = 48 1 1 2 D a = 1 2 1= 1(6 3) + 1( 3 + 2) + 2( 3 4) = 9 1 14 2 3 3 D a = 24 2 1 2 D b = 3 1 1 = 2( 3 + 2) 1( 9 + 1) + 2(6 1) = 2 + 8 + 10 1 2 3 D b = 16 2 1 1 D c = 3 2 1 = 2(4 + 3) + 1(6 1) + 1( 9 2) = 14 + 5 11 1 3 2 D c = 8 By Cramer s Rule, we have D a = a D = 24 48 D, b = b D = 16 48 D, c = c D = 8 48 a = 1 2, b = 1 1, c = 3 6 1 = 1 x 1 2, 1 y 1 = 1 1, 3 z = 1 6 x 1 = 2, y + 1 = 3, z = 6 x = 3, y = 4, z = 6 5. Solve the following equations for x, y, z, if sin x + cos y + tan z = 3, 2sin x + cos y + tan z = 4, 3sin x + 4cos y 2tan z = 5, where 0 x, y, z 90. Let sin x = a, cos y = b and tan z = c given equations become a + b + c = 3 2a + b + c = 4 3a + 4b 2c = 5 1 1 1 D = 2 1 1 = 1(2 4) 1( 4 3) + 1(8 3) = 6 + 7 + 5 3 4 2 D = 6 29

Target Publications Pvt. Ltd. Basic Mathematics Basic Physics (F.Y.Dip.Sem.-1) MSBTE 3 1 1 D a = 4 1 1 = 3(2 4) 1( 8 5) + 1(16 5) = 18 + 13 + 11 5 4 2 D a = 6 1 3 1 D b = 2 4 1 = 1( 8 5) 3( 4 3) + 1(10 12) = 13 + 21 2 3 5 2 D b = 6 1 1 3 D c = 2 1 4 = 1(5 16) 1(10 12) + 3(8 3) = 11 + 2 + 15 3 4 5 D c = 6 By Cramer s Rule, we have D a = a D = 6 6, b = D b D = 6 6, c = D c D = 6 6 a = 1, b = 1, c = 1 sin x = 1, cos y = 1, tan z = 1 sin x = sin 90, cos y = cos 0, tan z = tan 45.[ 0 x, y, z 90] x = 90, y = 0, z = 45 6. Find x, y, z, if 5 e x + 4 y 3 z = 1, 4 e x + 3 y 2 z = 2 and 3 e x 2 y + z = 3. Let e x = a, y = b, z = c given equations become 5a + 4b 3c = 1 4a + 3b 2c = 2 3a 2b + c = 3 5 4 3 D = 4 3 2 = 5(3 4) 4(4 + 6) 3(8 9) = 5 40 + 51 3 2 1 D = 6 1 4 3 D a = 2 3 2 = 1(3 4) 4(2 + 6) 3(4 9) = 1 32 + 39 3 2 1 D a = 6 5 1 3 D b = 4 2 2 = 5(2 + 6) 1(4 + 6) 3(12 6) = 40 10 18 3 3 1 D b = 12 30