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University
MRCETCourse
B.Tech 1st YearMALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (An Autonomous Institution – UGC, Govt.of India) Recognizes under 2(f) and 12(B) of UGC ACT 1956 (Affiliated to JNTUH, Hyderabad, Approved by AICTE –Accredited by NBA & NAAC -“A” Grade -ISO 9001:2015 Certified)
MATHEMATICS -I B.Tech – I Year – I Semester DEPARTMENT OF HUMANITIES AND SCIENCES MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD)
Preface for Engineering Mathematics Courses Education in mathematics forms the basis of s cience and engineering from undergraduate to graduate level, because engineering and science are largely based on mathematical modelling. The level and the quality of mathematics education sets the level of the education as a whole.
Our aim is to develop a complete program for mathematics education in science and engineering from basic undergraduate to graduate education. This includes several courses like o Mathematics I o Mathematics II
o Mathematics III o Probability and Random Processes o Probability and Statistics These courses are introduced at different levels/semesters of the engineering program some being common and some based on the branch/specialization chosen by the student.
The purpose of these courses is to arm the student with the necessary ideas and methods, so that when mathematical elements appear in other courses and research work, one can tackle them with confidence, possibly with further independent study into specialized areas. Its major role is to summarize, crystallize, enhance and give a forward orientation to the mathematical methods taught in undergraduate curriculum, with projections to future
requirements.
Common Features of the Mathematical Courses:
The courses are based on a synthesis of mathematics, computation and application.
The courses are designed basing on new interests and needs of the current scenario, giving a new united presentation from the start based on constructive mathematical methods including a computational methodology.
These courses are designed at different levels of ambition co ncerning both mathematical analysis and computation, while keeping a common basic core.
These courses increase the motivation of the student by applying mathematical methods to interesting and important concrete problems right from their introduction.
Whil e emphasis may be put on problem solving, these courses gives theoretical and computational methods and builds confidence.
The course contains most of the traditional material from basic courses in analysis, linear algebra, applied mathematics and higher e ngineering mathematics.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) Emphasis is put on giving the student a solid understanding of basic mathematical concepts.
The student acquires solid skills of implementing and applying the computational methods learnt.
The idea is that making the student comfort able with both advanced mathematical concepts and modern computational techniques, will open a wealth of possibilities of applying mathematics to problems of real interest.
Mathematics I : Course Description This course is about the mathematics that is mo st widely used i n the engineering core subjects. Mathematics I provide an introduction to linear algebra, multi -variable calculus, ordinary differential equations (ODEs) and Laplace transforms. Topics include the solutions for the system of linear equation s, Eigen values and Eigen vectors of a matrix, Extreme
values of functions of two variables with / without constraints, first order, first degree differential equations and their applications, higher order differential equations and differential equations with initial conditions using Laplace Transformation.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY B. TECH - I- YEAR - I-SEM L T/P/D C 3 1/ -/- 4
(R22A002 3)MATHEMATICS -I (common to all branches) Course Objectives: To learn 1. The concept of a Rank of the matrix and applying the concept to know the consistency and solving the system of linear equations.
2. The c oncept of Eigen values, Eigen vectors and Diagonolisation.
3. The maxima and minima of functions of several variables.
4. The A pplications of first order or dinary differential equations.
5. The methods to solve higher order differential equations.
UNIT I: Matrices Introduction ,Types of matrices ,R ank of a matrix - Echelon form and Normal form, Consistency of system of linear equations (Homogeneous and Non -Homogeneous) -Gauss elimination method and Gauss -Siedel iteration method.
UNIT I I: Eigen values and Eigen vectors Linear dependence and independence of vectors, Eigen values and Eigen vectors and their properties, Diagonalisation of a matrix. Cayley -Hamilton theorem(without proof), finding inverse and power of a matrix by Cayley -Hamilton Theorem; Quadratic forms and Nature of the Quadratic Forms; Reduction of Quad ratic form to canonical forms by Orthogonal
Transformation .
UNIT I II:Multi Variable Calculus ( Differentiation) Functions of two variables, Limit, Continuity,Partial derivatives, Total differential and differentiability, Derivatives of composite and impl icit functions , Jacobian -functional dependence and independence, Maxima and m inima and saddle points, Method of Lagrange multipliers ,Taylors theorem for two variables.
UNIT IV: First Order Ordinary Differential Equations Exact, Equations reducible to exact form , Applications of first order differential equations - Newton’s l aw of cooling, Law of natural growth and decay,Equations not of first degree -Equations solvable for p, equations solvable for y, equations solvable for x and Clairaut’s type.
UNIT V : Differential Equations of Higher Order Linear differential equations of second and higher order with constant coefficients: Non-homogeneous term of the type f(x) = eax, sinax, cosax, xn, eax V and xn V - Method of variatio n of parameters, Equations red ucible linear ODE with constant coefficients -Cauchy’s Euler equation and Legendre’s equation.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) Text Books i) Higher Engineering Mathematics by B V Ramana ., Tata McGraw Hill.
ii) Higher Engineering Mathematics by B.S. Grewal, Khanna Publishers.
iii)Advanced Engineering Mathematics by Kreyszig ,John Wiley & Sons .
Reference Books i)Advanced Engineering Mathematics by R.K Jain & S R K Iyenger, Narosa Publishers.
ii) Ordinary and Partial Differential Equations by M.D. Raisinghania, S.Chand Publishers iii)Engineering Mathematics by N.P Bali and Manish Goyal.
Course Outcomes : After learning the concepts of this paper the student will be able to 1. Analyze the solution of the system of linear equations and to find the Eigen values and Eigen vectors of a matrix.
2. Reduce the quadratic form to canonical form using orthogonal transformations.
3. Find the extreme values of functions of two varia bles with / without constraints .
4. Solve first order , first degree differential equation s and their applicati ons.
5. Solve higher order differential equation s.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) INDEX UNIT -I : Matrices UNIT -II : Eigen values and Eigen vectors
UNIT -III : Multi Variable Calculus ( Differentiation) UNIT -IV : First Order Ordinary Differential Equations UNIT -V : Differential Equations of Higher Order MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD)
Mathematics -I : CO -PO Mapping CO Statement PO1 PO2
PO3 PO4 PO5 PO6 PO7
PO8 PO9 PO10 PO11 PO12
C102.1 2 2 - 2 2 - - - - - - 1 C102.2 2 2 - 1 2 - - - - - - 1 C102.3 3 2 - 2 2 - - - - - - 2 C102.4 3 3 - 2 2 - - - - - - 2 C102.5 3 2 - 2 1 - - - - - - 1
C102 2.6 2.2 0 1.8 1.8 0 0 0 0 0 0 1.4 MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 1 UNIT -I MATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 2 MATRICES Introduction : The influence of Matrices in mathematical world is spread wide because it provides an important base to many of the principles and practices. The origin of mathematical matrices lies with the study of systems of simultaneous linear equations. Some
of the things Matrices is used for are to solve systems of linear format, to find least -square best fit lines to predict future outcomes or find trends, to encode and decode messages .
There are many uses of matr ices in Engineering such as Graph theory, Linear combinations of quantum states in Physics, Computer animation, F or writing secret messages and Cryptography.
Basic Definitions :
Matrix : A matrix is a two dimensional array of numbers or expressions arranged in a set of rows and columns. An 𝑚×𝑛 matrix 𝐴 has 𝑚 rows and 𝑛 columns and is written Eg: 𝐴=
mn m mnna a aa a aa a a.......... .......... ............... .......... ............................
2 12 12 211 12 11m x n = [𝑎ij ]𝑚𝑥𝑛 wher e 1≤𝑖≤𝑚,1≤𝑗≤𝑛.
Where 𝑎𝑖𝑗′𝑠 are scalars.
Order of the Matrix: The number of rows and columns represents the order of the matrix. It is denoted by 𝑚𝑥𝑛 , where 𝑚 is number of rows and 𝑛 is number of columns.
Types of Matrices:
Row Matrix: A Matrix having only one row is called a “Row Matrix”.
Eg:
31321x Column Matrix: A Matrix having only one column is called a “Column Matrix” .
Eg:
13211x
Null Matrix: A= ijmxna such that 𝑎𝑖𝑗=0 𝑖 and 𝑗. Then 𝐴 is called a “Zero Matrix”. It is denoted by 𝑂𝑚𝑥𝑛 .
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 3 Eg: O 2x3= Rectangular Matrix: If A=ijmxna , and 𝑚
𝑛 then the matrix 𝐴 is called a “Rectangular Matrix” Eg :
43 221 1 is a 2x3 matrix Square Matrix: If 𝐴=ijmxna and 𝑚 = 𝑛 then 𝐴 is called a “Square Matrix”.
Eg :
2211 is a 2x2 matrix Lower Triangular Matrix: A square Matrix nXn ijnxnAa is said to be lower Triangular
if ija= 0 if 𝑖 <𝑗 i.e. if all the elements above the principle diagonal are zeros.
Eg: is a Lower triangular matrix .
Upper Triangular Matrix: A square Matrix ijnxnAa is said to be upper triangular i f ija= 0 if 𝑖>𝑗. i.e. all the elements below the principle diagonal are zeros.
Eg:
is an Upper triangular matrix Triangular Matrix: A square matrix which is either lower triangular or upper triangular is called a triangle matrix.
