CU-BCA-SEM-I-Mathematics (1)

3.3 TRANSPOSE OF A MATRIX Transpose of a Matrix Let A = [aij]m x n, be a matrix of order m x n. A matrix obtained by interchanging the rows and columns of A is called the transpose of A and is denoted by A’ or AT. A’ = AT = [aij]n x m Example: 1 2 A  3 5  1 3 6  Let 6 7  then transpose of matrix (AT) is 2 5 7 Properties of Transpose 1. (A’)’ = A 51 CU IDOL SELF LEARNING MATERIAL (SLM)

2. (A + B)’ = A’ + B’ 3. (AB)’ = B’A’ 4. (KA)’ = kA’ 5. (AN)’ = (A’)N 6. (ABC)’ = C’ B’ A’ Transpose Conjugate of a Matrix A matrix A is called transpose conjugate then the transpose of the conjugate of a matrix A and it is denoted by A0 or A*. i.e., (A’) = A’ = A0 or A* Properties of Transpose Conjugate of a Matrix (i) (A*)* = A (ii) (A + B)* = A* + B* (iii) (kA)* = kA* (iv) (AB)* = B*A* (v) (An)* = (A*)n 3.4 WORKED EXAMPLES Example: 1 Solution: 52 CU IDOL SELF LEARNING MATERIAL (SLM)

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Example: 2 Solution: 54 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 3 55 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: Example: 4 Solution: 56 CU IDOL SELF LEARNING MATERIAL (SLM)

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Example: 5 Solution: Example: 6 58 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: Example: 7 59 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: Example: 8 Solution: 60 CU IDOL SELF LEARNING MATERIAL (SLM)

3.5 SUMMARY ● The maximum number of linearly independent column vectors or row vectors in a matrix is called its rank. ● Only if the matrix is non-singular can the columns (rows) of a square matrix be demonstrated to be linearly independent. In other words, any non-singular matrix of order n has a rank of n. ● The transposed matrix's row and column dimensions are the inverse of the original matrix's dimensions. 61 CU IDOL SELF LEARNING MATERIAL (SLM)

● The product of a square matrix multiplied by a column matrix comes naturally in linear algebra for solving linear equations, and represents linear transformations. ● A rectangular arrangement of numbers or phrases that can be used for a variety of purposes, including changing coordinates in geometry. 3.6 KEYWORD • Rank of a matrix :the number of linearly independent rows or columns in the matrix • Minor: a minor of a matrix A is the determinant of some smaller square matrix, cut down from A by removing one or more of its rows and columns. • Transpose : The transpose of a matrix is obtained by changing its rows into columns and its columns into rows • Conjugate : A conjugate matrix of a matrix is obtained by replacing each term with its complex conjugate. • Transpose conjugate: If A is an m×n matrix, then the Conjugate Transpose of A is obtained by taking the complex conjugate of each entry in A and then transposing A 3.7 LEARNING ACTIVITY 1. Find the rank of matrix  5 6  . 7 8 ___________________________________________________________________________ _____________________________________________________________________  2 1 1  3 1  5 2. Find the rank of 1 1 1  . ___________________________________________________________________________ _____________________________________________________________________  1 1 1  1 1 1 3. Find the transpose of matrix  1 0 1 . ___________________________________________________________________________ _____________________________________________________________________ 62 CU IDOL SELF LEARNING MATERIAL (SLM)

 1 2  2  4 3 4  4. Find the rank of   2 4  4 ___________________________________________________________________________ _____________________________________________________________________ 3 1  5 1 1  2 1  5 5. Find the rank of 1 5  7 2  . ___________________________________________________________________________ _____________________________________________________________________ 3.8 UNIT END QUESTIONS A. Descriptive Questions Short Questions 1. Find the rank of matrix 13 95 2. Find the rank on  5 7  5 7 1 2 3 2 3 4 3. Find the rank of  3 5 7 . A   2 53 B   1 0  A  BT  AT  BT 7 2 4 4. Let and verify that 5. Find the rank of 11 2 0  2 3 Long Questions 0 1 5  2 4  6 1. Find the rank of 1 1 5  . 63 CU IDOL SELF LEARNING MATERIAL (SLM)

