35113923_BCA114

96 Mathematics 9.5 Summary There are a number of basic operations that can be applied to modify matrices, called matrix addition,Subtraction of Matrix, scalar multiplication, matrix multiplication. If A = [aij] and B = [bij] are two matrix of the same order, m × n, then the sum of the two matrices A and B is another matrix C = [cij] of the same order m × n, where cij = aij + bij.    If A = aij and B = bij are two matrices of the same order, m × n, then the subtraction of the  two matrices A and B is another matrix C = cij of the same order m × n, where, cij = aij – bij If a matrix A = [aij] of order m × n is to be multiplied by a scalar k, where k is not = 0, then the scalar multiplication KA of the matrix A is obtained by multiplying every element of A by scalar constant K. Hence, KA = [kaij]. In order to multiply two matrices, we have to learn the method of multiplying a row by a column first. If A is a row matrix and B is column matrix, A × B can be found provided the number of column of A = the number of rows of B. 9.6 Key Words/Abbreviations  A × A = A2  A+ B= B+ A  AB ≠ BA in matrix 9.7 Learning Activity 1. If A 1 2 B  1 0  , find (i) A + B, (ii) A – B and (iii) A × B. 4 5 and 2    3    2. If A  7 6 and B  2 1 , find X such that 2A – 3B + 5I = 8X.    1 0 3 4    CU IDOL SELF LEARNING MATERIAL (SLM)

 97  Matrices 2 9.8 Unit End Questions (MCQ and Descriptive) A. Descriptive Type Questions Exercise 1. If A = 1 2 and B = 1 2 , find 2A – 3B.    2 1 3 4   (Ans.: 5  24) 3  2. If A =  1 1 and B =  1 0 and I is the unit matrix of order 2, find B – 4A + 2I.  1 1  2 1 (Ans.: 1  2 41) 3. If A = 9 1 and B =  1 5 , find the matrix X such that 3A + 5B + 2X = 0.    13 4 3 7 (Ans.:  16 1347)  47 / 2   2 0  3 5 , find the matrix X such that 2A – 3B + 5X = 0. 4. If A = and B =   7   1 3 2 (Ans.: =  13/ 5  47 / 2  337)  3 1  52, B = 1 2 4 and C =  8 2 4 5. If A = 3 1 3  2 3 , verify the following:  8 4  7 (a) A + B = B + A (b) A + (B + C) = (A + B) + C (c) 3(A + B – C) = 3A + 3B – 3C (d) (A + B)' = A' + B' CU IDOL SELF LEARNING MATERIAL (SLM)

98 Mathematics 6. (a) If A= 1 24, find the matrix B such that A + B = 0. 3   1  2 (Ans.:  3 )    (b) If A =  1 24 2 and B =  1  2 , find the matrix X such that 2X + 3A – 4B  3 1 0  3 0 = 0.    1 5 ) (Ans.:  9 2  6   x y Find x,y andz if (c) y  z = 3 11. z  2x y  x 1 (Ans.: x = 1, y = 2, z = 3)  If A = 1 7 , find the matrices.   (d) 2 , B = 2   3 7 3 9 (i) 4A, (ii) –5B, (iii) A + B, (iv)A – B, (v) A' – B' (Ans.: (i) =  4 8 10  35 (iii) = 3 9 , (iv) = 1  25,  1  02) 1 2 , (ii) = ,   5 (v) = 28  15  45 6 16   0  6. 3 1 and B = 5  3 , find the matrix: (a) A + 3B and (b) 4A – 7B. Also, If A = 0   1 2 show that (A2+ 3B)' = A' + 3B'. 18  8 , ( b) =  23 25 (Ans.: (a) =   22 4   7   3  7. If A = 2 I1, 1(b),AB–=B+32 0 and I is the identity matrix of order 2, find the matrices (a) A + 0 2 2A – B + I. B– 2I and (c) 4 1  1 1 , (c) =  2  23) (Ans.: (a) = , (b) = 1  2  2   2 0 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 2 99  8. 2 4 21 and B = 1 4 1 If A = 0 1 2 0 2, find the matrix A + 2B – 5I, where, I is the 3 2 1 2 1 3 identity matrix of order 3. 1 12 3 (Ans.: =  4 4 6)     7 4 2  9. If A = 01 2 46, B = 53 5 69 and C = 38 6 75, find the matrix 3A – B + 2C. 5 0 4       2 7 9 4 2 0 4 0 1  4 13 16 (Ans.: =  1 23 23)  18 19 29  1 1 2 3 5 , prove that (2A + 3B) – 4C = 10. If A = and C = 1 2 4 31, B = 1 3 2 5 1 0 2 4 2A + (3B – 4C). 11. If A = 1 2 and B  2 3 , find AB and BA. Also, check whether they are equal. Also,    1 4 4 1 show that (AB)' = B'A'. 10 75, A  5 1162, AB ≠ BA) (Ans.: = AB = 5 18  2 1  1 2 12. If A = 5,  1 and C  , show that A (B + C) = AB + AC.  3 B   3 1 1 4  0  13. If A = 51 2 23 , B  43 1 25 and C  04 1 22 , verify that: 0 2 3       1 1 1 2 0 3 1  2 3 (i) A (B + C) = AB + AC (ii) A(BC) = (AB) C CU IDOL SELF LEARNING MATERIAL (SLM)