Principal Diagonal of a Matrix: In a square matrix, the set of all 𝑎𝑖𝑗, for which 𝑖 = 𝑗 are called principal diagonal elements. The line joining the principal diagonal elements is calle d principal diagonal.
Note: Principal diagonal exists only in a square matrix.
Diagonal elements in a matrix: 𝐴= [𝑎ij]nxn, the elements 𝑎𝑖𝑗 of 𝐴 for which 𝑖 = 𝑗.
i.e. 𝑎11,𝑎22….𝑎nn are called the diagonal elements of 𝐴 Eg: A= diagonal elements are 1, 5, 9 000000
6370 25004
2 005 40831
987654321MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 4 Diagonal Matrix: A Square Matrix is said to be diagonal matrix, if ija = 0 for 𝑖
𝑗 i.e. all the elements except the principal diagonal elements are zeros.
Note: 1. Diagonal matrix is both lower and upper triangular.
2. If 𝑑1,𝑑2…….𝑑n are the diagonal elements in a diagonal matrix it can be represented as diag 12,, .,n d d d Eg : 𝐴 = diag (3,1, -2)= Scalar Matrix: A diagonal matrix whose leadin g diagonal elements are equal is called a
“Scalar Matrix”. Eg : 𝐴= Unit/Identity Matrix: If A = ijnxna such that 𝑎𝑖𝑗=1 for 𝑖 = 𝑗, and 𝑎𝑖𝑗=0 for ij then A is called a “Identity Matrix” or Unit matrix. It is denoted by
nI Eg: I2 = 1001 ; I3=
Trace of Matrix: The sum of all the diagonal elements of a square matrix 𝐴 is called Trace of a matrix 𝐴, and is denoted by Trace A or 𝑡𝑟 𝐴.
Eg : A = then 𝑡𝑟 𝐴 = 𝑎+𝑏+𝑐 Singular & Non Singular Matrices: A square matrix 𝐴 is said to be “Singular” if the determinant of │𝐴│= 0, Otherwise A is said to be “Non -singular”.
Note: 1. Only non -singular matr ices possess inverse.
2. The product of non -singular matrices is also non -singular.
Inverse of a Matrix: Let 𝐴 be a non -singular matrix of order 𝑛 if there exist a matrix 𝐵 such that 𝐴𝐵=𝐵𝐴=𝐼 then 𝐵 is called the inverse of 𝐴 and is d enoted by 𝐴−1 If inverse of a matrix exist, it is said to be invertible.
Note: 1. The necessary and sufficient condition for a square matrix to posses inverse is that |𝐴|≠0.
2 .Every Invertible matrix has unique inverse .
3. If 𝐴,𝐵 are two invertible square matrices then 𝐴𝐵 is also invertible and
2 00010003
200020002
100010001
cfgf bhghaMATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 5 1 11AB B A 4.
1d 0etwhereAdjAAAdetA , Theorem: The inverse of a Matrix if exists is Unique.
Note: 1. (𝐴-1)-1 = 𝐴 2. 𝐼-1 = 𝐼 Theorem: If 𝑨,𝑩 are invertible matrices of the same order, then (i). (𝐴𝐵)-1 = 𝐵-1𝐴-1 (ii). (𝐴𝑇)-1 = (𝐴−1)𝑇 Sub Matrix: - A matrix obtained by deleting some of the rows or columns or both from the
given matrix is called a sub matrix of the given matrix.
Eg: Let 𝐴 = . Then is a sub matrix of 𝐴 obtained by deleting first row and 4th column of 𝐴.
Minor of a Matrix: Let 𝐴 be an 𝑚𝑥𝑛 matrix. The determinant of a square sub matrix of 𝐴 is called a minor of the matrix.
Note: If the order of the square sub matrix is ‘ 𝑡’ then its determinant is called a minor of order ‘ 𝑡’.
Eg: 𝐴 = be a 4x3 matrix Here 𝐵 = [2131] is a sub -matrix of order ‘2’ = 2-3 = -1 is a minor of order ‘2 And 𝐶 = [211
312567] is a sub -matrix of order ‘3’ det 𝐶 = 2(7−12)−1(21−10)+(18−5) = −9 Properties of trace of a matrix: Let A and B be two square matrices and be any scalar
1) 𝑡𝑟 (𝐴) = (𝑡𝑟 𝐴) ; 2) 𝑡𝑟(𝐴+𝐵) = 𝑡𝑟𝐴 + 𝑡𝑟𝐵 ; 3) 𝑡𝑟 (𝐴𝐵) = 𝑡𝑟(𝐵𝐴) Idempotent Matrix: A square matrix A Such that 𝐴2=𝐴 then 𝐴 is called “Idempotent Matrix”.
Eg: 𝐴 = [1001]
1 5435 10987 651325431098x
765321213112
BMATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 6 Involutary Matrix: A square matrix 𝐴 such that 𝐴2 = 𝐼 then A is called an Involutary Matrix.
Eg: 𝐴 = [0110] Nilpotent Matrix: A square matrix 𝐴 is said to be Nilpotent if there exis ts a positive integer 𝑛 such that 𝐴n = 0 here the least 𝑛 is called the Index of the Nilpotent Matrix.
Eg: 𝐴 = [1000] Transpose of a Matrix: The matrix obtained by interchanging rows and columns of the given matrix A is called as transpose of the given matrix 𝐴. It is denoted by TA or A’
Eg: A = [1234] Then AT = [1324] Properties of transpose of a matrix: If A and B are two matrices andTA ,
TB are their transposes then 1) TTAA ; 2)T TTA B A B ; 3)
T TKA KA ; 4) T TTAB B A Symmetric Matrix: A square matrix A is said to be symmetric if TAA If
Tij jinxnA a then A a nxn where ij jiaa Eg: is a symmetric matrix Skew -Symmetric Matrix: A square matrix A is said to be Skew symmetric If
TAA .
If Tij ji ij jinxn nxnA a then A a where a a .
Eg : is a skew – symmetric matrix Note : All the principle diagonal elements of a skew symmetric matrix are always zero.
Since 𝑎ij = −𝑎ij 𝑎ij = 0 Theorem: Every square matrix can be expressed uniquely as the sum of symmetric and skew symmetric matrices.
Proof: Let 𝐴 be a square matrix, A = 12AA = 12TTA A A A =
1122TTA A A A = 𝑃 + 𝑄, where P =12TAA ; Q =
12TAA Thus every square matrix can be expressed as a sum of two ma trices.
cfgf bhgha
000c bc ab aMATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 7 Consider
1 1 12 2 2TTTT T T T TP A A A A A A =12TAA =P, since
TPP , P is symmetric Consider 1 1 12 2 2T
TTT T T T TQ A A A A A A = - 12TAA = - Q SinceTQQ , 𝑄 is Skew -symmetric.
To prove the representation is unique: Let 𝐴= 𝑅+𝑆1 be the representation, where 𝑅 is symmetric and 𝑆 is skew symmetric. i.e.
,TTR R S S Consider T T T TA R S R S R S 2 11 2 22TTA A S S A A Q
Therefore every square matrix can be expressed as a sum of a symmetric and a skew symmetric matrix Ex. Express the given matrix A as a sum of a symmetric and skew symmetric matrices where 𝑨=2 4 9
14 7 139 5 11
Soluti on:
2 14 34 7 59 3 11TA
4 10 12 2 5 6110 14 18 5 7 9 ;212 18 22 6 9 11TTA A P A A P is symmetric 0 18 6 0 9 3
118 0 8 9 0 4 ;26 8 0 3 4 0TA A Q A A Q is skew symmetric Now 𝐴 =𝑃+𝑄 = 2 5 6
5 7 96 9 11 +
0 9 39 0 43 4 0
Orthogonal Matrix: A square matrix A is said to be an Orthogonal Matrix if 𝐴𝐴T=𝐴T𝐴=𝐼.
Note: 1. If A, B are orthogonal matrices, then AB and BA are orthogonal matrices.
2. Inverse and transpose of an orthogonal matrix is also an orthogonal matrix.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 8 Result: If A, B are orthogonal matrices, each of order n then AB and BA are orthogonal matrices.
Result: The inverse of an orthogonal matrix is orthogonal and its transpose is also orthogonal Solved Problems :
1. Show that 𝑨 = is orthogonal.
Sol: Given A = then AT = Consider A.AT = = = A is orthogonal matrix.
2. Prove that the matrix is orthogonal.
Sol: Given A = Then AT = Consider A .AT = = = = I ⇒ 𝐴.𝐴T = 𝐼 Similarly 𝐴T .𝐴 = I Hence 𝐴 is orthogonal matrix
3. Determine the values of a, b, c when is orthogonal.
Sol: - For orthogonal matrix AAT = I So, AAT =
cos sinsin cos cos sinsin cos
cos sinsin cos
cos sinsin cos
cos sinsin cos
2 22 2sin cos sin cos cos sinsin cos sin cos sin cos
I1001
1 2221 2221
31
1 2221 222131
1 2221 222131
1 2221 222191
1 2221 2221
90009000991
100010001
cb ac bacb20
cb ac bacb20Icc cb bbaa
20MATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 9 = I = Solving 2b2-c2 =0, a2-b2-c2 =0 We get c = a2 =b2+2b2 =3b2 𝑎 =
From the diagonal elements of I 4𝑏2+𝑐2= 1 4𝑏2+2𝑏2=1 (since c2=2b2) b = a = = ; b = ; c = = 4. Is matrix 2 3 1
4 3 13 1 9 Orthogonal?
