Mathematics-I PDF - B.Tech 1st Year

MALLA 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...

MALLA 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 (R18A0021) Mathematics -I Course Objectives: To learn 1. The concept of rank of a matrix which is used to know the consistency of system of linear equations and also to find the eigen vectors of a given matrix. 2. Finding maxima and minima of functions of several variables. 3. Applications of first order ordinary differential equations. ( Newton’s law of cooling, Natural growth and decay) 4. How to solve first order linear, non linear partial differential equations and also method of separation of variables technique to solve typical second order partial differential equations. 5. Solving differential equations using Laplace Transforms. UNIT I: Matrices Introduction, types of matrices-symmetric, skew-symmetric, Hermitian, skew-Hermitian, orthogonal, unitary matrices. Rank of a matrix - echelon form, normal form, consistency of system of linear equations (Homogeneous and Non-Homogeneous). Eigen values and Eigen vectors and their properties (without proof), Cayley-Hamilton theorem (without proof), Diagonalisation. UNIT II:Functions of Several Variables Limit continuity, partial derivatives and total derivative. Jacobian-Functional dependence and independence. Maxima and minima and saddle points, method of Lagrange multipliers, Taylor’s theorem for two variables. UNIT III: Ordinary Differential Equations First order ordinary differential equations: Exact, equations reducible to exact form. Applications of first order differential equations - Newton’s law of cooling, law of natural growth and decay. 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 variation of parameters. UNIT IV: Partial Differential Equations Introduction, formation of partial differential equation by elimination of arbitrary constants and arbitrary functions, solutions of first order Lagrange’s linear equation and non-linear equations, Charpit’s method, Method of separation of variables for second order equations and applications of PDE to one dimensional (Heat equation). UNIT V: Laplace Transforms Definition of Laplace transform, domain of the function and Kernel for the Laplace transforms, Existence of Laplace transform, Laplace transform of standard functions, first shifting Theorem, Laplace transform of functions when they are multiplied or divided by “t”, Laplace transforms of derivatives and integrals of functions, Unit step function, Periodic function. Inverse Laplace transform by Partial fractions, Inverse Laplace transforms of functions when they are multiplied or divided by ”s”, Inverse Laplace Transforms of derivatives and integrals of functions, Convolution theorem, Solving ordinary differential equations by Laplace transforms. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 1  MATHEMATICS - I 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)Advanced Engineering Mathematics by Michael Green Berg, Pearson 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.Find the extreme values of functions of two variables with / without constraints. 3.Solve first and higher order differential equations. 4.Solve first order linear and non-linear partial differential equations. 5.Solve differential equations with initial conditions using Laplac DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 2  MATHEMATICS - I UNIT-I DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 3  MATHEMATICS - I MATRICES Matrix : A system of mn numbers real (or) complex arranged in the form of an ordered set of ‘m’ rows, each row consisting of an ordered set of ‘n’ numbers between [ ] (or) ( ) (or) || || is called a matrix of order m xn. a11 a12.........a1n  a   21 a12.........a 2 n  Eg: .........................  = [aij ]mxn where 1≤i≤m, 1≤j≤n.   .......................  a m1 a m 2.......a mn    mxn 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: 1 2 31x3 Column Matrix: A Matrix having only one column is called a “Column Matrix”. 