 5 3 0   1 2  4 2. Find the rank of   2  4 8  . 1 2 1 3  2 4 1  2 3. Find the rank of  3 6 3  7 0 1 2 1 1 2 3 2 4. Find the rank of  3 1 1 3 1 1 0 1 2 3 A  2 1 3 B  2 1 3 1 2 1 and  0 1 1 Find AT and BT & verify that 5. Let A  BT  AT  BT . B. Multiple Choice Questions  0 5  7   5 0 11  1. The matrix  7 11 0  is a. A skew-symmetric matrix b. a symmetric matrix c. a diagonal matrix d. an upper triangular matrix  5 2 x   y 2 3 2. If the matrix  4 t  7 is a symmetric matrix, then find the value of x, y and t respectively. a. 4, 2, 3 b. 4, 2, -3 c. 4, 2, -7 d. 2, 4, -7 64 CU IDOL SELF LEARNING MATERIAL (SLM)

 6 8 5 A  4 2 3 3. If 9 7 1 is the sum of a symmetric matrix B and skew-symmetric matrix C, then B is  0 1 2 A   1 0 3 4. If   2  3 0 , then A + 2AT equals a. A b. -AT c. AT d.2A2 A   3 3 x  12 2x  x  5. If is a symmetric matrix, then x = a. 4 b. 3 c.-4 d. -3 Answers 1-a, 2-b, 3-a. 4-c, 5-c 3.9 REFERENCES References book ● Vittal, P.R, “Allied Mathematics”, Reprint,Margham Publications, Chennai. ● Venkata chalapathy, S.G, “Allied Mathematics”, Margham Publications, Chennai. Textbook references 65 CU IDOL SELF LEARNING MATERIAL (SLM)

● Singaravelu, A. “Allied Mathematics”, Meenakshi Agency, Chennai. ● N. Herstein, Topics in Algebra, John Wiley and Sons, 2015. ● Gilbert Strang, Introduction to linear algebra, Fifth Edition, ANE Books, 2016. 66 CU IDOL SELF LEARNING MATERIAL (SLM)

UNIT - 4: MATRIX4 STRUCTURE 4.0 Learning Objectives 4.1 Introduction 4.2 Adjoint of matrix 4.3 Inverse of a matrix 4.4 Example Problems 4.5 Summary 4.6 Keywords 4.7 Learning Activity 4.8 Unit End Questions 4.9 References 4.0 LEARNING OBJECTIVES After studying this unit, you will be able to: ● Explain matrix's inverse and describing its features. ● Describe the inverse definition of a matrix commutatively, and the multiplication must operate in both directions. ● Describe a matrix must be square to be invertible, because the identity matrix must also be square. ● Demonstrate how to calculate determinants using minor and cofactor matrices. 4.1 INTRODUCTION The transpose of the cofactor matrix of that particular matrix is called theadjoint of a matrix. Consider A be given a matrix, the Adjoint is denoted by adj (A). Otherwise, the inverse of a matrix A is that matrix which when multiplied by the matrix A give an identity matrix. The inverse of a Matrix A is denoted by A-1. That is, multiplying a matrix by its inverse produces an identity matrix. Not all square matrices have an inverse matrix. If the determinant of the matrix is zero, then it will not have an inverse, and the matrix is said to be singular. Only non-singular matrices have inverses. 4.2 ADJOINT OF A MATRIX Adjoint of a Square Matrix: 67 CU IDOL SELF LEARNING MATERIAL (SLM)