Sol:- Given A=2 3 14 3 13 1 9
2 4 33 3 1
119 2 3 1 2 4 3
4 3 1 3 3 13 1 9 1 1 9 =
14 0 00 26 00 0 91
3I 3I
Matrix is not orthogonal.
Complex matrix: A matrix whose elements are complex numbers is called a complex matrix.
Conjugate of a complex matrix: A matrix obtained from A on replacing its elements by the corresponding conjugate complex numbers is called conjugate of a complex matrix. It is denoted by A If
,ij ijmxn mxnA a A a , where ija is the conjugate of ija .
Eg: If A= then = Note: If A and B be the conjugate matrices of A and B respectively, then (i)
A =A (ii) AB = A+B (iii)KA =
KA
2 2 2 2 2 2 2 22 2 2 2 2 2 2 22 2 2 2 2 2222 2 4c b a c b a c bc b a c b a c bc b c b c b
100010001
b2b361b321
61b2312 3 56 7 5i
ii A2 3 56 7 5i
ii MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 10 Transpose conjugate of a complex matrix: Transpose of conjugate of complex matrix is
called transposed conjugate of complex matrix. It is denoted by*A or A .
Note: If A and B be the transposed conjugates of 𝐴 and 𝐵 respectively, then (i) A = A (ii)
A B A B (iii) KA K A (iv) AB A B Hermitian Matrix: A square matrix 𝐴 is said to be Hermitian Matrix iff
AA .
Eg: A= then = and Aθ= Note: 1. In Hermitian matrix the principal diagonal elements are real.
2. The Hermitian matrix over the field of Real numbers is nothing but real symmetric matrix.
3. In Hermitian matrix A= ijnxna , ,ij jia a i j .
Skew -Hermitian Matrix: A square matrix A is said to be Skew -Hermitian Matrix iffAA.
Eg: Let A= then = and =-A A is skew -Hermitian matrix.
Note: 1. In Skew -Hermitian matrix the principal diagonal elements are either Zero or Purely Imaginary.
2. The Skew - Hermitian matrix over the field of Real numbers is nothing but real Skew - Symmetric matrix.
3. In Skew -Hermitian matrix A= ijnxna , ,ij jia a i j .
Unitary Matrix: A S quare matrix 𝐴 is said to be unitary matrix iff 1AA A A I or A A Eg:
Theorem1: Every square matrix can be uniquely expressed as a sum of Hermitian and skew – Hermitian Matrices.
Proof: - Let 𝐴 be a square matrix write 4 1 31 3 7ii
A4 1 31 3 7ii
4 1 31 3 7ii32
2iiii A32
2iiii
i ii iAT22 3TA
4 4242 461iiB
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 11 1 1 1(2 ) ( ) ( )2 2 211( ) ( ) .22A A A A A A A AA A A A A i eA P Q
Let 11;22P A A Q A A Consider
11( ) ( ) ( )22P A A A A A A P I.e.
, P P P is Hermitian matrix.
1 1 1( ) ( )2 2 2Q A A A A A A Q Ie , Q Q Q is skew – Hermitian matrix.
Thus every square matrix can be expressed as a sum of Hermitian & Skew Hermitian matrices.
To prove such representation is unique:
Let 𝐴 = 𝑅+𝑆------- (1) be another representation of 𝐴 where 𝑅 is Hermitian matrix & 𝑆 is skew – Hermitian matrix.
; R R S S Consider() A R S R S R S . Ie (2) A R A
11 2 2211 2 22A A R ie R A A PA A S ie S A A Q
Thus every square matrix can be uniquely expressed as a sum of Hermitian & skew Hermitian matrices.
Solved Problems :
1. If 𝑨=3 7 4 2 57 4 2 32 5 3 4iiii
ii then show that 𝑨 is Hermitian and 𝒊𝑨 is skew -Hermitian.
Sol: Given A=3 7 4 2 57 4 2 32 5 3 4iiii
ii then MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 12
3 7 4 2 57 4 2 32 5 3 4iiA i iii
And 3 7 4 2 57 4 2 32 5 3 4Tii
A i iii TAA Hence 𝐴 is Hermitian matrix.
Let B= iA i.e B=3 4 7 5 24 7 2 1 35 2 1 3 4i i i
i i ii i i then 3 4 7 5 2
4 7 2 1 35 2 1 3 4i i iB i i ii i i
3 4 7 5 24 7 2 1 35 2 1 3 4Ti i iB i i i
i i i Bi i ii i ii i i
4 31 2531 2 7425 74 3)1(
TB =-B 𝐵= 𝑖𝐴 is a skew Hermitian matrix.
2. If 𝑨 and 𝑩 are Hermitian matrices, prove that 𝑨𝑩−𝑩𝑨 is a skew -Hermitian matrix.
Sol: Given A and B are Hermitan matrices TAA And TBB ------------- (1) Now
T TAB BA AB BA TAB BA T T T T T TAB BA B A A B BA AB (By (1)) AB BA
Hence 𝐴𝐵−𝐵𝐴 is a skew -Hemitian matrix.
3. Show that 𝑨=a ic b idb id a ic 𝒊𝒔 unitary if and only if 𝒂2+𝒃2+𝒄2+𝒅2=𝟏 Sol: Given A=
a ic b idb id a ic Then a ic b idAb id a ic
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 13 Hence
icaidbidb icaA AT
icaidbidb icaica idbidb icaAA =2 2 2 2
2 2 2 200a b c da b c d AA I if and only if
2 2 2 21 a b c d 4. Given that A=0 1 21 2 0ii
, show that 1 AIAI is a unitary matrix.
Sol: we have 0 2121 0
1001iiAI 1 1 21 2 1ii And
0 2121 01001
iiAI =1 1 21 2 1ii
1 2121 11 411) (21iiiAI 1 1 2 1
1 2 16ii Let 1 AIAI B
1)21)(21( 21212121 )21)(21(1611 2121 11 2121 1
61i i i ii i i iiiiiB
4 4242 461iiBMATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 14 Now 4 2 4 12 4 46iBi and 4 2 4 1
2 4 46T iBi 4 2 4 12 4 436T iBBi
4 4242 4ii36 0 1 010 36 0 136I
1TBB i.e. 𝐵 is unitary matrix.
1 AIAIis a unitary matrix.
5. Show that the inverse of a unitary matrix is unitary.
Sol: Let A be a unitary matrix. Then AA I i.e 11AA I 11A A I
11A A I Thus 1A is unitary.
Rank of a Matrix:
Let 𝐴 be mxn matrix. If 𝐴 is a null matrix, we define its rank to be ‘ 0’. If 𝐴 is a non -zero matrix, we say that ‘ 𝒓’ is the rank of 𝐴 if i. Every (𝑟+1)𝑡ℎ order minor of 𝐴 is ‘0’ (zero) & ii. At least one 𝑟𝑡ℎ order minor of 𝐴 which is not zero.
It is denoted by (A) and read as rank of 𝐴.
Note: 1. Rank of a matrix is unique.
2. Every matrix will have a rank.
3. If 𝐴 is a matrix of order mxn, then Rank of 𝐴 ≤ 𝑚𝑖𝑛 (𝑚,𝑛) MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 15 4. If 𝜌 (𝐴) = 𝑟 then every minor of A of order 𝑟+1, or minor is zero.
5. Rank of the Identity matrix 𝐼𝑛 is 𝑛.
6. If 𝐴 is a matrix of order 𝑛 and 𝐴 is non -singular then 𝜌 (𝐴) = 𝑛 7. If 𝐴 is a singular matrix of order n then 𝜌(𝐴) < 𝑛 Important Note:
1. The rank of a matrix is ≤ 𝑟 if all minors of (𝑟+1)𝑡ℎ order are zero.
2. The rank of a matrix is ≥ 𝑟, if there is at least one minor of order ‘ 𝑟’ which is not equal to zero.
1. Find the rank of the given matrix Sol: Given matrix 𝐴 = 𝑑𝑒𝑡 𝐴 = 1(48−40)−2(36−28)+3(30−28)= 8−16+6 = −2 ≠ 0 We have minor of order 3 ∴𝜌 (𝐴) =3 2. Find the rank of the matrix
Sol: Given the matrix is of order 3x4 Its Rank ≤ min(3,4) = 3 Highest order of the minor will be 3.
Let us consider the minor Determinant of minor is 1( -49)-2(-56) + 3(35 -48) = -49+112 -39 = 24 ≠ 0.
Hence rank of the given matrix is ‘3’.
Elementary Transformations on a Matrix:
i). Interchange of 𝑖𝑡ℎ row and 𝑗𝑡ℎ row is denoted by 𝑅i ↔ 𝑅j (ii). If 𝑖𝑡ℎ row is multiplied with 𝑘 then it is denoted by 𝑅i 𝑘𝑅i (iii). If all the elements of 𝑖𝑡ℎ row are multiplied with 𝑘 and added to the corresponding elements of 𝑗𝑡ℎ row then it is denoted by 𝑅j 𝑅j +𝑘𝑅i Note: 1. The corresponding column transformations will be denoted by writing ‘ 𝑐’. i.e
𝑐i ↔𝑐j, 𝑐i 𝑘 𝑐j ,𝑐j 𝑐j + 𝑘𝑐i 2. The elementary operations on a matrix do not change its rank.