100 Mathematics 14. If A =  3 1 , find the matrix A2 – 2A. 1 1 (Ans.: =  2 2 )  2  2   1 2 1 15. Show that A3 – 3A2 – A + 9I= 0, where, A = 0 1 1. 3 1 1  16. Find AB and BA whenever they exist in each of the following cases: (a) 2 5 , B = 3 1 A=    1 3  2 5  (Ans.: = 11 1177) 1 1  2 A= (b)  23, B = 1 2 3 1 0 1 2  (Ans.: = AB = 2 7 12 , BA does not exist) 0    1   11 A = 1  (c)  0 2 , B = 3 1 1   0  0 1  (Ans.: = AB does not exist, BA = 5 3  02) 2 1  (d) A = 52 3 01, B = 10 2 03 1 2     3 2 1  1 1 2 2 (Ans.: = AB = 4198 131, BA = 2110 7 04)      39 8  3 8 3 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 2 101 17. If A = 2 1 , B = 2 5 7 and C  1 6 4    2   2 , verify that: (i) A(B + C) = AB 3 3  3 1  3  1 + AC and (ii) A(B – C) = AB– BC. 1 2 2 18. If A = 2 1 2, show that A2 – 4A is a scalar matrix. 2 2 1  19. If A =  2 4 , show that A2 is a null matrix. 1  2 2 4 4 20. If A = 4 2 4, show that A2 – 8A is a scalar matrix. 4 4 2  21. If A= 1 24, show that A2 – 5A – 2I is a zero matrix. 3 22. Show that the matrix A = 1 3 satisfies A2 – 4A+ 3I =0 where I = 1 01. 0 3 0  B. Multiple Choice/Objective Type Questions 1. 1 2  1 0  .   4 2 3  4   2 2 2 4   (a)   (b) 6 8 7 6   0 8  (c)   8 0 (d) None of the above 2. The order of matrix is never . (a) Zero (b) One (c) Three (d) None of the above CU IDOL SELF LEARNING MATERIAL (SLM)

102 Mathematics 3. We can multiply matrix iff . (a) Row of 1st matrix = Column of 2nd matrix (b) Square matrix (c) Column of 1st matrix = Row of 2nd matrix (d) None of the above 4. An identity matrix is not a scalar matrix. (a) False (b) True (c) Can't say (d) Both (a) and (b) 5. Order of matrix given by m × n is . (a) m → columns, n → rows (b) m = rows, n = columns (c) Row is given by m or n depending on which is bigger (d) None of these Answers: 1. (b), 2. (a), 3. (c), 4. (a), 5. (b) 9.9 References References of this unit have been given at the end of the book. CU IDOL SELF LEARNING MATERIAL (SLM)

UNIT 10 MATRICES 3 Structure: 10.0 Learning Objectives 10.1 Introduction 10.2 Multiplication of Two Matrices 10.3 Orthogonal Matrix 10.4 Minor of a Matrix 10.5 Rank of Matrix 10.6 Elementary Operations 10.7 Normal Form or Canonical Form 10.8 PAQ Normal Form 10.9 Square Matrix 10.10 Summary 10.11 Key Words/Abbreviations 10.12 Learning Activity 10.13 Unit End Questions (MCQ and Descriptive) 10.14 References 10.0 Learning Objectives After studying this unit, you will be able to:  Explain the transpose of a matrix.  Discuss how to find rank of a matrix.

104 Mathematics 10.1 Introduction In this chapter, we will come across orthogonal matrix. We also find rank of matrix using echelon and normal forms. A is said to be rank of matrix r, if there exists at least one non-zero minor of A of order r. 10.2 Multiplication of Two Matrices In order to multiply two matrices, we have to learn the method of multiplying a row by a column first. If A is a row matrix and B is column matrix, A × B can be found provided the number of column of A = the number of rows of B. Hence, if the order of A is 1 × m and the order of B is m × 1, only then we can find AB and the order of AB is 1 × 1. The element of AB is found as follows: Suppose A = [1 2 3] and B = 54   6 Then AB = [1 × 4 + 2 × 5 + 3 × 6] AB = [32] Hence, to find the element of AB, we find the product of the corresponding elements of A and B and add them. Example 1: If A = 3  5 and B = 1  2 , find AB and BA.  2 0   3 4 Solution: Both AB and BA can be formed and each is of order 2 × 2. AB = 3 5 1 2 2 0    34  (5)(3) (2)  ((05) )((44)) = ((32))((11))  (0) (3) (3) (2)  (2)   CU IDOL SELF LEARNING MATERIAL (SLM)