1  Eg: 1    2  3 x1 Null Matrix: A=  aij  mxn such that aij=0  i and j. Then A is called a “Zero Matrix”. It is denoted by Omxn. 0 0 0  Eg: O2x3=   0 0 0  Rectangular Matrix: If A=  aij  mxn , and m  n then the matrix A is called a “Rectangular Matrix” 1  1 2  Eg :  3 4 is a 2x3 matrix 2 Square Matrix: If A=  aij  mxn and m = n then A is called a “Square Matrix”. 1 1  Eg :   is a 2x2 matrix  2 2 Lower Triangular Matrix: A square Matrix AnXn   aij  nxn is said to be lower Triangular of aij = 0 if i j. i.e. all the elements below the principle diagonal are zeros. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 4  MATHEMATICS - I 1 3 8  Eg: 0 4  5 is an Upper triangular matrix 0 0 2  Triangle 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 called 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 1 2 3 Eg: A= 4 5 6 diagonal elements are 1, 5, 9  7 8 9  Diagonal Matrix: A Square Matrix is said to be diagonal matrix, if aij = 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  d1 , d2 ,., dn  3 0 0  Eg : A = diag (3,1,-2)= 0 1 0    0 0  2 Scalar Matrix: A diagonal matrix whose leading diagonal elements are equal is called a 2 0 0 “Scalar Matrix”. Eg : A= 0 2 0    0 0 2  Unit/Identity Matrix: If A =  aij  such that aij=1 for i = j, and aij=0 for i  j then A is nxn called a “Identity Matrix” or Unit matrix. It is denoted by In 1 0 0 1 0 Eg: I2 =   ; I3= 0 1 0 0 1  0 0 1  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. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 5  MATHEMATICS - I a h g Eg : A = h b f  then tr A = a+b+c  g f 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 matrices 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 denoted 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 | |≠. 2.Every Invertible matrix has unique inverse. 3. If A, B are two invertible square matrices then AB is also invertible and  AB  1  B 1 A1 AdjA 4. A1  where detA  0 , det A 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). (A1)-1 = (A-1)1 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. 1 5 6 7 Eg: Let A = 8 9 10 5 . Then 8 9 10 is a sub matrix of A obtained by deleting first   3 4 5    2 x3 3 4 5 1 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’. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 6  MATHEMATICS - I 2 1 1 3 1 2  Eg: A =  be a 4x3 matrix 1 2 3   5 6 7 Here B = [ ] is a sub-matrix of order ‘2’ B = 2-3 = -1 is a minor of order ‘2 And C = [ ] 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 = [ ] Involutary Matrix: A square matrix A such that A2 = I then A is called an Involutary Matrix. Eg: A = [ ] Nilpotent Matrix: A square matrix A is said to be Nilpotent if there exists a + ve integer n such that An = 0 here the least n is called the Index of the Nilpotent Matrix. Eg: A = [ ] 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 AT or A1 Eg: A = [ ] Then AT = [ ] Properties of transpose of a matrix: If A and B are two matrices and AT , B T are their transposes then 1)  AT   A ; 2)  A  B   AT  BT ; 3)  KA  KAT ; 4)  AB   BT AT T T T T Symmetric Matrix: A square matrix A is said to be symmetric if AT  A If A  aij  then AT  a ji  nxn where aij  a ji nxn a h g Eg: h b f  is a symmetric matrix   g f c  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 7  MATHEMATICS - I Skew-Symmetric Matrix: A square matrix A is said to be Skew symmetric If AT   A. If A  aij  then AT  a ji  where aij  a ji nxn nxn.  0 a  b Eg :  a 0 c  is a skew – symmetric matrix   b  c 0  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.  A  A =  A  AT  A  AT  = 1 1 Proof: Let A be a square matrix, A = 2 2 1 2  A  AT    A  AT  = P + Q, where P =  A  AT  ; Q =  A  AT  1 2 1 2 1 2 Thus every square matrix can be expressed as a sum of two matrices.   