12107443321
12107443321
5 07887654321
078765321
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 16 Equivalance of Matrices: If 𝐵 is obtained from 𝐴 after a finite number of elementary transformations on 𝐴, then 𝐵 is said to be equivalent to 𝐴.It is denoted as 𝐵~𝐴.
Note : 1. If 𝐴 and 𝐵 are two equivalent matrices, then rank 𝐴 = rank 𝐵.
2. If 𝐴 and 𝐵 have the same size and the same rank, then the two matrices are equivalent.
Elementary Matrix or E -Matrix: 𝐴 matrix is obtained from a unit matrix by a single elementary transfo rmation is called elementary matrix or E -matrix.
Notations: We use the following notations to denote the E -Matrices.
1) ijE Matrix obtained by interchange of 𝑖𝑡ℎ and 𝑗𝑡ℎ rows (columns).
2) ikE Matrix obtained by multiplying 𝑖𝑡ℎ row (column) by a non - zero number 𝑘.
3)ij kE Matrix obtained by adding 𝑘 times of 𝑗𝑡ℎ row (column) to 𝑖𝑡ℎ row (column).
Echelon form of a m atrix:
A matrix is said to be in Echelon form, if (i) Zero rows, if any exists, they should be below the non -zero row.
(ii) The first non -zero entry in each non -zero row is equal to ‘1’.
(iii) The number of zeros before the first non -zero element in a row is less than the number of such zeros in the next row.
Note : 1. The number of non -zero rows in echelon form of 𝐴 is the rank of ‘ 𝐴’.
1. The rank of the transpose of a matrix is the same as that of original matrix.
2. The condition (ii) is optional.
Eg: 1. is a row echelon form.
2. is a row echelon form.
Solved Problems :
1.Find the rank of the matrix 𝑨 = by reducing it to Echelon form.
Sol: Given 𝐴 = Applying row transformations on 𝐴.
0000110000 100001
000110131
131423732
131423732MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 17 R1 ↔ R 3 A ~
R2 → R 2 –3R1; R3→ R 3 -2R1 ~ R2 → R 2/7,R3→ R 3/9 ~ R3 → R 3 –R2 ~ This is the Echelon form of matrix 𝐴.
The rank of a matrix 𝐴= Number of non – zero rows =2 2.For what values of k the matrix has rank ‘3’.
Sol: The given matrix is of the order 4x4 If its rank is 3 det A =0 𝐴 = Applying R 2 → 4R 2-R1, R3 →4R 3 – kR1, R4 → 4R 4 – 9R1 We get A ~
Since Rank 𝐴 = 3 𝑑𝑒𝑡 𝐴 =0 4 1[(8−4𝑘) 3]−1(8−4𝑘) (4𝑘+27)] = 0 (8−4𝑘) (3−4𝑘−27) = 0
732423131
9 907 70131
110110131
000110131
3 992 2 201 1113 44kk
3 992 2 201 1113 44kk
3 27 4 008 384801 1 001 3 44
kk k k03 27 4 08 38 481 1 0
kk k kMATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 18 (8−4𝑘)(−24−4𝑘) =0 (2−𝑘)(6+𝑘)=0
𝑘 =2 𝑜𝑟 𝑘 = −6 3).Find the rank of the matrix using echelon form 2 1 3 54 2 1 38 4 7 13
8 4 3 1A Sol: Given
2 1 3 54 2 1 38 4 7 138 4 3 1A
By applying 2 2 1 3 3 1 4 4 1 2 ; 4 ; 4 R R R R R R R R R 2 1 3 5
0 0 5 7~0 0 5 70 0 15 21 3 12
1 2 3 ,,1 1 3R RRR R R 2 1 3 50 0 5 7~0 0 5 70 0 5 7
3 3 2 4 4 2 , R R R R R R 2 1 3 5
0 0 5 7~0 0 0 00 0 0 0
⇒ A is in echelon form ∴ Rank of A = 2 4).Find the rank of the matrix 𝐴 = 1 2 0 12 1 1 0
3 3 1 11 1 1 1 by reducing into echelon form.
Sol: By applying 2 2 1 3 3 1 4 4 1 2 ; 3 ; R R R R R R R R R A ~1 2 0 10 3 1 20 3 1 2
0 3 1 2 3 3 2R R R
A ~1 2 0 10 3 1 20 0 0 00 3 1 2
34RR A ~
1 2 0 10 3 1 20 3 1 20 0 0 0
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 19
3 3 2R R R A ~1 2 0 10 3 1 200000000
Clear it is in echelon form, rank of A = 2
Normal form/Canonical form of a Matrix:
Every non -zero Matrix can be reduced to any one of the following forms.
0; 0 ; ;0 0 0rrrrIIII Known as normal forms or canonical forms by using Elementary row or column or both transformations where
rI is the unit matrix of order ‘ 𝑟’ and ‘ 𝑂’ is the null matrix.
Note: 1.In this form “the rank of a matrix is equal to the order of an identity matrix.
2. Normal form another name is “canonical form” Solved Problems :
1. By reduci ng the matrix into normal form, find its rank.
Sol: Given A = 𝑅2 → 𝑅2 – 2𝑅1; 𝑅3 → 𝑅3 – 3𝑅1 A ~ R3 → R 3/-2 A ~ R3 → R 3+R2 A ~ c2→ c 2 - 2c1, c3→c 3-3c1, c4→c 4-4c1 A ~
10 50334124321
10 50334124321
22 4 6 05 2 3 04 3 21
11 2 305 2 3 04 3 21
6 0 005 2 3 04 3 21
6 0 005 2 3 00 0 0 0MATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 20 c3 → 3 c 3 -2c2, c4→3c 4-5c2 A ~ c2→ c 2 /-3, c 4→c 4/18 A ~ c4 ↔ c 3 A~ This is in normal form [I 3 0], ∴ Hence Rank of A is ‘3’.
2).Find the rank of the matrix A =1 1 1 11 2 3 42 3 5 53 4 5 8
by reducing into canonical form or normal form.
Sol: Given A =1 1 1 11 2 3 42 3 5 53 4 5 8
By applying 2 2 1 3 3 1 4 4 1 , 2 , 3 R R R R R R R R R
1 1 1 10 1 2 5~0 1 3 70 7 8 5
3 3 2 4 4 2 ,7 R R R R R R 1 1 1 10 1 2 5~0 0 1 20 0 6 30
4 4 3 6 R R R1 1 1 10 1 2 5~0 0 1 20 0 0 18
4418RR1 1 1 1
0 1 2 5~0 0 1 20 0 0 1
18 0 000 0300 0 01
1 0 000 0100 0 01
0 1 000 0100 0 01MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 21 Apply 2 2 1 3 3 1 4 4 1 ,, C C C C C C C C C
1 0 0 00 1 2 5~0 0 1 20 0 0 1
3 3 2 4 4 2 2 ; 5 C C C C C C 1 0 0 00 1 0 0~0 0 1 20 0 0 1
4 4 3 2 CCC
1 0 0 00 1 0 0~0 0 1 00 0 0 1
Clearly it is in the normal form 4I Rank of 𝐴 = 4
3).Define the rank of the matrix and find the rank of the following matrix 2 1 3 54 2 1 38 4 7 138 4 3 1
Sol: Let 𝐴=
2 1 3 54 2 1 38 4 7 138 4 3 1
2 2 13 3 1
4 4 1244R R RR R RR R R
2 1 3 50 0 5 7~0 0 5 70 0 15 21A
3 3 24 4 2 3R R R
R R R2 1 3 50 0 5 7~0 0 0 00 0 0 0A
It is in echelon form. So, rank of matrix = no. of non zero rows in echelon form.
( ) 2 Rank A4).Reduce the matrix A to normal form and hence find its rank 2 1 3 40 3 4 12 3 7 5
2 5 11 6A
Sol: Given 2 1 3 40 3 4 12 3 7 52 5 11 6A
MATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 22 1112CC 1 1 3 40 3 4 1~1 3 7 5
1 5 11 6A
3 3 24 4 1R R RR R R
1 1 3 40 3 4 1~0 2 4 10 4 8 2A
2 2 3R R R 1 1 3 4
0 1 0 0~0 2 4 10 4 8 2A
3 3 24 4 224R R R
R R R1 0 0 00 1 0 0~0 0 4 10 0 8 2A
4 4 3 2 R R R10000 1 0 0~0 0 4 10 0 0 0A
4 4 34 C C C
10000 1 0 0~0 0 4 10 0 0 0A
3313CC
1 0 0 00 1 0 0~0 0 1 00000A
⇒ 30~00IA
This is in normal form. Thus Rank of matrix = Order of identify matrix.
Rank( ) 3A 5). Reduce the matrix A =0 1 2 24 0 2 6
2 1 3 1 into canonical form and then find its rank.