 105  Matrices 3 = 212 246    BA = 1 2 3 5 3 4    20  = ((13))((33))  (2)(2) (1) (5)  ((40))((02)) (4) (2) (3) (5)     = 171 155      1 2  Example 2: Let A =  13 and C = 3 B= 21 1  1 32 . , 0 0 2 0  2     3 1 2  1 4  Show that (AB)C = A(BC). Solution: A is a 2 × 3 matrix; B is a 3 × 3 matrix. AB exists is order 2 × 3. Also C is of order 3 × 2. (AB)C exist and is order 2 × 2. Let us find (AB)C. 1 2 3 1 1 1 2 0 2 2 0 3  AB =  3 1 2 = 12  4  9 1 0  3 1  6  46  144 2 161  0  6 202 2  0  4        14 2 11  1 23 (AB)C =  4  0  4 4 6 1  = 14  0 11 42  4 2444 3 82  4  0  6 12  8 2  28  Now, B is order of 3 × 3 and C is of order 3 × 2. CU IDOL SELF LEARNING MATERIAL (SLM)

106 Mathematics BC exist and is of order 3 × 2. Also, A is of order 2 × 3 and BC is of order 3 × 2. A(BC) exist and is of order 2 × 2. Let us find A(BC). BC = 21 1 31  1 23 0 0     3 1 2 1 4 = 21  13  63 2  4 = 21 181   12     3  2  9  2  8  1 15  1 2 3  12 181 A(BC) =    2  2 0  1 15 = 2  2  3 1  36  3405 3 82 4  0  2 20 2  28  (AB)C = A(BC) Hence proved. Example 3: If A =  3 1 , find A2 – 5A+ 7I.  1 2  Solution: A2 = AA =  3 1  3 1 1 2 1 2  =  9 1 3  2  8 5  3  2 1 4  5 3  A2 – 5A+ 7I = 8 53 5 3 21 7 1 0  5 1 0 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 107  Matrices 3 = 8 5  15 105 7 0  5 3 0 7  5   =  8 15  7 55 70 0 00.  5  5  0 3 10  0  Example 4: If A = 1 3 find A2 – 2A + 3I where I is the identity matrix of order 2. 2 1  Solution: A2 = A × A, is a matrix of order 2. = 1 3   1 3     2 1 2 1   11  3 2 12333111 7 6 = 2 1  1 2 4 7 2 6 3I  3 1 0 3 0 2A = ,        4 2 0 1 0 3  So, the matrix A2 – 2A+ 3I = 7 6  2 6  3 0       4 7 4 2 0 3  = 7  2  3 6  6  30 8 0 4  4  0 7  2  0 8  The result matrix can be written as 8I. Example 5: If A = 3 7 and B =  2 5 , find (i) (A + B) (A – B), (ii) A2 – B2.  1  1 4 0  Solution: 7   4 (i) Consider A + B = 3  2 5 = 5 12 1 1 0 0 4 CU IDOL SELF LEARNING MATERIAL (SLM)