T Consider PT   1  A  AT    1  A  AT   1 AT   AT  = 1  A  AT  =P, since PT  P , T T 2  2 2 2 P is symmetric   T Consider QT    A  AT     A  AT   AT   AT  = - 1  A  AT  = - Q 1 1 T 1 T 2  2 2 2 Since QT  Q , Q is Skew-symmetric. To prove the representation is unique: Let A= R+S 1 be the representation, where R is symmetric and S is skew symmetric. i.e. RT  R, S T   S Consider AT   R  S   RT  S T  R  S  2  T 1   2   A  AT  2S  S  1 2  A  AT   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  2 4 9  where A= 14 7 13    9 5 11 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 8  MATHEMATICS - I  2 14 3  Solution: A   4 7 5  T  9 3 11  4 10 12  2 5 6  A  A  10 14 18   P   A  A   5 7 9  ; P is symmetric   T 1 T 2 12 18 22  6 9 11  0 18 6  0 9 3 A  A  18 0 8   Q   A  A    9 0 4  ; Q is skew  symmetric   T 1  2  6 8 0  3 4 0  2 5 6  0 9 3  Now A =P+Q =  5 7 9  +  9 0 4      6 9 11  3 4 0  Orthogonal Matrix: A square matrix A is said to be an Orthogonal Matrix if AAT=ATA=I, Similarly we can prove that A=A-1; Hence A is an orthogonal matrix. 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. 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 : cos  sin   1. Show that A =  cos   is orthogonal.  sin  cos  sin   cos   sin   Sol: Given A =  AT =  cos   cos   then  sin  sin  cos  sin   cos   sin   Consider A.AT =   sin  cos   sin  cos    cos 2   sin 2   cos  sin   cos  sin   1 0 =  =  I  sin  cos   cos  sin  cos 2   sin 2   0 1   A is orthogonal matrix.  1 2 2 2. Prove that the matrix 1 2 1 2  is orthogonal. 3  2 2  1 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 9  MATHEMATICS - I  1 2 2   1 2 2 Sol: Given A = 2 1 2  1 Then AT = 1 2 1 2  3  3  2 2  1 2 2  1  1 2 2   1 2 2 9 0 0 1 0 0 Consider A.AT = 1 2 1 2  2 1 2  = 1 0 9 0  = 0 1 0  = I 9    9    2 2  1 2 2  1 0 0 9 0 0 1  ⇒ A.AT = I Similarly AT.A = I Hence A is orthogonal matrix 0 2b c  3. Determine the values of a, b, c when a b  c  is orthogonal.   a  b c  Sol: - For orthogonal matrix AAT = I 0 2b c  0 a a  So, AAT = a b  c 2b b  b  I     a  b c  c  c c   4b 2  c 2 2b 2  c 2  2b 2  c 2   2  1 0 0 2b  c a2  b2  c2 a2  b2  c2  =I= 0 1 0  2    2b 2  c 2 a 2  b 2  c 2 a 2  b 2  c 2    0 0 1  Solving 2b2-c2 =0, a2-b2-c2 =0 We get c =  2b a2 =b2+2b2 =3b2  a =  3b From the diagonal elements of I 1 4b2+c2= 1  4b2+2b2=1 (since c2=2b2)  b =  6 1 1 1 a =  3b =  ; b= ; c =  2b =  2 6 3 2 3 1  4. Is matrix  4 3 1  Orthogonal?   3 1 9   2 3 1   2 4 3 Sol:- Given A=  4 3 1    3 3 1       3 1 9   1 1 9   2 3 1   2 4 3 14 0 0      4 3 1  3 3 1  =  0 26 0   I3        3 1 9  1 1 9   0 0 91 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 10  MATHEMATICS - I      I3  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 A   aij  , A  aij  , where aij is the conjugate of aij. mxn mxn  2  3i 5   2  3i 5  Eg: If A=   then A =   6  7i 5  i  6  7i 5  i  Note: If A and B be the conjugate matrices of A and B respectively, then (i)  A  =A (ii) A  B = A + B (iii)  KA  = K A 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 A  A.  4 1  3i   4 1  3i   4 1  3i  Eg: A=   then A =   and Aθ=  1  3i 7  1  3i 7  1  3i 7  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=  aij  , aij  a ji i, j. nxn Skew-Hermitian Matrix: A square matrix A is said to be Skew-Hermitian Matrix iff A   A.  3i  2  i   3i 2  i  Eg: Let A=   then A =  3i  2  i 2  i i  and A T  2  i i    2  i i       A =-A  A is skew-Hermitian matrix. T Note: 1. In Skew-Hermitian matrix the principal diagonal elements are either Zero or Purely Imaginary. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 11  MATHEMATICS - I 2. The Skew- Hermitian matrix over the field of Real numbers is nothing but real Skew - Symmetric matrix. 3. In Skew-Hermitian matrix A=  aij  , aij  a ji i, j. nxn Unitary Matrix: A Square matrix A is said to be unitary matrix iff AA  A A  I or A  A1 1   4  2  4i   6 2  4i  4  Eg: B Theorem1: Every square matrix can be uniquely expressed as a sum of Hermitian and skew – Hermitian Matrices. 1 1 1 A(2 A)  ( A  A)  ( A  A  A  A ) Proof: - Let A be a square matrix write 2 2 2 1 1 A  ( A  A )  ( A  A )i.eA  P  Q 2 2 Let P  1 2  A  A  ; Q   A  A  1 2 1  1 Consider P   ( A  A )   ( A  A )  ( A  A )  P 2  2 I.e. P  P, P is Hermitian matrix.  1  Q   ( A  A )    A  A    ( A  A )  Q 1 1 2  2 2 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 A  R  A      (2) DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 12  MATHEMATICS - I 1   2   A  A 1 2  A  A   P  2 R ie R  1   2   A  A  2S ie S   A  A   Q 1 2 Thus every square matrix can be uniquely expressed as a sum of Hermitian & skew Hermitian matrices. Solved Problems :  3 7  4i 2  5i  1) If A=  7  4i 2 3  i  then show that A is Hermitian and iA is skew-   2  5i 3  i 4  Hermitian.  