Sol: Apply12CC 1 0 2 2~ 0 4 2 61 2 3 1A
3 3 1R R R1 0 2 2
~ 0 4 2 60 2 1 3A
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 23 3 3 1 4 4 1 2 ; 2 C C C C C C 1 0 0 0~ 0 4 2 6
0 2 1 3A 2
22RR1 0 0 0~ 0 2 1 30 0 0 0A
23CC1000~ 0 1 2 3
0 0 0 0A3 3 2 4 4 2 2 ; 3 C C C C C C
1 0 0 0~ 0 1 0 00000A
Which is in the normal form2000I ,
(A) = 2 System of linear equations:
In this chapter we shall apply the theory of matrices to study the existence and nature of solutions for a system of m linear equations in ‘n’ unknowns.
The system of m linear equations in ‘n’ unknowns 1, 2, 3 n x x x x given by 11 1 12 2 13 3 121 1 22 2 23 3 21 1 2 2 3 3.. 0
.. 0.. 0(1)nnnnm m m mn na x a x a x a xa x a x a x a x
a x a x a x a x
The above set of equations can be written in the Matrix form as 𝐴 𝑋 = 𝐵 11 12 121 22 2
12nnn n nna a aa a aa a a
1
2.
.
nxxx
1
2. (2).
mbbb
A-Coefficient Matrix; X -Set of unknowns; B -Constant Matrix
Homogeneous Linear Equations: If 𝑏12 ........... 0m bb then 𝐵 = 0 Hence eq uatio n (2) Reduces to 𝐴𝑋 = 0 which are known as homogeneous linear equations Non-Homogeneous Linear equations:
It at least one of 𝑏1,𝑏2−−−−𝑏m is non zero. Then 𝐵0, the system Reduces to 𝐴𝑋 = 𝐵 is known as Non -Homogeneous Linear equations.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 24 Solution : A set of numbers12,n x x x which satisfy all the equations in the system is known as solution of the system.
Consistent: If the system possesses a solution then the system of equations is said to be consistent.
Inconsistent: If the system has no solution then the system of equations is sa id to be Inconsistent.
Augmented Matrix: A matrix which is obtained by attaching the elements of 𝐵 as the last column in the coefficient matrix 𝐴 is called Augmented Matrix. It is denoted by [𝐴|𝐵]11 12 13 1 121 22 23 2 21 2 3 3:
:
:nnm m m mna a a a bA B C a a a a ba a a a b
1. If ( / ) (A)AB , then the system of equations 𝐴𝑋 = 𝐵 is consistent (solution exits).
a). If ( / ) (A)AB = 𝑟 = 𝑛 (no. of unknowns) system is consistent and have a unique solution b). If ( / ) (A)AB = 𝑟 < 𝑛 (no. of unknowns) then the system of equations 𝐴𝑋 = 𝐵
will have an infinite no. of solutions . In this case (𝑛−𝑟) variables can be assigned arbitrary values.
2. If ( / ) (A)AB then the system of equations 𝐴𝑋=𝐵 is inconsistent (no solution).
In case of homog eneous system 𝐴𝑋 = 0, the system is always consistent.
(or) x 1 = 0, x 2 = 0, -------- , xn = 0 is always the solution of the system known as the” zero solution “.
Non-trivial solution:
If ( / B) (A)A = r < n (no. of unknowns) then the system of equations 𝐴𝑋 = 0 will have an infinite no. of non zero (non trivial) solutions. In this case (𝑛−𝑟) variables can be assigned arbitrary values.
Also we use some direct methods for solving the system of equ ations.
Note: The direct methods are Cramer’s rule, Matrix Inversion, Gaussian Elimination, Gauss Jordan, Factorization Tridiagonal system. These methods will give a unique solution.
Procedure to solve AX = B (Non Homogeneous equations) Let us first cons ider 𝑛 equations in n unknowns ie. 𝑚=𝑛 then the system will be of the form MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 25 11 1 12 2 13 3 1
21 1 22 2 23 3 21 1 2 2 3 3.. 0.. 0.. 0(1)nnnn
m m m mn na x a x a x a xa x a x a x a xa x a x a x a x
The above system can be written as 𝐴𝑋 = 𝐵 --------- (1)
Where 𝐴 is an 𝑛𝑛 matrix.
Solving 𝑨𝑿 = 𝑩 using Echelon form:
Consider the system of m equations in n unknowns given by 11 1 12 2 13 3 121 1 22 2 23 3 21 1 2 2 3 3.. 0.. 0
.. 0(1)nnnnm m m mn na x a x a x a xa x a x a x a xa x a x a x a x
We know this system (1) can be we write as 𝐴𝑋 = 𝐵 The augmented matrix of the above system is [A / B] = The system AX = B is consistent if 𝜌 (𝐴) = 𝜌[𝐴/𝐵] i). 𝜌 (𝐴) = 𝜌 [𝐴/𝐵] = 𝑟 < 𝑛 (no. of unknowns).Then there is infinite no. of solutions.
ii). 𝜌 (𝐴) = 𝜌 [𝐴/𝐵] = number of unknowns then the system will have unique solution.
iii). 𝜌 (𝐴) ≠ 𝜌 [𝐴/𝐵] the system has no solution.
Solved Problems :
1). Show that the equations 𝒙+𝒚+𝒛 = 𝟒,𝟐𝒙+𝟓𝒚−𝟐𝒛 =𝟑,𝒙+𝟕𝒚−𝟕𝒛 =𝟓 are not consistent.
Sol: Write given equations is of the form 𝐴𝑋 = 𝐵 i.e; Consider the Augment matrix is [𝐴 /𝐵] [𝐴/𝐵] = 11 12 1 121 22 2 2
12nnm m mn ma a a ba a a ba a a b
534
7 712 521 11zyx
57 7132 524 1 11MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 26
Applying R 2 →R 2-2R1 and R 3 → R 3-R1, we get [𝐴/𝐵] ~ Applying R 3→ R 3-2R2, we get [𝐴/𝐵] ~ ∴𝜌 (𝐴) =2 𝑎𝑛𝑑 𝜌 (𝐴/𝐵) =3 The given system is inconsistent as 𝜌 (𝐴) ≠ 𝜌[𝐴/𝐵].
2). Show that the equations given below are consistent and hence solve them 𝑥−3𝑦−8𝑧 = −10,3𝑥+𝑦−4𝑧 =0,2𝑥+5𝑦+6𝑧 =3 Sol: Matrix notation is Augmented matrix [A/B] is [A/B] = R2 → R 2-3R1 R3 → R 3 -2R1 ~
R2 →1/10 R 2 ~ R3 →R 3-11R 2 ~ This is the Echelon form of [A/B] ∴ρ (A) =2, ρ (A/B) =3 ρ (A) ≠ ρ[A/B].
The given system is inco nsistent.
3).Find whether the following equations a re consistent, if so solve 𝒙+𝒚+𝟐𝒛=𝟒 ,𝟐𝒙−𝒚+𝟑𝒛=𝟗,𝟑𝒙−𝒚−𝒛=𝟐 Sol: We write the given equations in the form 𝐴𝑋=𝐵 i.e;
18 605 4 304 1 11
11 0 005 4 304 1 11
3010
6 524 138 3 1zyx
3 6 520 4 1310 8 3 1
23 22 11003 20 10 010 8 3 1
23 22 1103 2 1 010 8 3 1
10 0 003 2 1 010 8 3 1
2941 1 331 221 1
zyxMATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 27 The Augmented matrix [𝐴/𝐵] = Applying R 2→ R 2-2R1 and R 3→R 3-3R1, we get [𝐴/𝐵] ~
Applying R 3 →3R 3-4R2, we get [𝐴/𝐵] ~ this matrix is in Echelon form. ρ(A) = 3 and ρ(A/B) = 3 Since 𝜌 (𝐴) =𝜌 [𝐴/𝐵]. ∴ The system of equations is consistent.
Here the number of unknowns is 3 Since 𝜌 (𝐴) =𝜌 [𝐴/𝐵] = number of unknowns The system of equations has a unique solution 𝑊𝑒 ℎ𝑎𝑣𝑒
−17𝑧 = −34 𝑧 = 2 −3𝑦−𝑧 =1 −3𝑦 =𝑧+1 −3𝑦 =3 𝑦= −1 𝑎𝑛𝑑 𝑥+𝑦+2𝑧 =4 𝑥=4−𝑦−2𝑧 =4+1−4=1 𝑥=1,𝑦=−1,𝑧=2 is the solution.
4). Show that the equations 𝒙+𝒚+𝒛=𝟔,𝒙+𝟐𝒚+𝟑𝒛=𝟏𝟒,𝒙+𝟒𝒚+𝟕𝒛=𝟑𝟎 are consistent and solve them.
Sol: We write the given equations in the form AX=B i.e.
The Augmented matrix [A/B] = Applying R 2→ R 2-R1 and R 3→R 3-R1, we get [A/B] ~
211 3931 2421 1
107 4 011 3 04 2 1 1
34 17 001 1 3 04 2 11
3414
17 0 01 3 021 1zyx
301467 413211 11zyx
307411432 16 11 1
24630821 06 11 1MATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 28 Applying R 3 →R 3-3R2, we get [A/B] ~ This matrix is in Echelon form. ρ(A) = 2 and ρ(A/B) = 2 Since 𝜌 (𝐴) =𝜌 [𝐴/𝐵].