Consider A = [aij]m x n be a square matrix of order n and Cij be the cofactor of aij in the determinant |A| then the adjoint of A and it is denoted by adj (A), is defined as the transpose of the matrix, formed by the cofactors of the matrix. Properties of Adjoint of a Square Matrix If A and B are square matrices of order n, then (a) A (adj A) = (adj A) A = |A|I adj (A’) = (adj A)’ adj (AB) = (adj B) (adj A) adj (kA) = kn – 1(adj A), k ∈ R adj (Am) = (adj A)m adj (adj A) = |A|n – 2 A, A is a non-singular matrix. |adj A| =|A|n – 1 ,A is a non-singular matrix. |adj (adj A)| =|A|(n – 1)2 , A is a non-singular matrix. Adjoint of a diagonal matrix is a diagonal matrix. 4.3 INVERSE OF MATRIX Inverse of a Square Matrix If A be a square matrix of order n, then a square matrix B such that AB = BA = I is called inverse of A and it is denoted by A-1. i.e., AA-1 = A-1A = 1 Properties of Inverse of a Square Matrix 1. Square matrix A is invertible if and only if |A| ≠ 0 2. (A-1)-1 = A 3. (A’)-1 = (A-1)’ 4. (AB)-1 = B-1A-1 In general (A1A1A1 … An)-1 = An-1An – -1 … A3-1A2-1A1-1 1 5. If a non-singular square matrix A is symmetric, then A-1 is also symmetric. 6. |A-1| = |A|-1 7. AA-1 = A-1A = I 8. (Ak)-1 = (A-1)Ak for k ∈ N Elementary Transformation Any one of the following operations on a matrix is called an elementary transformation. 68 CU IDOL SELF LEARNING MATERIAL (SLM)

1. Interchanging any two rows (or columns), denoted by Ri←→Rj or Ci←→Cj 2. Multiplication of the element of any row (or column) by a non-zero quantity and denotedbyRi → kRi or Ci → kCj 3. Addition of constant multiple of the elements of any row to the corresponding element of any other row, denoted byRi → Ri + kRj or Ci → Ci + kCj Equivalent Matrix ▪ Two matrices A and B are said to be equivalent, if one can be obtained from the other by a sequence of elementary transformation. ▪ The symbol≈ is used for equivalence. Relation Between Adjoint and Inverse of a Matrix Consider a matrix B, and another matrix C such that, B × C = C × B = I, then C is called as the inverse of B. When a number is multiplied by its reciprocal, then we will get 1. Same way, when we multiply a matrix by its inverse, we will get an Identity matrix. The inverse of a matrix is usually used to find the solution to a system of linear equations. Determinants and adjoints are used to find the inverse of a square matrix. The inverse of B is denoted by B-1. The relationship between the adjoint adj(B) and the inverse of a matrix B-1 is represented as B-1 = (1/|B|) × adj(B). 4.4 EXAMPLE PROBLEMS Example: 1 Solution: 69 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 2 Solution: 70 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 3 Solution: Example: 4 Solution: 71 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 5 Solution: 72 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 6 Solution: Example: 7 73 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: Example: 8 Solution: 74 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 9 Solution: Example: 10 75 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: Example: 11 Solution: 76 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 12 Solution: Example: 13 77 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: 78 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 14 Solution: 79 CU IDOL SELF LEARNING MATERIAL (SLM)

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4.5 SUMMARY ● In a system of linear equations, a matrix is commonly employed to represent the coefficients, and the determinant can be used to solve those equations. ● If the determinant of a matrix is not zero, it has a unique inverse. ● Write the m matrix on the left and the m identity matrix on the right to determine the inverse of a m matrix: Then row reduce the matrix to get a mm identity matrix on the left, which is reduced row-echelon form. ● On the right, the new mm matrix is the multiplicative inverse of the original matrix. ● Assume A=[aij] is an n-dimensional square matrix. The adjoint of a matrix A is hence the transpose of A's cofactor matrix. 4.6 KEYWORDS • Adjoint : The adjoint of a matrix A is the transpose of the cofactor matrix of A • Inverse: The inverse of matrix is another matrix, which on multiplying with the given matrix gives the multiplicative identity. 81 CU IDOL SELF LEARNING MATERIAL (SLM)

• Invertible: capable of being inverted or subjected to inversion an invertible matrix. • Singular : A matrix is said to be singular if and only if its determinant is equal to zero. • Non-singular : A non-singular matrix is a square one whose determinant is not zero 4.7 LEARNING ACTIVITY 1 0 1 3 4 5  1. Find inverse of matrix 0  6  7 ___________________________________________________________________________ _____________________________________________________________________  2 4 3 1 2 3 2. Find the determinant of  4 5 1 ___________________________________________________________________________ _____________________________________________________________________ 3. Find the inverse of matrix  1 23  0 ___________________________________________________________________________ _____________________________________________________________________  2 1 1  1 2 5 A  0 1 0  B   3 2 1 1 3 1 and  1 1 1 prove that AB1  B 1 A1 4. Let ___________________________________________________________________________ _____________________________________________________________________ 5. Find the adjoint of  2 1  . 4 5 ___________________________________________________________________________ _____________________________________________________________________ 82 CU IDOL SELF LEARNING MATERIAL (SLM)