3 7  4i 2  5i  Sol:  Given A= 7  4i 2 3  i  then   2  5i 3  i 4   3 7  4i 2  5i   3 7  4i 2  5i   A   7  4i 2   3  i  And  A   7  4i T 2 3  i   2  5i 3  i 4   2  5i 3  i 4    T A  A Hence A is Hermitian matrix. Let B= iA  3i 4  7i 5  2i  i.e B=  4  7i 2i 1  3i  then   5  2i 1  3i 4i   3i 4  7i 5  2i   B   4  7i 2i 1  3i   5  2i 1  3i 4i   3i 4  7i 5  2i   3i 4  7i  5  2i   B  1  3i   (1)  4  7i  2i  1  3i    B T   4  7i 2i    5  2i 1  3i 4i   5  2i 1  3i 4i    T  B =-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   A  A And  B   B ------------- (1) T T DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 13  MATHEMATICS - I Now  AB  BA  AB  BA   T T   T  AB  BA     BA   B   A   A  B  T T T T T T  AB  BA  AB (By (1))    AB  BA Hence AB-BA is a skew-Hemitian matrix.  a  ic b  id  3). Show that A=  is unitary if and only if a2+b2+c2+d2=1 b  id a  ic   a  ic b  id  Sol: Given A=  b  id a  ic   a  ic b  id  Then A    b  id a  ic   a  ic b  id  T  Hence A  A   a  ic   b  id  a  ic  b  id   a  ic b  id   AA   b  id a  ic   b  id a  ic   a 2  b2  c 2  d 2 0  =   0 a 2  b2  c 2  d 2   AA  I if and only if a 2  b 2  c 2  d 2  1  0 1  2i  , show that I  AI  A is a unitary matrix. 1 4)Given that A=    1  2i 0  1 0  0 1  2i  Sol: we have I  A     0 1  1  2i 0   1 1  2i   1  And 1  2i 1 0  0 1  2i  I  A   0 1  1  2i 0  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 14  MATHEMATICS - I  1 1  2i  =  1  2i 1  1  1  1  2i   ( I  A) 1    1  4i  1 1  2i 2 1  1 1 1  2i    6 1  2i 1  Let B  I  AI  A 1 1 1  1  2i   1  1  2i  1 1  (1  2i)( 1  2i)  1  2i  1  2i  B       6 1  2i 1  1  2i 1  6  1  2i  1  2i (1  2i)(1  2i)  1 1   4  2  4i  B 6 2  4i  4  1  4 2  4i  1  4 2  4i    T Now B      4  and B 6  2  4i 4  6  2  4i 1  4 2  4i   4 2  4i    T   2  4i  4  36  2  4i 4  B B   1 36 0  1 0   I 36  0 36  0 1   B T  B 1 i.e. B is unitary matrix.  I  AI  A is a unitary matrix. 1 5) Show that the inverse of a unitary matrix is unitary.  Sol: Let A be a unitary matrix. Then AA  I i.e  AA   1  I 1 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 15  MATHEMATICS - I   A  A1  I 1   A1  A1  I  1 Thus A 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) 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  (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 2 3 1. Find the rank of the given matrix 3 4 4    7 10 12 1 2 3 Sol: Given matrix A = 3 4 4    7 10 12 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 1 2 3 4 2. Find the rank of the matrix 5 6 7 8    8 7 0 5  Sol: Given the matrix is of order 3x4 Its Rank ≤ min(3,4) = 3 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 16  MATHEMATICS - I Highest order of the minor will be 3. 1 2 3  Let us consider the minor 5 6 7    8 7 0  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. 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 transformation is called elementary matrix or E-matrix. Notations: We use the following notations to denote the E-Matrices. 1) Eij Matrix obtained by interchange of ith and jth rows (columns). 2) Ei k  Matrix obtained by multiplying ith row (column) by a non- zero number k. 3) Eij k  Matrix obtained by adding k times of jth row (column) to ith row (column). Echelon form of a matrix: 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. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 17  MATHEMATICS - I 1 0 0 0 Eg: 1. 0 1 0 0 is a row echelon form.  0 0 1 1   0 0 0 0  1  3  1 2. 