The system of equations is consistent. Here the no. of unknowns are 3 Since rank of 𝐴 is less than the no. of unknowns, the system of equations will have infinite number of solutions in terms of n -r=3-2=1 arbitrary constant.
The given system of equations reduced form is 𝑥+𝑦+𝑧=6……… (1), 𝑦+2𝑧=8…….(2) Let 𝑧=𝑘, put 𝑧=𝑘 𝑖𝑛 (2) we get 𝑦=8−2𝑘 Put 𝑧=𝑘 𝑦=8−2𝑘 in (1), we get 𝑥=6−𝑦−𝑧=6−8+2𝑘=−2+𝑘
∴ 𝑥=−2+𝑘,𝑦=8−2𝑘,𝑧=𝑘 is the solution, where k is an arbitrary constant.
5). Show that23x y z ;3 2 1x y z ;2 2 3 2; 1x y z x y z are consistent and solve them
Sol: The above system in matrix notation is 314 3 4 11 2 1 33 1 2 12 2 3 2
1 1 1 1xyzA X B
The Augmented matrix is 1 2 1 33 1 2 12 2 3 2
1 1 1 1AB 2 2 13 3 1
4 4 132R R RR R RRRR
1 2 1 30 7 5 8~0 6 5 00 3 2 4
2 2 3R R R 1 2 1 30 1 0 4~0 6 5 4
0 3 2 4
0000821 06 11 1
0860 002101 11zyx
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 29 3 3 14 4 233R R R
R R R 1 2 1 30 1 0 4~0 0 5 200 0 2 8
3
35RR 1 2 1 30 1 0 4~0 0 1 40 0 2 8
4 4 3 2 R R R 1 2 1 3
0 1 0 4~0 0 1 40 0 0 0
∴𝜌 (𝐴) = 3 = 𝜌 (𝐴/𝐵 ∴𝜌 (𝐴) =𝜌 (𝐴/𝐵) = No. of unknowns = 3 The given system has unique solution.
The systems of equations equivalent to given system are 2 3 4; 48 4 3 4; 43 4 11, 4, 4.x y z y z
x y zxx y z
6). Solve 3; 3 5 2 8; 5 3 4 14 x y z x y z x y z Sol: - 1 1 1 3
3 5 2 85 3 4 14xyz
Augmented Matrix is AB 1 1 1 33 5 2 8
5 3 4 14 2 2 13 3 13
5R R RR R R1 1 1 3~ 0 8 1 1
0 8 1 1 3 3 2R R R1 1 1 3
~ 0 8 1 10 0 0 0 2 A AB
Number of unknowns (3) The system has infinite number of solutions.
3, 8 1 8 1 x y z y z y z MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 30 1 1 24 1 8 23 738 8 8 8k k k k kLet z k y and x k 23 7 23
8 8 8118 8 801kxkX y X
zk where k is any real number.
7). Find whether the following system of equations is consistent. If so solve them.
2 2 2, 3 2 5, 2 5 3 4, 4 6 0.x y z x y z x y z x y z Sol: In Matrix form it is 1 2 2 23 2 1 52 5 3 4
1 4 6 0xyzAX B
2 2 13 3 14 4 132R R RR R R
R R R1 2 2 20 8 5 1~0 9 1 8
0 2 4 2 2 2 3R R R
1 2 2 20 1 4 7~0 9 1 80 2 4 2
3 3 24 4 292R R RR R R
1 2 2 20 1 4 7~0 0 37 550 0 12 16
4414RR~
1 2 2 20 1 4 70 0 37 550 0 3 4
4 4 337 3 R R R~ 1 2 2 2
0 1 4 70 0 37 550 0 0 17
is in echelon form 34A and AB A AB .∴The given system is in consistent.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 31 8). Discuss for what values of λ, μ the simultaneous equations 𝒙+𝒚+𝒛 = 𝟔,𝒙+𝟐𝒚+𝟑𝒛 =𝟏𝟎,𝒙+𝟐𝒚+𝝀𝒛 = 𝝁 have (i). No solution
(ii). A unique solution (iii). An infinite number of solutions.
Sol: The matrix form of given system of Equations is A X = = B The augmented matrix is [A/B] = R2 → R 2 – R1, R3 → R 3-R1 [A/B] ~ R3 →R 3 – R2 ~ Case (i): let 𝜆 ≠ 3 the rank of 𝐴 = 3 and rank [𝐴/𝐵] = 3
Here the no . of unknowns is ‘3’ ∴𝜌 (𝐴) =𝜌 (𝐴/𝐵)= No. of unknowns The system has unique solution if λ≠3 and for any value of ‘μ’.
Case (ii): Suppose 𝜆 =3 and 𝜇≠10.
We have 𝜌 (𝐴) = 2,𝜌 (𝐴/𝐵) = 3 The system has no solution.
Case (iii): Let 𝜆 =3 and 𝜇=10.
We have 𝜌 (𝐴) = 2,𝜌 (𝐴/𝐵) = 2 Here 𝜌 (𝐴) = 𝜌 (𝐴/𝐵) ≠ No. of unknowns =3 The system has infinitely many solutions.
9). Find the values of a and b for which the equations x +𝒚+𝒛=𝟑; 𝒙+𝟐𝒚+𝟐𝒛=𝟔;𝒙+𝒂𝒚+𝟑𝒛=𝒃 have (i) No solution (ii) A unique solution (iii) Infinite no of solutions.
Sol: The above system in matrix notation is 1 1 1 31 2 2 613xy
a z b
106213 211 1 1zyx
21103 2161 1 1
6 1 104 2 1 06 1 1 1
10 3 004 2 106 1 1 1
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 32 Augmented matrix 1 1 1 3
1 2 2 613ABab
2 2 13 3 1R R RR R R1 1 1 3
~ 0 1 1 30 1 2 2ab
3 3 2R R R 1 1 1 3~ 0 1 1 30 3 0 9ab
1 1 1 3For a 3 & b 9 ~ 0 1 1 30 0 0 0
23 A AB ⇒ It has infinite no of solutions.
1 1 1 33 & ~ 0 1 1 30 3 0 9For a b any valueab
3 A AB ⇒ It has a unique solution.
1 1 1 33& 9 ~ 0 1 1 30 0 0 9For a bb
23A AB ⇒ Inconsistent no solution 10). Solve the following system completely.
21; 2 4 ; 4 10 x y z x y x x y z Sol: The above system in matrix notation is 21 1 1 11 2 41 4 10x
yzA X B
Augmented Matrix is 21 1 1 1
1 2 41 4 10AB
2 2 13 3 1R R RR R R21 1 1 1
~ 0 1 3 10 3 9 1
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 33 3 3 2 3 R R R 21 1 1 1~ 0 1 3 1
0 0 0 3 2 Here
23A and AB ⇒ The given system of equations is consistent if 223 2 0 2 2 0 2 1 0 2, 1 Case (i): When 1 2 A AB
Number of unknowns.
The system has infinite number of solutions.
The equivalent matrix is 11110 1 3 00000
The equivalent systems of equations are 1; 3 0 x y z y z 3 ( 3 ) 1 2 1 1 2 Let z k y k and x k K x k x k
1 2 1 20 3 0 30 0 1xkX y k X kzk
where k is any arbitrary constant.
Case (ii): When 2 2 A AB no. of unknowns.
The system has infinite number of solutions.
The equivalent matrix is ~11110 1 3 10000
The system of equations equivalent to the given system is 𝑥 + 𝑦 + 𝑧 = 1; 𝑦 + 3𝑧 = 1 1 3 (1 3 ) 1 2 Let z k y k and x K k x k 0 2 0 2
1 3 1 30 0 1xkX y k X kzk
where k is an y arbitrary constant.
11). Show that the equations 3 4 5 ;4 5 6 ;5 6 7x y z a x y z b x y z c don’t have a solution unless 2. a c b solve equations when 𝒂=𝒃=𝒄= −𝟏.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 34 Sol: The Matrix notation is 3 4 54 5 6
5 6 7xaybzcA X B
Augment Matrix is 3 4 54 5 6
5 6 7aAB bc
2 2 13 3 13435R R RR R R
3 4 5~ 0 1 2 3 40 2 4 3 5abaca
3 3 2 2 R R R3 4 5~ 0 1 2 3 4
0 0 0 3 6 3abaa b c
Here 23A and AB The given system of equations is consistent if 3 6 3 0 3 3 6 2a b c a c b a c b
Thus the equations don’t have a solution unless 2 , 1 a c b when a b c The equivalent matrix is 3 4 5 10 1 2 1
0 0 0 0 2 A AB No. of unknowns.
The system has infinite number of solutions. The system of equations equivalent to the given system 3 4 5 1; 2 1 2 1x y z y z y z 1 2 3 4 8 5 1 1 Let z k y k and x k k x k
1 1 11 2 1 20 0 1xkX y k kzk
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 35 Linearly dependent set of vectors: A set { 𝑥1,𝑥2,−−−−−−−−−−−−−−𝑥r} of 𝑟
vectors is said to be a linearly dependent set, if there exist 𝑟 scalars 𝑘1,𝑘2−−−𝑘r not all zero, such that 𝑘1𝑥1+𝑘2𝑥2+−−−−−−−−+𝑘r𝑥r = 0 Linearly independent set of vectors: A set { 𝑥1,𝑥2,−−−−−−−−−−−−−−𝑥r} of 𝑟 vectors is said to be a linearly independent set, if 𝑘1𝑥1+𝑘2𝑥2+−−−−−−−−+𝑘r𝑥r = 0
then 𝑘1 =0,𝑘2 = 0−−−−−−𝑘r =0.