  108 Mathematics (A – B) = 3 7   2 5 = 1 2 ...(1) 1 4 1 0 2 4  (A + B) × (A – B) = 5  12  1 2    0 4 2 4  = 5  24 101648 8      = 29 58    8 (ii) A2 = A × A 3 7  3 1   6 7  1 4 1 4  = 9  7 21  2168 16 49 3  4 7  7 23  B2 = B × B =  2 1  05 2 5 1 0 =  4  5 10  0  2  0  5  0 =   1 10  2  5  Now, A2 – B2 = 16  7  4239  1 10  2  5 = 17 39 ...(2)  9 28.    10.3 Orthogonal Matrix For a orthogonal matrix if A.A = I, then matrix A is called orthogonal matrix. Exercise 3: Check given matrix are orthogonal CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 109 cos θ  sin θ 0 A = sin θ cos θ 0  0 0 1   cos   sin  0 Solution: Given, A = sin  cos  0  0 0 1   cos  sin  0 A =  sin  cos  0  0 0 1  For orthogonal matrix, A.A = I cos   sin  0  cos  sin  0 cos  sin  0 0  sin  cos  0  0 1  0 0 1  cos2   sin 2   0 cos sin   sin  cos  0  0  0  = sin  cos   cos sin   0 sin 2   cos2   0 0  0  0 0  0  0 0  0  1  0 0 0  1 0 0 = 0 1 0 = I 0 0 1  Hence, A.A = I given matrix is orthogonal. Exercise 4: Express the matrix A as the sum of a symmetric and a skew-symmetric matrix when  4 2  3 A=  1 3  6.   5 0  7 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  110 Solution:   4 2  3   4 1  5 Given: A =  1 3  6 and A =  1 3 0    5 0  7  3  6  7   8 3  8 A + A =  3 6  6   8  6  4   0 1 2 A – A = 1 0  6  2 6 0  A = 1 (A + A) + 1 (A – A) 22 8 3  8 1 0 1 2 1   2 1 0 = 3 6  6 2  8 6  6 + 6 0  4  2    10.4 Minor of a Matrix In a square matrix, each element possesses its own minor. The minor is defined as a value obtained from the determinant of a square matrix by deleting out a row and a column corresponding to the element of a matrix. Given a square matrix A, by minor of an element [aij], we mean the value of the determinant obtained by deleting the ith row and jth column of A matrix. It is denoted by Mij. In order to find the minor of the square matrix, we have to erase out a row and a column one by one at the time and calculate their determinant, until all the minors are computed. The following are the steps to calculate minor from a matrix: 1. Hide ith row and jth column one by one from given matrix, where i refer to m and j refers to n that is the total number of rows and columns in matrices. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 111 2. Evaluate the value of the determinant of the matrix made after hiding a row and a column from Step 1. Minor of 3×3 Matrix a b c Consider the 3*3 matrix A  d e f  We had to hide the first row and column in order to g h i find the minors of matrices.  e f = ei– hf h M11 i     d f = di– fg g M12 i     d e – g M13   = dh eg  h   b c = bi – ch h M21 i     a c = ai– cg g M22 i     a b = ah – bg g M23 h     b c – e M31   = bf ce  f   a c – d M32   = af cd  f   a b – d e M33  = ae bd CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  112 2 – 1 3 Example: Consider the 3*3 matrix A  0 4 2 1 –1 – 2  Solution: We first calculate minor of element 2. Since it is (1,1) element of A, we delete first row and first column, so that determinant of remaining array is 4 2  (4*–2) – (2*–1) = –8+2= –6 = M11 – 1 – 2 = Since –1 is (1,2) element, we delete first row and second column. The determinant of 0 2  0*–2 – (2*1) = –2 = M12 remaining array 1 – 2 = 0 4  – 4 = –4 = M13 The minor of 3 is 1 –1 = 0 The minor of 0 is – 1 3  (–1)(–2) – (3)(–1) = 2 + 3 = 5 = M21 – 1 – 2 = 2 3 (2)(–2) – (3)(1) = –4–3 = –7 = M 22 The minor of 4 is 1 – 2=  The minor of 2 in (2,3) place in 2 – 1 (2)(–1) – (1)(1) = –2+1 = –1 = M23 1 – 1 = – 1 3 The minor of 1 is 4 2 = (–1)(2) – (3)(4) = –2–12 = –14 = M31  The minor of (–1) is 2 3 = (4)–0 = 4 = M32 0 2    The minor of (–2) is 2 – 1 (2)(4)–0 = 8 = M33 0  = 4  CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 113 Minor of 2×2 Matrix For a 2*2 matrix, calculation of minors is very simple. Let us consider a 2×2 matrix A  a b c d. We had to hide the first row and column to find the minors of matrices.   – – – d= d M11   – – = c M12  – c   – b – – = b M21   – M22  a –  = a –   Example: Consider the matrix P  2 6 – 4 7. For finding minor of 2 we delete first row and first column.  – 2 6 Solution: – 4 7. So that remaining array is |7| = 7 = M11  Similarly, minors of 6, –4 and 7 will be –4,6,2 respectively. 10.5 Rank of Matrix Let A be a m × n matrix. Then, A is said to be a matrix of rank r, if: 1. There exists at least one non-zero minor of A of order r, and 2. Each minor of A of order greater than r is zero, i.e., each minor of order (r + 1) or higher is zero or vanishes, we generally denoted rank of matrix by (A). Let A be a square matrix of order m × n. Then, rank of Matrix A is: CU IDOL SELF LEARNING MATERIAL (SLM)