0 1 1  is a row echelon form.   0 0 0  Solved Problems : 2 3 7 1. Find the rank of the matrix A = 3  2 4  by reducing it to Echelon form.   1  3  1  2 3 7  Sol: Given A = 3  2 4  Applying row transformations on A.   1  3  1  1  3  1 R1 ↔ R3 A ~ 3  2 4 2 3 7 1  3  1  R2 → R2 –3R1; R3→ R3 -2R1 ~ 0 7 7 0 9 9  1  3  1 R2 → R2/7,R3→ R3/9 ~ 0 1 1  0 1 1  1  3  1 R3 → R3 –R2 ~ 0 1 1  0 0 0  This is the Echelon form of matrix A. The rank of a matrix A= Number of non – zero rows =2 4 4  3 1  1 1 1 0  2. For what values of k the matrix  has rank ‘3’. k 2 2  2   9 9 k 3  Sol: The given matrix is of the order 4x4 If its rank is 3  det A =0 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 18  MATHEMATICS - I 4 4  3 1  1 1 1 0  A=  k 2 2  2   9 9 k 3  Applying R2 → 4R2-R1, R3 →4R3 – kR1, R4 → 4R4 – 9R1 4 4 3 1  0 0 1  1  We get A ~  0 8  4k 8  3k 8  k    0 0 4k  27 3 Since Rank A = 3  det A =0 0 1 1  4 8  4k 8  3k 8  k  0 0 4k  27 3  1[(8-4k) 3]-1(8-4k) (4k+27)] = 0  (8-4k) (3-4k-27) = 0  (8-4k)(-24-4k) =0  (2-k)(6+k)=0  k =2 or k = -6 2 1 3 5 4 2 1 3  3). Find the rank of the matrix using echelon form A 8 4 7 13    8 4 3 1 2 1 3 5 4 2 1 3  Sol: Given A 8 4 7 13    8 4 3 1 2 1 3 5  0 0 5 7  By applying R2 R2  2R1 ; R3 R3  4R1 ; R4 R4  4 R1 ~ 0 0 5 7    0 0 15 21 2 1 3 5 R R R 0 7  R1 1 , R2 2 , R3 3 ~ 0 5 1 1 3 0 0 5 7   0 0 5 7 2 1 3 5 0 7  R3 R3  R2 , R4 R4  R2 ~ 0 5 0 0 0 0   0 0 0 0 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 19  MATHEMATICS - I ⇒ A is in echelon form ∴ Rank of A = 2 1 2 0 1 4). Find the rank of the matrix A = 2 1 1 0  by reducing into echelon form.  3 3 1 1    1 1 1 1 1 2 0 1  Sol: By applying R2 R2  2R1; R3 R3  3R1; R4 R4  R1 A ~ 0 3 1 2 0 3 1 2    0 3 1 2  1 2 0 1  R3 R3  R2 A ~ 0 3 1 2  0 0 0 0    0 3 1 2  1 2 0 1  R3  R4 A ~ 0 3 1 2  0 3 1 2    0 0 0 0  1 2 0 1 R3 R3  R2 A ~ 0 3 1 2  0 0 0 0   0 0 0 0 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.  Ir 0 I  0 ; Ir 0 ;  r  ;  I r  Known as normal forms or canonical forms by using Elementary  0 0 row or column or both transformations where I r 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 2 3 4  1. By reducing the matrix 2 1 4 3  into normal form, find its rank. 3 0 5 10 1 2 3 4  Sol: Given A = 2 1 4 3  3 0 5 10 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 20  MATHEMATICS - I 1 2 3 4  R2 → R2 – 2R1; R3 → R3 – 3R1  A ~ 0  3  2 5  0  6  4  22 1 2 3 4  R3 → R3/-2 A ~ 0  3  2  5 0 3 2 11  1 2 3 4  R3 → R3+R2  A ~ 0  3  2  5 0 0 0 6  0 0 0 0  c2→ c2 - 2c1, c3→c3-3c1, c4→c4-4c1  A ~ 0  3  2  5 0 0 0 6  1 0 0 0 c3 → 3 c3 -2c2, c4→3c4-5c2 A ~ 0  3 0 0  0 0 0 18 1 0 0 0 c2→ c2 /-3, c4→c4/18 A ~ 0 1 0 0  0 0 0 1  1 0 0 0 c4 ↔ c3 A~ 0 1 0 0  0 0 1 0  This is in normal form [I3 0], ∴ Hence Rank of A is ‘3’. 1 1 1 1 2). Find the rank of the matrix A = 1 2 3 4  by reducing into canonical form or  2 3 5 5     3 4 5 8  normal form. 1 1 1 1 Sol: Given A = 1 2 3 4    2 3 5 5     3 4 5 8  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 21  MATHEMATICS - I 1 1 1 1  By applying R2 R2  R1 , R3 R3  2R1 , R4 R4  3R1 0 1 2 5  ~  0 1 3 7    0 7 8 5  1 1 1 1  R3 R3  R2 , R4 R4  7 R2 0 1 2 5  ~ 0 0 1 2    0 0 6 30  1 1 1 1  R4 R4  6R3 0 1 2 5  ~ 0 0 1 2    0 0 0 18 1 1 1 1 R 0 R4 4 1 2 5 18 ~ 0 0 1 2    0 0 0 1 1 0 0 0 0 2 5 Apply C2 C2  C1 , C3 C3  C1 , C4 C4  C1 1 ~ 0 0 1 2    0 0 0 1 1 0 0 0 0 0 0  C3 C3  2C2 ; C4 C4  5C2 1 ~ 0 0 1 2    0 0 0 1 1 0 0 0 0 0  C4 C4  2C3 ~ 1 0 0 0 1 0   0 0 0 1 Clearly it is in the normal form  I 4   Rank of A = 4 3). Define the rank of the matrix and find the rank of the following matrix 2 1 3 5 4 2 1 3   8 4 7 13    8 4 3 1 2 1 3 5  2 1 3  Sol: Let A=  4 8 4 7 13    8 4 3 1 R2 R2  2 R1 2 1 3 5  0 0 5 7  R3 R3  4 R1 A~  0 0 5 7  R4 R4  4 R1   0 0 15 21 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 22  MATHEMATICS - I 2 1 3 5 R3 R3  R2 0 0 5 7  R4 R4  3R2 A~  0 0 0 0   0 0 0 0 It is in echelon form. So, rank of matrix = no. of non zero rows in echelon form.  Rank  ( A)  2 2 1 3 4 0 3 4 1  4). Reduce the matrix A to normal form and hence find its rank A 2 3 7 5   2 5 11 6 2 1 3 4  1  Sol: Given A  0 3 4 2 3 7 5   2 5 11 6 1 1 3 4 1 0 3 4 1  C1 C1 A~  2 1 3 7 5   1 5 11 6 1 1 3 4 R3 R3  R2 0 3 4 1  R4 R4  R1 A~  0 2 4 1   0 4 8 2 1 1 3 4 0 0  R2 R2  R3 A~  1 0 0 2 4 1   0 4 8 2 1 0 0 0 R3 R3  2 R2 0 1 0 0  R4 R4  4 R2 A~  0 0 4 1   0 0 8 2 1 0 0 0 0 0  R4 R4  2R3 A~  1 0 0 0 4 1   0 0 0 0 1 0 0 0 0 0  C4 4C4  C3 A~  1 0 0 0 4 1   0 0 0 0 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 23  MATHEMATICS - I 1 0 0 0 1 0 1 0 0  I 0 C3 C3 A~  ⇒ A~  3 3 0 0 1 0 0 0   0 0 0 0 This is in normal form. Thus Rank of matrix = Order of identify matrix.  