Lecture notes on basic Mathematics

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MATHEMATICS I & II DIPLOMA COURSE IN ENGINEERING FIRST SEMESTER A Publication under Government of Tamilnadu Distribution of Free Textbook Programme ( NOT FOR SALE ) Untouchability is a sin Untouchability is a crime Untouchability is inhuman DIRECTORATE OF TECHNICAL EDUCATION GOVERNMENT OF TAMIL NADUSEMESTER I MATHEMATICS – I UNIT – I DETERMINANTS 1.1 Definition and expansion of determinants of order 2 and 3 Properties of determinants Cramer’s rule to solve simultaneous equations in 2 and 3 unknowns-simple problems. 1.2 Problems involving properties of determinants 1.3 Matrices Definition of matrix. Types of matrices. Algebra of matrices such as equality, addition, subtraction, scalar multiplication and multiplication of matrices. Transpose of a matrix, adjoint matrix and inverse matrix-simple problems. 1.1. DETERMINANT The credit for the discovery of the subject of determinant goes to the German mathematician, Gauss. After the introduction of determinants, solving a system of simultaneous linear equations becomes much simpler. Definition: Determinant is a square arrangement of numbers (real or complex) within two vertical lines. Example : a b 1 1 is a determinant a b 2 2 Determinant of second order: a b The symbol consisting of 4 numbers a, b, c and arranged in c d two rows and two columus is called a determinant of second order. The numbers a,b,c, and d are called elements of the determinant The value of the determinant isΔ =ad- bc 1Examples: 2 3 1. = (2) (1) – (5) (3) = 2 – 15 = -13 5 1 4 6 2. = (4) (-5) – (6) (3) = -20 – 18 = -38 3− 5 Determent of third order: a b c 1 1 1 The expression a b c consisting of 2 2 2 a b c 3 3 3 nine elements arranged in three rows and three columns is called a determinant of third order The value of the determinant is obtained by expanding the determinant along the first row b c a c a b 2 2 2 2 2 2 Δ =a -b + c 1 1 1 b c a c a b 3 3 3 3 3 3 =) a (b c - b c ) - b (a c - a c ) + c (a b - a b 1 2 3 3 2 1 2 3 3 2 1 2 3 3 2 Note: The determinant can be expanded along any row or column. Examples: = 1(1− 8)− 2(2− 20)+ 3(4− 5) 1 2 3 = 1(−7)− 2(−18)+ 3(−1) (1) 2 1 4 =−7+ 36− 3 5 2 1 =−10+ 36= 26 = 3(6+ 4)+ 1(−2+ 3) 3 0 1 = 3(10)+ 1(1) (2) 2− 3 4 = 30+ 1 1−1− 2 = 31 Minor of an element Definition : Minor of an element is a determinant obtained by deleting the th th row and column in which that element occurs. The Minor of I row J Column element is denoted by m ij 2Example: 1−1 3 0 4 2 11 5− 3 0 4 Minor