4.8 UNIT END QUESTIONS A. Descriptive Questions Short Questions 1. Find the determinant of  1 2  . 3 6  2 4 5 1 2 0 2. Find the determinant of matrix  1 4 1 3. Find the inverse of matrix  1 2  4 5 4. Find the inverse of  1 22  1 5. Define the inverse of matrix. Long Questions  1 1 2  1 2 3 1. Using elementary transformation, find the inverse of matrix  3 1 1  2 1 1   6 7  5 2. Find the inverse of matrix  5  4 3  1 2 4 5  2 0 3. Find the inverse of matrix 3 1 6  8 0 1 4 5 6  4. Find the inverse of  2 1 7  1 1 2 2 1 2 5. Find the inverse of matrix  2 1 1 B. Multiple Choice Questions 83 CU IDOL SELF LEARNING MATERIAL (SLM)

A   1 3  2 7 1. Find the inverse of the matrix , using elementary row transformation. 2. 3. 4. 5. 84 CU IDOL SELF LEARNING MATERIAL (SLM)

Answers 1-a, 2-d, 3-a. 4-a, 5-d 4.9 REFERENCES References book ● Vittal, P.R, “Allied Mathematics”, Reprint,Margham Publications, Chennai. ● Venkata chalapathy, S.G, “Allied Mathematics”, Margham Publications, Chennai. Textbook references ● Singaravelu, A. “Allied Mathematics”, Meenakshi Agency, Chennai. ● N. Herstein, Topics in Algebra, John Wiley and Sons, 2015. ● Gilbert Strang, Introduction to linear algebra, Fifth Edition, ANE Books, 2016. 85 CU IDOL SELF LEARNING MATERIAL (SLM)

UNIT - 5: MATRIX 5 STRUCTURE 5.0 Learning Objectives 5.1 Introduction 5.2 Eigen values and Eigen vectors 5.2.1 Rule to find the Eigen Values and Eigen Vectors 5.3 Properties of Eigen values and Eigen vectors 5.4 Problems in Eigen values and Eigen vectors 5.5 Summary 5.6 Keywords 5.7 Learning Activity 5.8 Unit End Questions 5.9 References 5.0 LEARNING OBJECTIVES After studying this unit, you will be able to: ● Construct, and give examples of mathematical expressions that involve vectors, matrices, and linear systems of linear equations. ● Describe the mathematical expressions to compute quantities that deal with linear systems and Eigen value problems. ● Identify to mathematical statements and expressions (for example to assess whether a particular statement is accurate or to describe solutions of systems in terms of existence and uniqueness). ● Write logical progressions of precise mathematical statements to justify and communicate your reasoning. ● Apply linear algebra concepts to model, solve, and analyze real-world situations 5.1 INTRODUCTION We will study the behavior of linear endomorphisms of R-vector spaces, i.e., R-linear transformations T: V → V, by studying subspaces E ⊆ V which are preserved via scaling by the endomorphism: T(x) = λx for all x ∈ E. 86 CU IDOL SELF LEARNING MATERIAL (SLM)

Such a subspace is called an Eigen space of the endomorphism T, associated to the number λ, which is called an Eigen value. A nonzero vector x such that T(x) = λx for some number λ is called an eigenvector. “Eigen-” is a German adjective which means “characteristic” or “own”. Henceforth, we’ll bandy the prefix “Eigen-” about without apology, whenever we refer to objects which arise from Eigen spaces of some linear endomorphism. Understanding Eigen data associated to a linear endomorphism T is among the most fruitful ways to analyze linear behavior in applications. There are three primary, non-exclusive themes, which I name principal directions, characteristic dynamical modes, and spectral methods. Principal directions arise whenever an eigenvector determines a physically/geometrically relevant axis or direction. Characteristic dynamical modes arise in dynamical problems, whenever a general solution is a superposition (i.e., a linear combination) of certain characteristic solutions. Spectral methods involve studying the eigenvalues themselves, as an invariant of the object to which they are associated. 5.2 EIGEN VALUES ANDEIGEN VECTORS Definition: Let A = [aij] be a square matrix of order n. If there exists a non-zero column matrix X and a scalar λ, such that AX = λX then λ is called an Eigen value of the matrix A and X is called the Eigen vector corresponding to the Eigenvalue λ. Steps to find the Eigenvalues and the corresponding eigenvectors of a square matrix A. Let λ be an Eigenvalue of A and X be the corresponding Eigenvector. Then by Definition, AX = λX = λIX, where I is the identity matrix of order n. ⇒ (A − λI) X= 0 (1) This represents a system of linear homogeneous equations in x1, x2, ….,xn. The homogeneous system has a non-trivial solution if |A − λ I| = 0 (2) The equation |A − λ I| = 0 is called the characteristic equation of the matrix A. When we solve the characteristic equation, we get n values for λ. These n roots of the characteristic equation are called the characteristic roots or latent roots or Eigen values of A. 87 CU IDOL SELF LEARNING MATERIAL (SLM)