Linear combination of vectors:
A vector x which can be expressed in the form 𝑥 = 𝑘1𝑥1+𝑘2𝑥2+−−−−−+𝑘r𝑥r is said to be a linear combination of { 𝑥1,𝑥2,−−−−−−−−−−𝑥r} here 𝑘1,𝑘2−−−𝑘r are any scalars.
Linear dependence and independence of Vectors:
Solved Problems :
1). Show that the vectors (𝟏,𝟐,𝟑),(𝟑,−𝟐,𝟏),(𝟏,−𝟔,−𝟓) from a linearly dependent set.
Sol: The Given Vector The Vectors X 1, X 2, X3 from a square matrix.
1 3 1 1 3 12 2 6 2 2 63 1 5 3 1 5Let A Then A
= 1(10+6)−2(15−1) + 3(−18+2) = 16+32−48 = 0 The given vectors are linearly dependent |𝐴| = 0 2). Show that the Vector 𝑿1=(𝟐,𝟐,𝟏),𝑿2=(𝟏,𝟒,−𝟏) 𝒂𝒏𝒅 𝑿3=(𝟒,𝟔,−𝟑) are linearly dependent.
Sol: Given Vectors 𝑿1=(𝟐,𝟐,𝟏),𝑿2=(𝟏,𝟒,−𝟏) 𝒂𝒏𝒅 𝑿3=(𝟒,𝟔,−𝟑)The Vectors 𝑋1,𝑋2,𝑋3 form a square matrix.
2 1 4 2 1 42 4 6 2 4 61 1 3 1 1 3Let A Then A
5611233213 2 1 X X X
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 36 = 2(−12+6) + 2(−3+4)+1(6−16) = −20≠ 0 The given vectors are linearly dependent |A|≠0 Consistency of system of Homogeneous linear equations:
A system of m homogeneous linear equations in n unknowns, namely 11 1 12 2 13 3 121 1 22 2 23 3 21 1 2 2 3 3.. 0.. 0
.. 0(1)nnnnm m m mn na x a x a x a xa x a x a x a xa x a x a x a x
i.e. Here 𝐴 is called Co –efficient matrix.
Note: 1. Here 𝑥1= 𝑥2 = −−−−−−−−− 𝑥n = 0 is called trivial solution or zero solution of 𝐴𝑋 = 0 2. A zero solution always linearly dependent.
Theorem: The number of linearly independent solutions of the linear system 𝐴𝑋 = 0 is (𝑛−𝑟),𝑟 being the rank of the matrix 𝐴 and 𝑛 being the number of variables.
Note: 1.if 𝐴 is a non -singular matrix then the linear system 𝐴𝑋 = 0 has only the zero solution.
2. The system 𝐴𝑋 =0 possesses a non -zero solution if and only if 𝐴 is a singular matrix.
Working rule for finding the solutions of the equation 𝑨𝑿 = 𝟎 (i). Rank of 𝐴 = No. of unknowns i.e. 𝑟 = 𝑛 ∴ The given system has zero solution.
(ii). Rank of A < No of unknowns (𝑟<𝑛) and No. of equations < No. of unknowns (𝑚<𝑛) then the system has infinite no. of solutions.
Note: If 𝐴𝑋 =0 has more unknowns than equations the system always has infinite solutions.
Solved Problems :
1). Solve the system of equations 𝒙+𝟑𝒚−𝟐𝒛 = 𝟎,𝟐𝒙−𝒚+𝟒𝒛 = 𝟎,𝒙−𝟏𝟏𝒚 +𝟏𝟒𝒛 = 𝟎
0...00...
............
213 2 12 23 22 211 13 12 11nmn m m mnnxxxa a a aa a a aa a aa
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 37 Sol: We write the given system is AX = 0 i.e.
R2 →R 2 -2R1; R3→R 3-R1 A ~ R3 →R 3 -2R2 A ~ The Rank of the 𝐴 = 2 𝑖.𝑒.𝜌 (𝐴) = 2 < No. of unknowns = 3 We have infinite No. of solution Above matrix can we write as 𝑥+3𝑦−2𝑧 = 0 −7𝑦+8𝑧 = 0, 0 = 0
𝐿𝑒𝑡 𝑧 = 𝑘 𝑡ℎ𝑒𝑛 𝑦=8/7𝑘 & 𝑥= −10/7 𝑘 Giving different values to 𝑘, we get infinite no. of values of 𝑥,𝑦,𝑧.
2). Find all the non -trivial solution2 3 0;3 2 0; 4 5 0.x y z x y z x y z Sol: In Matrix form it is 2 1 3 03 2 1 0
1 4 5 00xyzAX
2 1 3 0
3 2 1 01 4 5 0The Augumented matr AOix131 4 5 0
~ 3 2 1 02 1 3 0RR 2 2 1
3 3 11 4 5 03~ 0 14 14 020 7 7 0R R RR R R 3 3 21 4 5 02 ~ 0 14 14 0 0 0 0 0. it is echelon form R R R
The Rank of the A = 2 i.e. ρ(A) = 2 < No. of unknowns = 3 Hence the system has non trivial solutions. From echelon form, reduced equations are 4 5 0 14 14 0x y z and y z
4 5 0 . Let z k then y k and x k k x k
0004 11 14 1 22 31zyx
1614 087 02 31
0 0 087 02 31
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 38 Thus, the solution set is 11.
1xkX y k k Kzk
3). Show that the only real number λ for which the system 𝒙+𝟐𝒚+𝟑𝒛 = 𝝀𝒙,𝟑𝒙+𝒚+𝟐𝒛= 𝝀𝒚,𝟐𝒙+𝟑𝒚+𝒛 = 𝝀𝒛, has non -zero solution is 6 and solve them.
Sol: Above system can we expressed as AX = 0 i.e.
Given system of equations possess a non –zero solution i.e. ρ (A) < no. of unknowns.
For this we must have det A = 0 1 2 33 1 2 02 3 1
1 1 2 36663 1 2 0
2 3 1R R R R
1 1 1(6 ) 3 1 2 02 3 1
𝐶2→𝐶2−𝐶1,𝐶3→𝐶3−𝐶1 (6− λ )|1 0 03−2− λ −12 1 −1− λ |=0 (6−𝜆)[(−2− 𝜆)(−1− 𝜆)+1] =0 (6− 𝜆) (𝜆2+3 𝜆+3) = 0
𝜆 = 6 only real values.
When λ = 6, the given system becomes R2 → 5R 2+3R 1, R3→5R 3+2R 1 ~ R3 →R 3+R2 ~ −5𝑥+2𝑦+3𝑧 = 0 𝑎𝑛𝑑 −19𝑦+19𝑧 = 0 𝑦 = 𝑧
00013 22 1 332 1
zyx
0005 322 5 3325zyx
00019 19019 19 0325
zyx
0000 0 019 19 0325zyxMATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 39
𝐿𝑒𝑡 𝑧 = 𝑘 𝑦 = 𝑘 𝑎𝑛𝑑 𝑥 = 𝑘.
∴ 𝑇ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑥 = 𝑦 = 𝑧 = 𝑘.
Gauss elimination method:
This method of solving a system of 𝑛 liner equations in 𝑛 unknowns consists of eliminating the coefficients in such a way that the system reduces to upper triangular system which may be solved by backward substitution. We discuss the method here for 𝑛 = 3. The method is analogous for 𝑛>3.
Consider the system 𝑎11𝑥1+𝑎12𝑥2+𝑎13𝑥3=𝑏1 𝑎21𝑥1+𝑎22𝑥2+𝑎23𝑥3=𝑏2 ……………… (1) 𝑎31𝑥1+𝑎32𝑥2+𝑎33𝑥3=𝑏3 The augmented matrix of this system is
[𝐴,𝐵]~[𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33𝑏1𝑏2𝑏3] …………………..(2)
Performing 𝑅2→𝑅2−𝑎21𝑎11𝑅1 and 𝑅3→𝑅3−𝑎31𝑎11𝑅1 , we get [𝐴,𝐵]~[𝑎11 𝑎12 𝑎130 𝛼22 𝛼23
0 𝛼32 𝛼33𝑏1𝛽2𝛽3] …………………….(3) Where 𝑎22=𝑎22−𝑎12(𝑎21𝑎11) ; 𝑎23=𝑎23−𝑎13(𝑎21
𝑎11) 𝑎32=𝑎32−𝑎12(𝑎31𝑎11) ; 𝑎33=𝑎33−𝑎13(𝑎31𝑎11) 𝛽2=𝑏2−(𝑎21
𝑎11)𝑏1; 𝛽3=𝑏3−(𝑎31𝑎11)𝑏1 Here we assume 𝑎11≠0 We call −𝑎21𝑎11,−𝑎31
𝑎11 as multipliers for the first stage. 𝑎11 is called first pivot.