114 Mathematics 1. (A) = n if A  0, i.e., A is non-singular 2. (A) < n if A = 0, i.e., A is singular 1 72. Exercise 5: Find Rank of A = 4 Solution: Given, A is square matrix A = 7 – 8 = –1  0 So, rank A = order of 2 rank A = 2  (A) = 2 Exercise 6: Find the rank of matrix A. A = 2 1 20 5 4 6. 1 0 Solution: Given, Matrix 3 × 3 and A = 76  0 Therefore, Rank A = 3, (A) = 3 Exercise 7: Find rank of A= 2 3 1 4 0 6  Solution: Since, A is of 2 × 3 i.e., Rank A  2 Rank A  Min (2, 3) 23 Now, A = 40 =2×0–3×4 = –12  0 Rank A = Rank A1 = 2 (A) = 2 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 115 10.6 Elementary Operations There are three kinds of elementary matrix operations. 1. Interchange two rows (or columns). 2. Multiply each element in a row (or column) by a non-zero number. 3. Multiply a row (or column) by a non-zero number and add the result to another row (or column). When these operations are performed on rows, they are called elementary row operations; and when they are performed on columns, they are called elementary column operations. For example, suppose you want to interchange rows 1 and 2 of Matrix A. To accomplish this, you could premultiply A by E to produce B, as shown below. R1 <--> R2 = 01 135 10 246 E A R1 <--> R2 = 0+ 2 0+ 4 0+ 6 0+ 1 0+ 3 0+ 5 R1 <--> R2 = 246 =B 13 5 Here, E is an elementary operator. It operates on A to produce the desired interchanged rows in B. What we would like to know, of course, is how to find E. Read on. How to Perform Elementary Row Operations To perform an elementary row operation on a A, an r × c matrix, take the following steps. 1. To find E, the elementary row operator, apply the operation to an r × r identity matrix. CU IDOL SELF LEARNING MATERIAL (SLM)

116 Mathematics 2. To carry out the elementary row operation, premultiply A by E. We illustrate this process below for each of the three types of elementary row operations.  Interchange two rows: Suppose we want to interchange the second and third rows of A, a 3 × 2 matrix. To create the elementary row operator E, we interchange the second and third rows of the identity matrix I3. 100 100 010 ~ 001 001 010 I3 E  Then, to interchange the second and third rows of A, we premultiply A by E, as shown below. R2 <--> R3 = 100 01 001 23 010 45 E A R2 <--> R3 = 1*0 + 0*2 + 0*4 1*1 + 0*3 + 0*5 0*0 + 0*2 + 1*4 0*1 + 0*3 + 1*5 0*0 + 1*2 + 0*4 0*1 + 1*3 + 0*5 R2 <--> R3 = 01 45 23 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 117  Multiply a row by a number: Suppose we want to multiply each element in the second row of Matrix A by 7. Assume A is a 2 × 3 matrix. To create the elementary row operator E, we multiply each element in the second row of the identity matrix I2 by 7. 10 10 ~ 01 07 I2 E  Then, to multiply each element in the second row of A by 7, we premultiply A by E. 7R2 --> R2 = 10 012 07 345 EA 7R2 --> R2 = 1*0 + 0*3 1*1 + 0*4 1*2 + 0*5 0*0 + 7*3 0*1 + 7*4 0*2 + 7*5 7R2 --> R2 = 012 21 28 35  Multiply a row and add it to another row: Assume A is a 2 × 2 matrix. Suppose we want to multiply each element in the first row of A by 3; and we want to add that result to the second row of A. For this operation, creating the elementary row operator is a two- step process. First, we multiply each element in the first row of the identity matrix I2 by 3. Next, we add the result of that multiplication to the second row of I2 to produce E. 10 1 0 10 ~ 0 + 3*1 ~ 01 1 + 3*0 31 I2 E CU IDOL SELF LEARNING MATERIAL (SLM)

118 Mathematics  Then, to multiply each element in the first row of A by 3 and add that result to the second row, we premultiply A by E. 1 0 01 3R1 + R2 --> R2 = 3 1 23 EA 3R1 + R2 --> R2 = 1*0 + 0*2 1*1 + 0*3 3*0 + 1*2 3*1 + 1*3 3R1 + R2 --> R2 = 01 How to Perform Elementary Column Operations 26 To perform an elementary column operation on A, an r × c matrix, take the following steps. 1. To find E, the elementary column operator, apply the operation to an c×c identity matrix. 2. To carry out the elementary column operation, postmultiply A by E. Let's work through an elementary column operation to illustrate the process. For example, suppose we want to interchange the first and second columns of A, a 3 × 2 matrix. To create the elementary column operator E, we interchange the first and second columns of the identity matrix I2. 10 01 ~ 01 10 I2 E Then, to interchange the first and second columns of A, we postmultiply A by E, as shown below. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 119 C1 <--> C2 = 01 01 23 10 45 E A C1 <--> C2 = 0*0 + 1*1 0*1 + 1*0 2*0 + 3*1 2*1 + 3*0 4*0 + 5*1 4*1 + 5*0 10 C1 <--> C2 = 3 2 54 Note that the process for performing an elementary column operation on an r × c matrix is very similar to the process for performing an elementary row operation. The main differences are:  To operate on the r × c matrix A, the row operator E is created from an r × r identity matrix; whereas the column operator E is created from an c × c identity matrix.  To perform a row operation, A is premultiplied by E; whereas to perform a column operation, A is postmultiplied by E Problem 1: Assume that A is a 4 × 3 matrix. Suppose you want to multiply each element in the second column of matrix A by 9. Find the elementary column operator E. Solution: To find the elementary column operator E, we multiply each element in the second column of the identity matrix I3 by 9. 10 0 100 010~ 090 00 1 001 I3 E CU IDOL SELF LEARNING MATERIAL (SLM)