substitute the value of λ in the equations (2), we get a non-zero (non- trivial) solution of X. X is called the invariant vector or latent vector or Eigen vector of A corresponding to the Eigen value λ. 5.2.1 Rule to Find The Eigen Values And Eigen Vectors Given a square matrix A, write the characteristic polynomial (A− λI) X= 0 Write the characteristic equation |A− λI| = 0 Solve the characteristic equation |A− λI| = 0 for the eigen values λ For each eigen value λ, solve the homogeneous system of equations (A− λI) X= 0 Note: The Eigen vector corresponding to an Eigen value is not unique. 1. If all the Eigen values λ1,λ2,…,λnof a matrix A are distinct, then the corresponding Eigenvectors are linearly independent. 2. If two or more Eigen values are equal, then it is Eigenvectors may be linearly independent or linearly dependent. 5.3 PROPERTIES OFEIGENVALUES Properties of Eigen values: • The sum of the Eigen values of a matrix A is equal to the sum of the principal diagonal elements of A.(The sum of the principal diagonal elements is called the Trace of the matrix.) • The product of the Eigen values of a matrix A is equal to |A|. • A square matrix A and its transpose AT have the same Eigen values. • If |A| = 0, i.e. A is a singular matrix, at least one of the Eigen values of A is zero and conversely. • The Eigen values of a triangular matrix are just the elements in the main diagonal. • If λ is an Eigen value of a non-singular matrix A, then λ-1 is an Eigen value of A-1. • If λ be an Eigen value of a matrix A, then λ m is an Eigen value of Am and if k is a scalar kλ is an Eigen value of kA. • The Eigen values of are also symmetric matrix (i.e.a symmetric matrix with real elements) are real. • The eigenvectors corresponding to distinct Eigen values of a real symmetric matrix are orthogonal. 88 CU IDOL SELF LEARNING MATERIAL (SLM)

5.4 PROBLEMS IN EIGEN VALUS AND EIGEN VECTORES Example: 1 A3 A  3 2 1 2 Obtain the Eigen value of where Solution: The characteristic equation of A is A  I  0 . A  3 2 1 2 Given 3 2 0 2  A  I  1  (3  )(2  )  2  0 6  2  3  2  2  0 2  5  4  0 ( 1)(  4)  0   1,4. To find the corresponding Eigen vectors, consider the equation (A – λI)X = 0 3  2 2    x1   0  1   x2  When λ = 1 2 2  x1   0 1 x2  we have 1  2x1 + 2x2 = 0 x1 + x2 = 0 -x2⇒ X1  1 or  1  1   both equations are same as x1 + x2 = 0. ⇒x1 =  1  1 2  x1   0   2 x2  When λ = 4 we have  1  -x1 + 2x2 = 0 x1- 2x2 = 0 x1 2x2 X2  2 1 both equations are same as x1-2x2 = 0. = Example: 2 Find the Eigen values and Eigenvectors of the matrix 89 CU IDOL SELF LEARNING MATERIAL (SLM)