of 3 = = 0-44 = -44 11 5 − 1 3 Minor of 0 = = 3-15 = -12 5− 3 Cofactor of an element Definition : th th Co-factor of an element in i row,j column is the signed minor of th th I row J Column element and is denoted by A . ij i+j (i.e) A = (-1) m ij ij i+ j Thesignis attachedbythe rule (-1) Example 3− 2 4 2 1 0 7 11 6 2 0 1+2 3 Co-factor of -2 = (-1) = (-1) (12)= -12 7 6 − 2 4 3+1 4 Co-factor of 7= (-1) = (-1) (0-4)= -4 1 0 Properties of Determinants: Property 1: The value of a determinant is unaltered when the rows and columns are interchanged. a a a a b c 1 2 3 1 1 1 T (i.e) IfΔ= b b b andΔ = a b c , 1 2 3 2 2 2 c c c a b c 1 2 3 3 3 3 T thenΔ =Δ 3Property 2: If any two rows or columns of a determinant are interchanged the value of the determinant is changed in its sign. a a a b b b 1 2 3 1 2 3 IfΔ = b b b andΔ = a a a , 1 1 2 3 1 2 3 c c c c c c 1 2 3 1 2 3 thenΔ =-Δ 1 Note: R and R are interchanged. 1 2 Property 3: If any two rows or columns of a determinant are identical, then the value of the determinant is zero. a a a 1 2 3 (i.e) The value of a a a is zero Since R≡ R 1 2 1 2 3 c c c 1 2 3 Property 4: If each element of a row or column of a determinant is multiplied by any number K= 0, then the value of the determinant is multiplied by / the same number K. a b c 1 1 1 IfΔ = a b c 2 2 2 a b c 3 3 3 Ka Kb Kc 1 1 1 andΔ= a b c , 1 2 2 2 a b c 3 3 3 thenΔ =KΔ 1 Property 5: If each element of a row or column is expressed as the sum of two elements, then the determinant can be expressed as the sum of two determinants of the same order. 4a+ d b+ d c+ d 1 1 1 2 1 3 (i.e) IfΔ = a b c , 2 2 2 a b c 3 3 3 a b c d d d  1 1 1 1 2 3   thenΔ = a b c + a b c 2 2 2 2 2 2     a b c a b c 3 3 3  3 3 3 Property 6: If each element of a row or column of a determinant is multiplied by a constant K=/ 0 and then added to or subtracted from the corresponding elements of any other row or column then the value of the determinant is unaltered. a b c 1 1 1 LetΔ = a b c 2 2 2 a b c 3 3 3 a+ ma+ na b+ mb+ nb c+ mc+ nc 1 2 3 1 2 3 1 2 3 Δ = a b c 1 2 2 2 a b c 3 3 3 a b c ma mb mc na nb nc 1 1 1 2 2 2 3 3 3 = a b c + a b c + a b c 2 2 2 2 2 2 2 2 2 a b c a b c a b c 3 3 3 3 3 3 3 3 3 a b c a b c 2 2 2 3 3 3 =Δ +m a b c +n a b c 2 2 2 2 2 2 a b c a b c 3 3 3 3 3 3 Δ =Δ +m (0) +n(0) =Δ 1 Property 7: In a given determinant if two rows or columns are identical for x = a, then (x-a) is a factor of the determinant. 