Now applying 𝑅3→𝑅3−𝑎32𝑎22(𝑅2), we get [𝐴,𝐵]~[𝑎11 𝑎12 𝑎130 𝛼22 𝛼230 0 𝛾33𝑏1
𝛽2∆3] ……………………(4) Where 𝛾33=𝛼33−(𝛼32𝛼22)𝛼23; ∆3=𝛽3−(𝛼32
𝛼22)𝛽2 We have assumed 𝛼22≠0 Here the multiplier is −𝛼32𝛼22 and new pivot is 𝛼22.
The augmented matrix (4) corresponds to an upper triangular system which can be by backward substitution. The solution obtained is exact.
1.Solve the equations 𝟑𝒙+𝒚+𝟐𝒛=𝟑 ;𝟐𝒙−𝟑𝒚−𝒛=𝟑;𝒙+𝟐𝒚+𝒛=𝟒 using gauss elimination method .
Sol.The given system of equations can be written as AX = B Where A= [3 1 22−3−11 2 1] , X = [𝑥𝑦
𝑧] , B = [3−34] The Augmented matrix is MATHEMATICS - I
DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 40 [𝐴,𝐵] = [3 1 2 2−3−11 2 1 3 −3
4] ~[1 2 12−3−13 1 2 4 −3
3] (Applying 𝑅1↔ 𝑅3) ~[1 2 10−7−30−5−1 4 −11
−9] (Applying 𝑅2−2𝑅1 , 𝑅3−3𝑅1) ~[1 2 10−7−30 0 84 −11
−8] (Applying 7 𝑅3 −5𝑅2) This corresponds to upper triangular system which gives 8𝑧= −8 ⇒𝑧=−1 7𝑦+3𝑧=11⇒7𝑦=11−3𝑧=14 ⇒𝑦=2 𝑥+2𝑦+𝑧=4⇒𝑥=4−2𝑦−𝑧=4−4+1⇒𝑥=1
The solution is 𝑥=1,𝑦=2,𝑧=−1 2. Express the following system in matrix fo rm and solve by gauss elimination method 𝒙+𝒚+𝒛=𝟔 ; 𝟑𝒙+𝟑𝒚+𝟒𝒛=𝟐𝟎 ;𝟐𝒙+𝒚+𝟑𝒛=𝟏𝟑.
Sol. The Augmented matrix of the given equations is [𝐴,𝐵]=[111334213 6 20
13] Performing the row operations 𝑅2→𝑅2−3𝑅1 and 𝑅3→𝑅3−2𝑅1 , we get ~[1 1 10 0 10−116
21] Using the operation 𝑅2↔𝑅3 ,we get [𝐴,𝐵]~[1 1 10−110 0 1 6
12] This corresponds to upper triangular system which gives 𝑧= 2 −𝑦+𝑧=1⇒−𝑦=1−𝑧=−1 ⇒𝑦=1
𝑥+𝑦+𝑧=6⇒𝑥=6−𝑦−𝑧=6−1−2⇒𝑥=3 The solution is 𝑥=3,𝑦=1,𝑧=2 Gauss Seidel Iteration Method: This method is a iterative method and is used to find the solution of given system of linear equations.
Consider the system of equations are 11 1 12 2 13 3 121 1 22 2 23 3 21 1 2 2 3 3.. 0.. 0
.. 0(1)nnnnm m m mn na x a x a x a xa x a x a x a xa x a x a x a x
the above system of equations as diagonally dominant and rewrite the given system of equations are MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 41
𝑥1=1𝑎11[𝑏1−𝑎12𝑥2−𝑎13 𝑥3] 𝑥2=1𝑎22[𝑏2−𝑎21𝑥1−𝑎23𝑥3] 𝑥3=1
𝑎33[𝑏3−𝑎31𝑥1−𝑎32𝑥2] then the first approximation is 𝑥1(1)=1𝑎11[𝑏1−𝑎12𝑥2(0)−𝑎13 𝑥3(0)] 𝑥2(1)=1
𝑎22[𝑏2−𝑎21𝑥1(1)−𝑎23𝑥3(0)] 𝑥3(1)=1𝑎33[𝑏3−𝑎31𝑥1(1)−𝑎32𝑥2(1)] Let us assume the initial approximate solution be 𝑥2(0) , 𝑥3(0) is equal to zero and substituting in 𝑥1(1) then evaluating value of 𝑥1(1),Similarly fi nd 𝑥2(1) , 𝑥3(1).
Second approximation is 𝑥1(2)=1𝑎11[𝑏1−𝑎12𝑥2(1)−𝑎13 𝑥3(1)] 𝑥2(2)=1𝑎22[𝑏2−𝑎21𝑥1(2)−𝑎23𝑥3(1)]
𝑥3(2)=1𝑎33[𝑏3−𝑎31𝑥1(2)−𝑎32𝑥2(2)] Proceeding in the same way, we get successive iterations. then the (k+1)th iteations are given by 𝑥1(𝑘+1)=1
𝑎11[𝑏1−𝑎12𝑥2(𝑘)−𝑎13 𝑥3(𝑘)] 𝑥2(𝑘+1)=1𝑎22[𝑏2−𝑎21𝑥1(𝑘+1)−𝑎23𝑥3(𝑘)] 𝑥3(𝑘+1)=1𝑎33[𝑏3−𝑎31𝑥1(𝑘+1)−𝑎32𝑥2𝑘+(1)]
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 42 The iteration process is stopped when the desired order of approximation is reached or two successive iterations are nearly the same. The final values of x 1 ,x2 ,x3 so obtained constitute an approximate solution of the above system.
This method can be generalized to a system of ‘ n’ equations in ’ n ‘ unknowns. This method is known as G auss-Seidel iteration method.
1.Use Gauss -Seidel iteration method to solve the system of equations 𝟏𝟎𝒙 +𝒚+𝒛=𝟏𝟐 𝟐𝒙+𝟏𝟎𝒚 +𝒛=𝟏𝟑 𝟐𝒙+𝟐𝒚+𝟏𝟎𝒛 =𝟏𝟒 .
Sol: The given system of equations are 10𝑥+𝑦+𝑧=12 2𝑥+10𝑦+𝑧=13 2𝑥+2𝑦+10𝑧=14 The gi ven system is diagonally dominant and rewrite the given system of equations are First Iteration :
𝑥1(1)=110[12−𝑦(0)−𝑧(0)] …………(1) 𝑦1(1)=110[13−2𝑥11−𝑧(0)] …………(2) 𝑧1(1)=1
10[14−2𝑥1(1)−2𝑦1(1)]……..(3) we start iteration by taking 𝑦(0)=0 , 𝑧(0)=0 in (1) we get 𝑥1(1)=110[12−0−0]=1.2 𝑥1(1) =1.2
Putting 𝑥1(1) =1.2, 𝑧(0)=0 in (2) we get 𝑦1(1)=110[13−2(1.2)−0]=1.06 𝑦1(1) =1.06 Putting 𝑥1(1) =1.2 , 𝑦1(1) =1.06 in (3) ,we get
𝑧1(1)=110[14−2(1.2)−2(1.06)]=0.95 𝑧1(1) =0.95 Hence the first iteration values of 𝑥,𝑦,𝑧 are 𝑥1(1) =1.2, 𝑦1(1) =1.06, 𝑧1(1) =0.95.
MATHEMATICS - I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 43 Second iteration :
𝑥1(2)=110[12− 𝑦1(1)−𝑧1(1)] …………(4) 𝑦1(2)=110[13−2𝑥1(2)−𝑧1(1)]…….(5) 𝑧1(2)=1
10[14−2𝑥1(2)−2𝑦1(2)]……....(6) Putting 𝑦1(1),𝑧1(1) values in equation (4) ,then we get 𝑥1(2) =0.999 Putting 𝑥1(2) , 𝑧1(1) values in equation (5) ,then we get 𝑦1(2) =1.005
Putting 𝑥1(2) , 𝑦1(2) values in equation (6) , the we get 𝑧1(2) =0.999 Hence the second iteration values of x ,y, z are 𝑥1(2) =0.999 , 𝑦1(2) =1.005, 𝑧1(2) =0.999 .
Again taking 𝑥1(2) =0.999 , 𝑦1(2) =1.005, 𝑧1(2) =0.999 as the initial values ,we get 𝑥1(3)=110[12−1.005 −0.999 ] =0.999=1.00 𝑦1(3)=110[13−2.0−0.999 ]=1.00
𝑧1(3)=110[14−2−2]=1.00 Hence the 3rd Iteration values os x,y,z are 𝑥1(3)=1.00 , 𝑦1(3)=1.00 , 𝑧1(3)=1.00.
Similarly ,we find the 4th iteration values are 𝑥1(4)=1.00 ,𝑦1(4)=1.00 ,𝑧1(4)=1.00 We tabulate the results as follows Variable 1st Iteration 2nd Iteration 3rd Iteration 4th Iteration x 1.20 0.999 1.00 1.00 y 1.06 1.005 1.00 1.00
z 0.95 0.999 1.00 1.00 Hence the solution of the given system of equations is 𝑥=1 ,𝑦=1,𝑧=1 2.Use Gauss -Seidel iteration method to solve the system of equations 8x1-3x2+2x 3 =20 4x1+11x 2-x3=33
6x1+3x 2+12x 3=36