Rank  ( A)  3 0 1 2 2 5). Reduce the matrix A =  4 0 2 6  into canonical form and then find its rank.   2 1 3 1  1 0 2 2 Sol: Apply C1  C2 A ~ 0 4 2 6  1 2 3 1  1 0 2 2 R3 R3  R1 A ~ 0 4 2 6  0 2 1 3  1 0 0 0  C3 C3  2C1; C4 C4  2C1 A ~ 0 4 2 6  0 2 1 3 R2 1 0 0 0  R2 A ~ 0 2 1 3 2 0 0 0 0  1 0 0 0  C2  C3 A ~ 0 1 2 3 0 0 0 0  1 0 0 0  C3 C3  2C2 ; C4 C4  3C2 A ~ 0 1 0 0  0 0 0 0  I 0 Which is in the normal form  2   (A) = 2 0  , 0 Note:.If A is an mxn matrix of rank r, there exists non-singular matrices P and Q such that  I r 0 PAQ =   0 0  Suppose we want to find P and Q we have procedure. Let order of matrix ‘A’ is ‘3 i.e. A = I3 A I3 1 0 0 1 0 0 A = 0 1  0  A 0 1 0  0 0 1  0 0 1  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 24  MATHEMATICS - I Now we go on applying elementary row operations and column operations on the matrix A  I 0 (L.H.S) until it is reduced to the normal form  r  0 0  Every row operations will also be applied to the pre-factor of on R.H.S Every column operation will also be applied to the post –factor of on R.H.S. Solved Problems : 1 0  2 1. Find the non-singular matrices P and Q is of the normal form where A = 2 3  4  3 3  6  Sol: Write A = I3 A I3 1 0  2 1 0 0 1 0 0 ~ 2 3  4  = 0 1   0  A 0 1 0  3 3  6  0 0 1  0 0 1  1 0  2 1 0 0  1 0 0 R2 → R2-2R1, R3→R3-3R1 ~ 0 3 0  =  2 1 0 A 0 1    0  0 3 0   3 0 1 0 0 1  1 0  2 1 0 0 1 0 0 R3 → R3-R2, R2 →1/3 R2 ~ 0 1 0     2 / 3 1 / 3 0  A 0 1    0  0 0 0   1 1 1  0 0 1  1 0 0 1 0 0 1 0 2 c3 → c3-2c1 ~ 0 1 0  =   2 / 3 1 / 3 0 A 0 1    0 0 0 0  1 1 1  0 0 1  1 0 0 1 0 2 ~[  ]= PAQ where P =  2 / 3 1 / 3 0 Q = 0 1 0  1 1 1  0 0 1  2. Find the non-singular matrices P and Q such that the normal form of A is P A Q. 1 3 6  1 Where A = 1 4 5 1 . Hence find its rank.  1 5 4 3  Sol: we write A = I3A I4 1 3 6  1 1 0 0 1 0 0 0 0 1 0 0  ~ 1 4 5 1 =  0  A  1 0  0 0 1 0 1 5 4 3  0 0 1    0 0 0 1  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 25  MATHEMATICS - I 1 3 6 1  1 0 0 0 1 0 0  0 R2 →R2- R1, R3→R3-R1 ~ 0 1 1 2  =   1 1 0 A  1 0 0     0 0 1 0 0 2  2 4  1 0 1   0 0 0 1  1 3 6 1  1 0 0 0 1 0 0  0 0  R3 →R3-2R2 ~ 0 1  1 2 =  1 1 0 A  1 0    0 0 1 0 0 0 0 0   1  2 1   0 0 0 1  Applying c2→c2-3c1,c3 →c3-6c1, and c4→c4+c1, we get. 1 0 0 0 1 0 0 1  3  6 1 ~ 0 1  1 2 =  1 1 0  A 0 1 0 0     0 1 0  0 0 0  1  2 1  0 0    0 0 0 1  Applying c3→c3+c2 and c4→c4-2c2, we get. 1  3  9 7  1 0 0  0 1 0 1 1  2 0 0 ~ 0 1 =  1 1 0  A  0 0    0 0 1 0  0 0 0 0  1  2 1   0 0 0 1   I 0 1  3  9 7  1 0 0  0   2  = P A Q Where P =  1 1 0  Q=  1 1  2 0 0    0 0 1 0  1  2 1   0 0 0 1   I 0 Here A ~  2  , ∴ Hence ρ(A) =2 0 0  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 x1, x2, x3     xn given by a11 x1  a12 x2 ......  a1n xn  b1 a21 x1  a22 x2 ......  a2 n xn  b2 } --------------------------- (1).............................................. am1 x1  am 2 x2 ......  amn xn  bm The above set of equations can be written in the Matrix form as A X = B  a11 a12    a1n   x1   b1  a a22    a2 n  x  b   21  2  2        .   .  (2)              .  .   an1 an 2    ann   xn  bm  A-Coefficient Matrix; X-Set of unknowns; B-Constant Matrix Homogeneous Linear Equations: If b 1  b2 ...........  bm  0 then B = 0 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 26  MATHEMATICS - I Hence eqn (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 B  0, the system Reduces to AX = B is known as Non-Homogeneous Linear equations. Solutions: A set of numbers x1 , x2   xn 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 said 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]  a11 a12 a13   a1n : b1   A B   C   a21 a22 a23   a2 n : b2   am1 am 2 am3   amn : b3  1. If  ( A / B)   (A) , then the system of equations AX = B is consistent (solution exits). a). If  ( A / B)   (A) = r = n (no. of unknowns) system is consistent and have a unique solution b). If  ( A / B)   (A) = 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 / B)   (A) then the system of equations AX=B is inconsistent (no solution). In case of homogeneous system AX = 0, the system is always consistent. (or) x1 = 0, x2 = 0, --------, xn = 0 is always the solution of the system known as the” zero solution “. Non-trivial solution: If  ( A / B)   (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 equations. 