51 1 1 LetΔ = a b c 3 3 3 a b c 1 1 1 For a=b,Δ= b b c =0 C , and C are identical 1 2 3 3 3 b b c ∴ (a-b) is a factor ofΔ Notation : Usually the three rows of the determinant first row, second row and third row are denoted by R,R and R respectively and the 1 2 3 columns by C ,C and C 1 2 3 If we have to interchange two rows say R and R the symbol 1 2 double sided arrow will be used. We will write like this R↔ R it 2 2 should be read as “is interchanged with” similarly for columns C↔ C . 2 2 If the elements of R are subtracted from the corresponding 2 elements of R , then we writeR-R similarly for columns also. 1 1 2 If the elements of one column say C ,‘m’ times the element of 1 C and n times that of C are added, we write like this C→C+m C 2 3 1 1 2 +n C . Here one sided arrow is to be read as “is changed to” 3 Solution of simultaneous equations using Cramer’s rule: Consider the linear equations. a x+ b y= c 1 1 1 a x+ b y= c 2 2 2 a b 1 1 letΔ = a b 2 2 6c b 1 1 Δ= x c b 2 2 a c 1 1 Δ= y a c 2 2 Δ Δ y x Then x = and y = ,provided Δ≠ 0 ΔΔ x, y are unique solutions of the given equations. This method of solving the line equations is called Cramer’s rule. Similarly for a set of three simultaneous equations in x, y and z a 1 x+ b 1y+ c 1z= d 1 a 2 x+b 2y+ c 2z= d 2 and a 3 x+b 3y+ c 3z= d , the solution of the system of equations, 3 Δ ΔΔ y x z by cramer’s rule is given by, x= ,y = and z = , ΔΔΔ provided Δ≠ 0 where a b c d b c 1 1 1 1 1 1 Δ = a b cΔ= d b c 2 2 2 x 2 2 2 a b c d b c 3 3 3 3 3 3 a d c a b d 1 1 1 1 1 1 Δ= a d c andΔ= a b d y 2 2 2 z 2 2 2 a d c a b d 3 3 3 3 3 3 71.1 WORKED EXAMPLES PART – A x 2 1. Solve =0 x 3x Solution: x 2 =0 x 3x 2 3x− 2x= 0 x(3x− 2)= 0 2 x= 0 or x= 3 x 8 2. Solve =0 2 x Solution: x 8 =0 2 x 2 x− 16= 0 2 x= 16 x=±4 m 2 1 3. Find the value of ‘m’ when 3 4 2= 0 − 7 3 0 Solution: m 2 1 Given 3 4 2= 0 − 7 3 0 Expanding the determinant along,Rwehave 1 m(0-6)-2 (0+14) +1 (9+28) = 0 m(-6) -2 (14) +1 (37) = 0 8-6m -28 + 37= 0 -6m +9 = 0 -6m = -9 9 3 m= = 6 2 1 2 0 4. Find the Co-factor of element 3 in the determinant− 1 3 4 5 6 7 Solution: 1 0 2+2 Cofactor of 3 = A = (-1) 22 5 7 4 = (-1) (7-0) =7 PART – B 1. Using cramer’s rule, solve the following simultaneous equations x+y+ z=2 2x-y–2z = -1 x–2y–z=1 Solution: 1 1 1 Δ= 2− 1− 2 1− 2− 1 = 1 (1-4) -1 (-2+2) +1 (-4+1) = 1 (-3) -1 (0) +1 (-3) =-3-3 =-6≠ 0 2 1 1 Δ=− 1− 1− 2 x 1− 2− 1 9= 2 (1-4) -1 (1+2) +1 (2+1) = 2 (-3) -1 (3) +1 (3) =-6-3+3=-6 1 2 1 Δ = 2− 1− 2 y 1 1− 1 = 1 (1+2) -2 (-2+2) +1 (2+1) = 1 (3) -2 (0) +1 (3) =3+3 = 6 1 1 2 Δ = 2− 1− 1 z 1− 2 1 = 1(-1-2) -1 (2+1) +2 (-4+1) = -3-3-6 = -12 ∴ By Cramer’s rule, Δ Δ−6 6 y x x= = =1 y= = =-1 Δ− 6Δ− 6 Δ−12 z z=== 2 Δ− 6 