120 Mathematics 10.7 Normal Form or Canonical Form Ir 0 Let A be a matrix of m × n order with rank r. The matrix obtained in the form of    0 0 after applying a finite chain of E-operations in referred as normal form of Matrix A. Here, Ir is the rowed Identity matrix other normal forms are: Ir  or Ir : 0  OR I0r 00 I0r  Remark To reduce the matrix to normal form, apply row as well as column transformations. 1. First try to convert left and top element I and other first column element zero. 2. Try to convert diagonal element 1 non-diagonal element zero. 3. Continue the process till matrix reduces to normal form. 1 4 3 42 Exercise 8: Find the rank of matrix by reducing it to normal where A = 1 2 3 5. 2 6 7 Solution: We have 1 4 3 2 A = 1 2 3 4 2 6 7 5  By applying R2  R2 – R1 and R3  R3 – 2R1, we get 1 4 3 2  0  2 0 2 0  2 1 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 121  Matrices 3 By applying c2  c2 + 4c1, c3  c3 – 3c1 and c4  c4 – 2c1, we get 1 0 0 0  0  2 0 2 0  2 1 1  Now, applying c  1 c , we get 2 22 1 0 0 0  0 1 0 2 0 1 1 1 Applying c2  c2 – c3, we get 1 0 0 0  0 1 0 2 0 0 1 1 Applying c4  c4 – 2c2 , – 1 × c2, we get 1 0 0 0  0 1 0 0 0 0 1 1 Applying c4  c4 – c3, we get 1 0 0 0  0 1 0 0 0 0 1 0  [I3 : 0]  Hence, (A) = 3 Exercise 9: Reduce the following matrix to normal form and find its rank. CU IDOL SELF LEARNING MATERIAL (SLM)

122 Mathematics  6 1 3 8  6 1 A=  4 2 10 39 7  4 12 15 16 6 1 3 18  6  4 2  Solution: Given, A= 10 7  15 16 39 4 12 Applying c1  c2, we get  1 6 3 8 2 6 1  3 4 9 7 10  15 4 16 12  By applying c2  c2 – 6c1, c3  c3 – 3c1 and c4  c4 – 8c3, we get  1 0 0 0  2 8 0 17 3  8 0 17  17 4 8 0  Applying  1  c and  1  c , we get    2  4  8   17    1 0 0 0  2 1 0 1 3 1 0 1   4 1 0 1 Applying R4  R4 – R2, R4  R4 – 2R1, R3  R3 – R2 and R3  R3 – R1, we get  1 0 0 0  1 0 1 2 0 0 0  0   0 0 0 0 CU IDOL SELF LEARNING MATERIAL (SLM)

 123  Matrices 3 Applying c1  c1 – 2c2 and c4  c4 – c2, we get 1 0 0 0  0 0 0 1 0 0  0 0   0 0 0 0  [I2 : 0] is in the Normal form   (A) = 2  1 2 1 0 Exercise 10: Reduce the matrix A =  2 4 3 0 to canonical form (Normal) and find its  1 0 2 8 rank. Solution: We have,   1 2 1 0 A =  2 4 3 0  1 0 2 8  Applying C2  C2 – 2C1 and C3  C3 – C1, we get  1 0 0 0   2 8 5 0  1  2 1  8  Applying R2  R2 + 2R1 and R3  R3 – R1, we get 1 0 0 0  0 8 5 0 0  2 1  8  Applying C  1 C , we get 2 82 CU IDOL SELF LEARNING MATERIAL (SLM)