1 1 3 A  1 5 1 3 1 1 Solution: The characteristic equation of A is A  I  0 λ3 - S1λ2 + S2λ-S3 =0 where S1 Sum of main Diagonal elements=1+5+1 = 7 S2  Sum of the minors of the main diagonal elements 51 1 3 11 = 1 1 + 3 1 + 1 5 = (5-1) + (1-9) + (5-1) = 0 113 1 5 1  36 S3  Determinant value of A = 3 1 1 Characteristic equation is 3  72  0  36  0 Solving the Characteristic equation, the Eigenvalues–2, 3, 6 The Eigen values of A are λ = –2, 3, 6. To find the Eigen vectors: 1  1 3 x1  0     0  1 5 1   x2   Consider the equation (A – λI)X = 0 ⇒ 3 1 1 x3 0 3 1 3 x1  0 1 1   0 7  x2   Case (i) put λ = –2. 3 1 3x3  0 3x1 + x2 + 3x3 =0----------------(1) x1 + 7x2 + x3 =0----------------(2) 3x1 + x2 + 3x3 =0-----------(3) Consider equations (1) and (2) by using the rule of cross-multiplication, we have x1  x2  x3 x1  x2  x3 1 21 3  3 211 .  20 0 20 90 CU IDOL SELF LEARNING MATERIAL (SLM)

1   X1   0  The eigenvector corresponding to λ = – 2 is  1   2 1 3 x1  0    0  1 2 1  x2   Case (ii) put λ = 3.  3 1  2x3  0 -2x1 + x2 + 3x3 =0----------------(1) x1 + 2x2 + x3 =0----------------(2) 3x1 + x2 -3x3 =0----------------(3) Consider equations (1) and (2) and Solve by method of cross-multiplication, x1  x2  x3 we have  5 5  5 1 X 2  1  1  The eigenvector corresponding to λ = 3 is  5 1 3 x1  0    0  1 1 1  x2   Case (iii) put λ = 6.  3 1  5x3 0 -5x1 + x2 + 3x3 =0----------------(1) x1 - x2 + x3 =0----------------(2) 3x1 + x2 -5x3 =0----------------(3) Consider equations (1) and (2) and solving these equations by the rule of cross-multiplication, x1  x2  x3 we have 4 8 4 4 1 X 3  8 or 2 Thus the eigenvector corresponding to λ = 6 is 4 1 Example: 3 Find the eigen values and eigen vectors of the matrix 0 1 1 A  1 0 1 1 1 0 91 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: The characteristic equation of the matrix A is A  I  0 λ3 - S1λ2 + S2λ-S3 =0 where S1  sum of main Diagonal elements=0+0+0 = 0 S2  Sum of the minors of the main diagonal elements 01 01 01 = 1 0 + 1 0 + 1 0 = (0-1)+(0-1)+ (0-1) = -3 011 1 0 1  0(0  1)  1(0  1)  1(0  1)  2 110 S3  Determinant value of A = 3  02  3  2  0 Characteristic equation is Thus the eigen values of A are -1,-1,2 To find Eigen vectors: Solve (A-λI) X=0 0   1 1  x1   0  0 1  x 2   0  0   x 3    0 1    1 0  2 1 1 x1   0  2  x2   0  1 1 1  x3    0  (i) when λ= 2,  1  2  2x1  x2  x3  0 (1) x1  2x2  x3  0 (2) x1  x2  2x3  0 (3 Solving (1) and (2) by rule of cross multiplication, we get x1  x2  x3 1 2 1 2 41 x1  x2  x3 333 x1  x2  x3 111 92 CU IDOL SELF LEARNING MATERIAL (SLM)

1 1 .∙. The eigen vectors are given by X1 = 1 (ii) when λ=-1 1 1 1  x1  0 1 1   0 1  x2   1 1 1 x3  0 x1  x2  x3  0 (4) x1  x2  x3  0 (5) x1  x2  x3  0 (6) Sole the above the three equations reduce to one x1 + x2 + x3 = 0. There is one equation in three unknowns. Put x1  0 we get x2  x3  0 x2   x3  x2  x3 1 1 0   X2   1  Hence one Eigen vector is  1 To get another Eigen vector, Put x2  0 we get x1  x3  0 x1  x3  x1  x3 1 1 1   X3   0  Hence another Eigen vector is  1 Example: 4 Find the Eigen values and Eigen vectors of the matrix  2 2  3  1  6 A   2  1  2 0  Solution: The characteristic equation of the matrix A is A  I  0 93 CU IDOL SELF LEARNING MATERIAL (SLM)