2. Using Cramer’s rule solve: -2y+3z-2x+1=0 -x+y-z+5=0 -2z -4x+y = 4 Solution: Rearrange the given equations in order -2x-2y+3z = -1; -x+y-z = -5; -4x+y-2z =4 − 2− 2 3 Δ =− 1 1− 1 − 4 1− 2 10= -2(-2+1) +2 (2-4) + 3(-1+4) = -2 (-1) +2 (-2) +3 (3) = 2-4+9 =7 −1− 2 3 Δ =− 5 1−1 x 4 1− 2 = -1(-2+1) +2 (10+4) +3 (-5-4) = 1+28-27 =2 − 2− 1 3 Δ =− 1− 5− 1 y − 4 4− 2 = -2 (10+4) +1 (2-4) +3 (-4-20) = -2(14) +1 (-2) +3 (-24) = -28-2-72 =-102 − 2− 2− 1 Δ =− 1 1− 5 z − 4 1 4 = -2 (4+5) +2 (-4-20) -1 (-1+4) = -18-48-3 =-69 Δ Δ 2− 102Δ−69 y x z x== ,y== , and z== Δ 7Δ 7Δ 7 3. Using Cramer’s rule solve 2x-3y =5 x-8=4y 11Solution: 2x-3y = 5 x-4y = 8 2− 3 Δ = = (2) (-4) – (-3) (1) 1− 4 =-8 +3=-5 5− 3 Δ = = (5) (-4) – (-3) (8) x 8− 4 =-20 + 24=4 2 5 Δ = = 16 – 5= 11 y 1 8 By Cramer’s rule Δ 4 4 x x= = =- Δ− 5 5 Δ 11 11 y y= = =- Δ− 5 5 1.2 PROBLEMS INVOLVING PROPERTIES OF DETERMINANTS PART-A 1) Evaluate 20 11 31 11− 7 4 19 11 30 Solution: 20 11 31 Δ =− 11 7 4 19 11 30 31 11 31 = 4− 7 4 C→ C+ C 1 1 2 30 11 30 Δ =0 since C≡ C 3 1 122) Without expanding, find the value of 1− 2 3 1−1 2 3− 6 9 Solution: 1− 2 3 1− 2 3 LetΔ = 1−1 2 = 1−1 2 3− 6 9 3(1) 3(−2) 3(3) 1− 2 3 =3 1−1 2 1− 2 3 = 3 (0) = 0, since R≡ R 1 3 3) Evaluate 1 a b+ c 1 b c+ a 1 c a+ b Solution: 1 a b+ c LetΔ = 1 b c+ a 1 c a+ b 1 a+ b+ c b+ c = 1 a+ b+ c c+ a C→ C+ C 2 2 3 1 a+ b+ c a+ b 1 1 b+ c = (a+b+c) 1 1 c+ a 1 1 a+ b = (a+b+c) (0) = 0, since C≡ C 1 2 13x− y y− z z− x 4) Prove that y− z z− x x− y =0 z− x x− y y− z Solution: x− y y− z z− x L.H.S = y− z z− x x− y z− x x− y y− z x− y+ y− z+ z− x y− z z− x y− z+ z− x+ x− y z− x x− y C→ C+ C+ C 1 1 2 3 z− x+ x− y+ y− z x− y y− z 0 y− z z− x = 0 z− x x− y =0 =R.H.S 0 x− y y− z PART – B 2 1 x x 2 1) Prove that 1 y y = (x-y) (y-z) (z-x) 2 1 z z Solution: 2 1 x x 2 L.H.S = 1 y y 2 1 z z 2 2 0 x− y x− y 2 2 0 y− z y− z R→ R− R ,R→ R− R 1 1 2 2 2 3 2 1 z z 140 1 x+ y = (x-y) (y-z) 0 1 y+ z 2 1 z z 1 x+ y = (x-y) (y-z) (expanded along the first column) 1 y+ z =(x-y) (y-z) 1(y+z) –1(x+y) = (x-y) (y-z) (z-x) L.H.S = R.H.S 1 1 1 2) Prove that a b c = (a + b + c) (a-b) (b-c) (c-a) 3 3 3 a b c Solution: 1 1 1 L.H.S = a b c 3 3 3 a b c o o 1 Δ= a− b b− c c C→ C− C ,C→ C− C 1 1 2 2 2 3 3 3 3 3 3 a− b b− c c 0 0 1 Δ = a− b b− c c 2 2 2 2 3 (a− b)(a+ ab+ b ) (b− c)(b+ bc+ c ) c 0 0 1 Δ = (a-b) (b-c) 1 1 c 2 2 2 2 3 a+ ab+ b b+ bc+ c c 1 1 Δ = (a-b) (b-c) 2 2 2 2 a+ ab+ b b+ bc+ c (expanded along the first row ) 152 2 2 2 = (a - b) (b - c)b+ bc+ c−(a+ ab+ b) 2 2 2 2 = (a - b) (b - c) b + bc + c - a - ab - b 2 2 = (a - b) (b - c) bc + c - a - ab 2 2 = (a - b) (b - c) c - a +b(c - a) = (a - b) (b - c) (c + a) (c - a) +b (c - a) = (a - b) (b - c) (c - a) c + a + b= R.H.S 1+ a 1 1 3) Prove that 1 1+ a 1 =a² (3+a) 1 1 1+ a Solution: 1+ a 1 1 L.H.S = 1 1+ a 1 1 1 1+ a 3+ a 3+ a 3+ a 1 1+ a 1 R→ R+ R+ R 1 1 2 3 1 1 1+ a 1 1 1 = 3+ a1 1+ a 1 1 1 1+ a 0 0 1 =(3+a)− a a 1 C→ C - C , C→ C - C 1 1 2 2 2 3 0− a 1+ a − a a =(3+a) =(3+a) (a²-0) 0− a =a² (3+a) = R.H.S 16a− b− c 2a 2a 3 4) Prove that 2b b− c− a 2b = (a+b+c) 2c 2c c− a− b Solution: a− b− c 2a 2a L.H.S = 2b b− c− a 2b 2c 2c c− a− b a+ b+ c a+ b+ c a+ b+ c = 2b b− c− a 2b R→ R+ R+ R 1 1 2 3 2c 2c c− a− b 1 1 1 = (a+b+c) 2b b− c− a 2b 2c 2c c− a− b 0 0 1 C→ C - C , 1 1 2 = (a+b+c) a+ b+ c− (a+ b+ c) 2b C→ C - C 2 2 3 0 a+ b+ c c− a− b a+ b+ c− (a+ b= c) = (a+b+c) 0 (a+ b+ c) 2 = (a+b+c) (a+b+c) –0 3 = (a+b+c) =R.H.S 1+ a 1 1  1 1 1 5) Prove that 1 1+ b 1 =abc 1+++   a b c   1 1 1+ c Solution: 1 1 1 a(+ 1) a( ) a( ) a a a 1+ a 1 1 1 1 1 LetΔ = 1 1+ b 1 = b( ) b(+ 1) b( ) b b b 1 1 1+ c 1 1 1 c( ) c( ) c(+ 1) c c c 171 1 1 + 1 a a a 1 1 1 = abc+ 1 b b b 1 1 1 + 1 c c c 1 1 1 1 1 1 1 1 1 1+++ 1+++ 1+++ a b c a b c a b c 1 1 1 =abc+ 1 R→ R+ R+ R 1 1 2 3 b b b 1 1 1 + 1 c c c 1 1 1 1 1 1 1 1 1 =abc (1+++ )+ 1 a b c b b b 1 1 1 + 1 c c c 0 0 1 1 1 1 1 =abc (1+++ )− 1 1 C→ C - C , C→ C - C 1 1 2 2 2 3 a b c b 1 0− 1+ 1 c − 1 1 1 1 1 =abc (1+++ ) a b c 0− 1 1 1 1 =abc (1+++ ) 1-0 a b c 1 1 1 Δ =abc (1+++ ) a b c 181.3 MATRICES Introduction: The term matrix was first introduced by a French mathematician Cayley in the year 1857. The theory of matrices is one of the powerful tools of mathematics not only in the field of higher mathematics but also in other branches such as applied sciences, nuclear physics, probability and statistics, economics and electrical circuits. Definition: A Matrix is a rectangular array of numbers arranged in to rows and columns enclosed by parenthesis or square brackets. Example: 2 1 0     1 2 1.A= 2.B=− 5 6 7     3 4     1 0 8   Usually the matrices are denoted by capital letters of English alphabets A,B,C…,etc and the elements of the matrices are represented by small letters a,b,c,.etc. Order of a matrix If there are m rows and n columns in a matrix, then the order of thematrixismxnorm byn. a a a  1 2 3 Example: A=   b b b  1 2 3 A has two rows and three columns. We say that A is a matrix of order 2x3 Types of matrices Row matrix: A matrix having only one row and any number of columns is called a row matrix. Eg: A = (1 2 -3) 19

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