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) DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 27  MATHEMATICS - I Let us first consider n equations in n unknowns ie. m=n then the system will be of the form a11x1+a12x2+a13x3+…..+a1nxn =b1 a21x1+a22x2+a23x3+…..+a2nxn =b2 ………………………………….. an1x1+an2x2+an3x3+…..+annxn =bn The above system can be written as AX = B --------- (1) Where A is an n  n matrix. Solving AX = B using Echelon form: Consider the system of m equations in n unknowns given by a11x1+a12x2+a13x3+…..+a1nxn =b1 a21x1+a22x2+a23x3+…..+a2nxn =b2 ………………………………….. am1x1+am2x2+am3x3+…..+amnxn =bm We know this system can be we write as AX = B  a11 a12 a1n b1    a a22 a2 n b2  The augmented matrix of the above system is [A / B] =  21      am1 am 2 amn bm  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. 1 1 1   x   4 Sol: Write given equations is of the form AX = B i.e; 2 5  2  y   3      1 7  7   z  5  1 1 1 4 Consider the Augment matrix is [A /B]  [A/B] = 2 5 2 3 1 7 7 5  1 1 1 4  Applying R2 →R2-2R1 and R3 → R3-R1, we get [A/B] ~ 0 3 4  5 0 6 8 1  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 28  MATHEMATICS - I 1 1 1 4  Applying R3→ R3-2R2, we get [A/B] ~ 0 3 4  5 0 0 0 11  ∴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 1  3  8  x   10 Sol: Matrix notation is 3 1  4   y   0      2 5 6   z  3  1  3 8  10 Augmented matrix [A/B] is [A/B] = 3 1 4 0   2 5 6 3  1 3 8  10  R2 → R2-3R1 R3 → R3 -2R1 ~ 0 10 20 3 0   0 11 22 23 1 3 8  10  R2 →1/10 R2 ~ 0 1 2 3    0 11 22 23  1 3 8  10  R3 →R3-11R2 ~ 0 1 2 3    0 0 0  10 This is the Echelon form of [A/B] ∴ρ (A) =2, ρ (A/B) =3 ρ (A) ≠ ρ[A/B]. The given system is inconsistent. 3). Find whether the following equations are consistent, if so solve them. x+y+2z=4, 2x-y+3z=9, 3x-y-z=2 1 1 2   x   4 Sol: We write the given equations in the form AX=B i.e; 2  1 3   y   9  3  1  1  z  2 1 1 2 4  The Augmented matrix [A/B] = 2 1 3 9   3 1 1 2  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 29  MATHEMATICS - I 1 1 2 4  Applying R2→ R2-2R1 and R3→R3-3R1, we get [A/B] ~ 0  3 1 1   0  4  7  10  1 1 2 4  Applying R3 →3R3-4R2, we get [A/B] ~ 0  3 1 1   0 0  17  34  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 1 1 2   x  4  We have 0  3  1   y   1  0 0  17   z   34  -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. 1 1 1  x  6  Sol: We write the given equations in the form AX=B i.e. 1 2 3   y   14      1 4 7   z  30 1 1 1 6  The Augmented matrix [A/B] = 1 2 3 14    1 4 7 30  1 1 1 6  Applying R2→ R2-R1 and R3→R3-R1, we get [A/B] ~ 0 1 2 8    0 3 6 24  1 1 1 6  Applying R3 →R3-3R2, we get [A/B] ~ 0 1 2 8    0 0 0 0  This matrix is in Echelon form. ρ(A) = 2 and ρ(A/B) = 2 Since ρ (A) =ρ [A/B]. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 30  MATHEMATICS - I 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. 1 1 1  x  6  The given system of equations reduced form is 0 1 2   y   8      0 0 0   z  0 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 that x  2 y  z  3 ; 3 x  y  2 z  1 ; 2 x  2 y  3z  2; x  y  z  1 are consistent and solve them 1 2 1 3  3 1 2   x       y   1  Sol: The above system in matrix notation is  2 2 3    2    z  31   1 1 1  43  1 41 A X B 1 2 1 3   3 1 2 1  The Augmented matrix is  AB      2 2 3 2    1 1 1 1 R2 R2 3R1 1 2 1 3  0 7 5 8  R3 R3 2 R1 ~  0 6 5 0  R4 R4 R1   0 3 2 4  1 2 1 3  0 1 0 4  R2 R2  R3 ~  0 6 5 4    0 3 2 4  1 2 1 3  R3 R3 3R1 0 1 0 4  R4 R4 3R2 ~  0 0 5 20    0 0 2 8  1 2 1 3  R 0 1 0 4  R3 3 ~  5 0 0 1 4    0 0 2 8  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 31  MATHEMATICS - I 1 2 1 3    R4 R4  2R3 0 1 0 4  ~ 0 0 1 4    0 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 x  2y  z  3  y  4; z  4 x84  3 y  4; z  4 x  3  4  1  x  1, y  4, z  4. 