λ3 - S1λ2 + S2λ-S3 =0 where S1  Sum of main Diagonal elements =- 2+ 1+ 0 = -1 S2  Sum of the minors of the main diagonal elements 1 6 2 3 2 2 =  2 0 +  1 0 + 2 1 = (0-12)+(0-3)+ (-2-4) = -21 S3  Determinant value of A 2 2 3 2 1 6 1 2 0 =  (2)(0  12)  (2)(0  6)  (3)(4  1)  24  12  9  45 3  2  21  45  0 Characteristic equation is Thus the Eigen values of A are -3, -3, 5 To find Eigen vectors: solve (A-  I) X=0  2   2 3  x1   0  1  x  0  2 2 6  2    0   1 0   x3  (i) When λ=-3, 1 2  3 x1   0  4  6 x  0  2 2 3  2    0  1 x3  we get only one Independent equation from x1  2x2  3x3  0 (1) 2x1  4x2  6x3  0 (2)  x1  2x2  3x3  0 (3) from(1), (2)and (3) we get x1  2x2  3x3  0 Case1: putx1  0 we get x2  x3 32 0 3 .∙. The Eigen vectors are given by X1 = 2 Case 2 : put x2 =0 in x1  2x2  3x3  0 we get 94 CU IDOL SELF LEARNING MATERIAL (SLM)

x1  x3 31 3 0 .∙. The Eigen vectors are given by X2= 1  7 2  3 x1  0  6   0   2 4   x2   (ii) When λ=5 1  2  5x3  0  7x1  2x2  3x3  0 (4) x1  2x2  3x3  0 (5)  x1  2x2  5x3  0 (6) By using method of cross multiplication, we get from (4) and (5) x1  x2  x3  12  12  6  42 28  4 x1  x2  x3 1 2 1 1   X3   2  Hence the Eigen vectors are given by  1 ` Example: 5 Find the eigen values and eigen vectors of the matrix  7 2 0  A   2 6  2  0  2 5  Solution: The characteristic equation of the matrix A is A  I  0 λ3 - S1λ2 + S2λ-S3 =0 where S1  sum of main Diagonal elements =7+6+ 5 = 18 S2  Sum of the minors of the main diagonal elements 6 2 7 0 7 2 =  2 5 + 0 5 +  2 6 = (30-4)+(35-0)+ (42-4) = 99 S3  Determinant value of A 95 CU IDOL SELF LEARNING MATERIAL (SLM)

7 2 0  2 6  2  (7)(30  4)  (2)(10  0)  (0)(4  0) 0 2 5  182  20  162 3  2  21  45  0 Characteristic equation is Thus the eigen values of A are 3,6,9 To find Eigen vectors: solve (A- I) X=0 7   2 0 x1   0  6  x  0   2 2 2  2    0  0 5   x3  4 2 0 x1    0  2 3  2 x 2   0 (i) when λ=3,  0 2 2  x 3   0  we get only one Independent equation 4 x1  2 x2  0 x3  0 (1)  2 x1  3x2  2 x3  0 (2) 0 x1  2 x2  2 x3  0 (3) Solving (2) and (3) by rule of cross multiplication, we get x1  x2  x3 64 04 40 x1  x2  x3 244 x1  x2  x3 122 1 2 .∙. The eigen vectors are given by X1 = 2  1  2 0   x1  0  2 2   0 0   x2   (ii) when λ=6,  0  2 1 x3  0 x1  2x2  0x3  0 (4)  2x1  0x2  2x3  0 (5) 0x1  2x2  x3  0 (6) By using method of cross multiplication, we get from (5) and (6) 96 CU IDOL SELF LEARNING MATERIAL (SLM)