124 Mathematics   1 0 0 0  0  1 5 0 0 1 1  8  4  Applying C3  C3 – 5C2, we get   1 0 0 0  0 1 0 0 1 9 8  4 0   4  Applying R3  4R3, we get 1 0 0 0  0 1 0 0 0 1 9  32  Applying R3  R3 + R2, we get 1 0 0 0  0 1 0 0 0 0 9  32  1 Applying C3 × , we get 9 1 0 0 0  0 1 0 0 0 0 1  32 Applying C4  C4 + 32C3, we get 1 0 0 0  0 1 0 0 0 0 1 0 [I3 : 0] = (A) = 3 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 125 10.8 PAQ Normal Form  Ir 0 If A be m × n, matrix of rank r can be reduced to  0 from by E-operations. Since, 0 E-operations are equivalent to fore or post multiplication by the corresponding elementary matrices, therefore, E-matrix. Ir 0 PAQ =    0 0 where, P = pk, pk – 1 … p1 Q = Q1, Q2 … Qj Thus, we concluded that, for any matrix A of m × n order with rank r, there exist non- singular matrices P and Q such that PAQ = Ir 0 0 0.  To find P and Q: 1. We write A = Im A In, m = rows, n = columns unit matrix of order m × n. 2. We can Apply row operation to A on left and column operation to A on right. 3. When A to the left reduces to normal form, we have A = PAQ. Exercise 11: Find non-singular matrices P and such that PAQ is in the Normal form. 1 1 2 A = 1 2 3 0  1  1 1 1 2 Solution: Given, A = 1 2 3 0  1  1  We can write A = I3 A I3 CU IDOL SELF LEARNING MATERIAL (SLM)