6). Solve x  y  z  3; 3x  5 y  2 z  8; 5 x  3 y  4 z  14 1 1 1   x   3  Sol: - 3 5 2  y    8       5 3 4  z  14 1 1 1 3  Augmented Matrix is  AB  3 5 2 8  5 3 4 14  1 1 1 3  R2 R2  3R1 ~ 0 8 1 1 R3 R3  5R1 0 8 1 1 1 1 1 3  R3 R3  R2   ~ 0 8 1 1 0 0 0 0    A    AB  2  Number of unknowns (3)  The system has infinite number of solutions. x  y  z  3,  8 y  z  1  8 y  z  1 Let z  k  y  1 k and x  3 1  k   k  24  1  k  8k  23  7k 8 8 8 8  x   238  87 k   238   X   y    18  8k   X   81  where k is any real number.  z   0  k   1  7). Find whether the following system of equations is consistent. If so solve them. x  2 y  2 z  2, 3x  2 y  z  5, 2 x  5 y  3z  4, x  4 y  6 z  0. DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 32  MATHEMATICS - I 1 2 2  2  3 2 1   x       y   5  Sol: In Matrix form it is  2 5 3     4     z    1 4 6  0 AX  B 1 2 2 2  R2 R2  3R1 0 8 5 1 ~  R3 R3  2 R1 0 9 1 8    R4 R4  R1 0 2 4 2  1 2 2 2  0 1 4 7  ~  0 9 1 8   R2 R2  R3 0 2 4 2  1 2 2 2  0 1 4 7  ~ R3 R3  9 R2 0 0 37 55    R4 R4  2 R2 0 0 12 16  1 2 2 2  1 4 7  R4 1 R4 ~ 0 4 0 0 37 55    0 0 3 4  1 2 2 2  1 4 7  R4 37 R4  3R3 ~ 0 is in echelon form 0 0 37 55   0 0 0 17    A  3 and   AB  4    A    AB.∴The given system is in consistent. 8). Discuss for what values of , the simultaneous equations x+y+z = 6, x+2y+3z =10, x+2y+ z = have (i). No solution (ii). A unique solution (iii). An infinite number of solutions. 1 1 1  x  6  Sol: The matrix form of given system of Equations is A X = 1 2 3   y   10 = B     1 2    z     DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 33  MATHEMATICS - I 1 1 1 6 The augmented matrix is [A/B] = 1 2 3 10 1 2    1 1 1 6  R2 → R2 – R1, R3 → R3-R1 [A/B] ~ 0 1 2 4  0 1  1   6 1 1 1 6  R3 →R3 – R2 ~ 0 1 2 4   0 0   3   10 Case (i): let ≠ 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 ≠3 and for any value of ‘ ’. Case (ii): Suppose =3 and ≠10. We have ρ (A) = 2, ρ (A/B) = 3 The system has no solution. Case (iii): Let =3 and =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. 1 1 1   x   3 Sol: The above system in matrix notation is 1 2 2   y   6       1 a 3   z  b  1 1 1 3 Augmented matrix  AB  1 2 2 6 1 a 3 b  1 1 1 3  R2 R2  R1  ~ 0 1 1 3  R3 R3  R1 0 a  1 2 b  2  DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 34  MATHEMATICS - I 1 1 1 3  R3 R3  R2  ~ 0 1 1 3  0 a  3 0 b  9  1 1 1 3     For a  3 & b  9 ~ 0 1 1 3  0 0 0 0     A    AB  2  3 ⇒ It has infinite no of solutions. 1 1 1 3     For a  3 & b  any value ~ 0 1 1 3  0 a  3 0 b  9     A    AB  3 ⇒ It has a unique solution. 1 1 1 3     For a  3&b  9 ~ 0 1 1 3  0 0 0 b  9     A  2    AB  3 ⇒ Inconsistent  no solution 10). Solve the following system completely. x  y  z  1; x  2 y  4 x   ; x  4 y  10 z   2 1 1 1   x   1  1 2 4   y      Sol: The above system in matrix notation is      1 4 10   z   2  A X  B 1 1 1 1  Augmented Matrix is  AB   1 2 4   1 4 10  2  R2 R2  R1 1 1 1 1  ~ 0 1 3   1   R3 R3  R1 0 3 9  2  1 1 1 1 1    ~ 0 1 3  1  R3 R3  3R2 0 0 0  2  3  2  Here   A  2 and   AB  3 ⇒ The given system of equations is consistent if  2  3  2  0   2  2    2  0    2 1  0    2,  1 Case (i): When   1 DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 35  MATHEMATICS - I   A    AB  2  Number of unknowns.  The system has infinite number of solutions. 1 1 1 1  The equivalent matrix is 0 1 3 0    0 0 0 0 The equivalent systems of equations are x  y  z  1; y  3z  0  Let z  k  y  3k and x  (3k )  K  1  x  2k  1  x  1  2k  x  1  2k  1  2 X   y   0  3k   X  0  k  3 where k is any arbitrary constant.        z   0  k  0  1  Case (ii): When   2   A    AB  2  no. of unknowns.  The system has infinite number of solutions. 1 1 1 1  The equivalent matrix is ~ 0 1 3 1    0 0 0 0 The system of equations equivalent to the given system is x + y + z = 1; y + 3z = 1 Let z  k  y  1  3k and x  (1  3K )  k  1  x  2k  x   0  2k  0 2  X   y    1  3k   X  1   k  3        z   0  k  0  1  where k is any arbitrary constant. 11). Show that the equations 3x  4 y  5 z  a; 4 x  5 y  6 z  b;5 x  6 y  7 z  c don’t have a solution unless a  c  2b. solve equations when a=b=c= -1. 3 4 5  x  a  4 5 6  y   b  Sol: The Matrix notation is      5 6 7   z   c  A X  B 3 4 5 a  Augment Matrix is  AB  4 5 6 b   5 6 7 c  3 4 5 a  R2 3R2  4 R1 ~ 0 1 2 3b  4a   R3 3R3  5R1