x1  x2  x3 04 02 40 x1  x2  x3 4 2 4 x1  x2  x3 2 1 2 2   X2   1  Hence the corresponding eigen vector is  2  2  2 0  x1  0  2 2   0 3   x2   (iii)when λ=9,  0  2  4x3  0  2x1  2x2  0x3  0 (7)  2x1  3x2  2x3  0 (8) 0x1  2x2  4x3  0 (9) From (7 )and (8) we get x1  x2  x3 12  4 0  8 4  0 x1  x2  x3 8 8 4 x1  x2  x3 2 2 1 2 X 3   2 Hence the eigen vectors are given by  1  Example: 6 Verify that the sum of Eigen values of A equals the trace of A and that their product equals |A|, for the matrix 1 0 0  A  0 3 1 0 1 3  Solution: The characteristic equation of A is (l – λ)(λ2 – 6λ + 8) =0 The Eigenvalues of A are λ = 1, 2, 3. Sum of the Eigenvalues = 7. Trace of the matrix = 1 + 3 + 3 = 7 Product of the Eigenvalues = 8. |A| = 8. Hence the properties verified. Example: 7  2 0  1   A   0 2 0  Find the Eigenvalues and eigenvectors of (adj A), given that the matrix 1 0 2  97 CU IDOL SELF LEARNING MATERIAL (SLM)

Solution: The characteristic equation of A is (2 – λ)3 – (2 – λ) =0 i.e. (2 – λ) (λ2 – 4λ + 3) =0 The Eigen values of A are λ = 1, 2, 3. Case (i) λ = 1.  1 0 1 x1       0 1 0   x2   0 The eigenvector is given by 1 0 1 x3  1 x1  x2  x3 X1  0 1 0 1  1 Case (ii) λ = 2.  0 0 1 x1       0 0 0   x2   0 The eigenvector is given by 1 0 0 x3  i.e. –x3 = 0 and –x1 = 0.  x1 = 0 and x3 = 0 and x2 is arbitrary. Let x2 = 1 0 X 2  1  0 Case (iii) λ = 3. 1 0 1 x1       0 1 0   x2   0 The eigenvector is given by 1 0 1x3  x1  x2  x3 1   X3   0  1 0 1  1 The eigen values of A–1 are 1, 1, 1with the eigenvectors X , X , X. 23 1 2 3 Now adj A=A−1 A i.e., adj A = |A| · A–1 = 6A–1 (∵ A = 6 for the given matrix A) The eigen values of (adj A) are equal to 6 times those of A–1, namely, 6, 3, 2. The corresponding eigenvectors are X1, X2, X3 respectively. 98 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 8 Verify that the eigenvectors of the real symmetric matrix  3 1 1  A  1 5 1  1 1 3  are orthogonal pairs. Solution: The characteristic equation of A is λ3 – 11λ2 + 36λ – 36 =0 i.e. (λ – 2) (λ – 3) (λ – 6) = 0 The eigen values of A are λ = 2, 3, 6. Case (i) λ = 2.  1 1 1  x1   1 1   3   x2   0 The eigenvector is given by  1 1 1 x3  1   x1  x2  x3 X1   0  2 0  2  1 Case (ii) λ = 3.  0 1 1  x1  1 1   2   x2   0 x1 x2 x3 The eigenvector is given by  1 1  1 1 1 0 x3   1 X 2  1 1 Case (iii) λ = 6.  3 1 1  x1    1 1   1   x2   0 The eigenvector is given by  1 1  3x3  x1  x2  x3 1 2 4 2 X 3   2  1  1 11  0 X T  1 X2 1 0 Now 1 1 1 2  0 XT X 3  1 1 2  1  1 1  TX  1 2 0   0 3 X1 1 . Hence the eigen vectors are orthogonal in pairs. 99 CU IDOL SELF LEARNING MATERIAL (SLM)

Example: 9 1 1 1     1 1 1  Find the sum and product of the Eigen values of the matrix.  1 1 1 Solution: Sum of eigen values = Sum of the diagonal elements. = -1-1-1 = -3. The product of the eigen values of a matrix A is a equal to its determinant. 1 1 1 = 4. 1 1 1 .∙. Product of Eigen values = 1 1 1 Example: 10 For a given matrix A of order three, A  32 and two of its Eigen values are 8 and 2. Find the sum of the Eigen values. Solution: Let the three eigen values of A be λ1 ,λ2 , λ3. The product of three Eigen values is A λ1 λ2 λ3 = 32(given) 16λ3 = 32 λ3 = 2. Sum of Eigen values = 8+2+2 =12 Example: 11 3 2 Obtain the Eigen value of A3 where A  1 2 Solution: The Characteristic equation of A is A  I  0 . A  3 2 1 2 Given 3 2  A  I  1 2    0 100 CU IDOL SELF LEARNING MATERIAL (SLM)