126 Mathematics 11 1 23 = 01 0 00 A 01 0 00 2 1 1       0  1  1 0 0 1 0 0 1  Applying C2  C2 – C1 and C3  C3 – 2C1, we get 11 0 01 = 10 0 00 A 01 1  02 1 1 1       0  1  1 0 0 1 0 0 1  Applying R2  R2 – R1, we get 11 0 01 = 11 0 00 A 01 1  02 1 1 1       0  1  1  0 0 1 0 0 1  Applying C3  C3 – C2, we get 1 0 0  1 0 0 1 1 1  1 1 0 = 1 1 0 A 0 1 1  0 1 0  0 0 1 0 0 1  Applying R3  R3 + R2, we get 1 0 0  1 0 0 1 1 1 1 1 0 = 1 1 0 A 0 1 1 0 0 0  1 1 1 0 0 1  Thus, we have required form, where  1 0 0 1 1 1 P = 1 1 0 and Q = 0 1 1  1 1 1 0 0 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 127  Matrices 3 Exercise 12: Find the non-singular matrix P and Q such that PAQ is in Normal form. 1 1 1 A = 1 1 1 3 1 1 1 1 1 Solution: Given, A = 1 1 1 3 1 1  We can write, A = I3 A I3 1 1 1 1 0 0 1 0 0 1 1 1 = 0 1 0 A 0 1 0 3 1 1 0 0 1 0 0 1  Applying C2  C2 – C1, C3  C3 – C1, we get 1 0 0 1 0 0 1 1 1 1  2  2 = 0 1 0 A 0 1 0 3  2  2 0 0 1 0 0 1  Applying R2  R2 – R1, R3  R3 – 3R1, we get 01 0  02 = 11 0 00 A 01 1 10 2 1 1       0  2  2  0 0 1 0 0 1  Applying  1  C ,  1  C , we get   2   3  2   2    01 0 0 00 A  1  1  12 1 01 = 11 1 0  12      2 0 0  0 12 1 1  0 0 1 0 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  128 Applying C3  C3 – C2, we get 01 0 00 = 11 0 00 01 1 0   1   1   12 1  A 2 2  0 1 0  0 0 1 0 0  12  Applying R3 – R2, we get 01 0 00 = 11 0 00 A 1 1 0 1 0 0  12 1      2 2  0 0 0  1 1 1  0  0  1 2   I2 0 PAQ =  0 = Normal form 0  1 0 0 1 1 0 0 2  P = 1 0 and Q =   1 1 0  1  1122 2  1 1 0      10.9 Square Matrix A matrix whose number of rows is equal to number of columns, i.e., m = n is called square matrix.   c a b a b e.g., 1. c  , 2. d e f  , 3. 111  d 22 g h i3 3 10.10 Summary One more type of matrix is explained as Orthogonal Matrix - is a square matrix with real entries whose columns and rows are orthogonal unit vectors (that CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 129 is, orthonormal vectors). Equivalently, a matrix A is orthogonal if its transpose is equal to its inverse. Minor of a Matrix The minor is defined as a value obtained from the determinant of a square matrix by deleting out a row and a column corresponding to the element of a matrix. The following are the steps to calculate minor from a matrix: 1. Hide ith row and jth column one by one from given matrix, where i refer to m and j refers to n that is the total number of rows and columns in matrices. 2. Evaluate the value of the determinant of the matrix made after hiding a row and a column from Step 1. And are calculated as under for Minor of 3×3 Matrix Minor of 2x2 Matrix Each minor of A of order greater than r is zero, i.e., each minor of order (r + 1) or higher is zero or vanishes, we generally denoted rank of matrix by (A). The three kinds of elementary matrix operations are : 1. Interchange two rows (or columns). 2. Multiply each element in a row (or column) by a non-zero number. 3. Multiply a row (or column) by a non-zero number and add the result to another row (or column). 10.11 Key Words/Abbreviations  Types of problems in normal form of matrix 1. No. of equation = No. of unknowns 2. No. of equation > No. of unknowns 3. No. of equation < No. of unknowns  p(A) represents rank of matrix A  To find P and Q: We write A = Im A In, m = rows, n = columns unit matrix of order m×n.  In PAQ form, A be m×n, matrix of rank r can be reduced toE- Ir 0  0 from by 0 operations. CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  130 10.12 Learning Activity  1. Find the rank of given matrix A  5 4 . 3    1     2. Find the rank of given matrix B  3 6 3 1 2 1.      10.13 Unit End Questions (MCQ and Descriptive) A. Descriptive Type Questions Exercise 4 7 1 8 then find: (i) 3A + 2B and (ii) 2A – 4B. 1. If A = 1 9 and B = 7 2, 6 8 4 3 , then find: (i) A + 4B, (ii) 2A + B and (iii) A – 2B. 2. If A = and B =    7 1 3 4 3.  1  3 and B =  1 0 1 , then verify (AB) = BA. If A =  2 1 2 1 3 4. Express the matrix A as the sum of symmetric and skew symmetric, when 1 7 1 A = 2 3 4.  5 0 5 1 1 that A2 – 4A + 5I = 0. 5. Given that A = 2  show 3 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 3 131  1 1  1   3  62 2 1 6. Show that given matrix is orthogonal, when A =  3 6 0 .  1   1 1  3 6 2  7. Reduce the following matrix to the Normal form find its rank.  8 1 3 6 1 1 1 (a)  0 3 2 2 (b) 1 1 1    8 1  3 4  3 1 1    3 1 1 3   1 4 6 1 (c) 7 1 1 1 1 1  11 6  (d) 1 1 2 1 7 2 12 1 3 1 0 1 Ans: (a) (A) = 3 (b) (A) = 3 (c) (A) = 2 8. Find rank of the given below. 5 4 1 3  (b) 0 1 6 (c) 1 5 4 6 3 (a) 3 1 0 1 2 2 1 Ans: (a) (A) = 2 (b) (A) = 3 (c) (A) = 1 2 1 4 9. If A = 2  4 4, find P and Q. Two non-singular matrices such that PAQ = I when 9 7  5 I is unit matrix. 10. Find two non-singular matrix P and Q so that PAQ is in Normal form 3 2 1 5 A = 5 1 4  2. 1  4 11  19 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  132  4  9 9  1 Ans: P = 00 0  51 and Q = 0 121   371 1 177    21 7 7   1  2 / 3 1/3 0 0 1 0   0 0 0 1  B. Multiple Choice/Objective Type Questions 1. A square matrix whose determinant is zero is called a . (a) Triangular (b) Singular (c) Non-singular (d) Zero 2. The inverse of a square matrix can be found out only if it is . (a) Singular (b) Non-singular (c) Zero (d) Identity 3. The square matrix whose non-diagonal element are all zero and the diagonal element are all . (a) Scalar (b) Square (c) Diagonal (d) None of these 4. Which of the following is square matrix? (a) 2 1  2 0 2 (b) 1 0    3 1 (d) 0 2 (d) None of the above 6 2 5. The matrix of order 3 × 4 has rows and columns. (a) 4, 3 (b) 3, 4 (c) 2, 1 (d) None of these Answers: 1. (b), 2. (b), 3. (a), 4. (a), 5. (b) 10.14 References References of this unit have been given at the end of the book. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 133 UNIT 11 MATRICES 4 Structure: 11.0 Learning Objectives 11.1 Introduction 11.2 Transpose of a Matrix 11.3 Orthogonal Matrix 11.4 Minor and Co-factors of a Matrix 11.5 Adjoint of a Square Matrix 11.6 Inverse of a Matrix 11.7 Eigen Values and Eigen Vectors 11.7.1 Vectors 11.7.2 Vector Space 11.7.3 Linear Combination of Vectors 11.7.4 Inner Product of Two Vectors 11.7.5 Concept of Eigen Values and Eigen Vectors 11.7.6 Method of Finding Eigen Values 11.7.7 Cayley-Hamilton Theorem 11.7.8 Transpose of a Matrix 11.7.9 Symmetric Matrix 11.7.10 Skew Symmetric Matrix CU IDOL SELF LEARNING MATERIAL (SLM)

134 Mathematics 11.7.11 Hermitian Matrix 11.7.12 Skew Hermitian Matrix 11.7.13 Unitary Matrix 11.7.14 Orthogonal Matrix 11.7.15 Diagonalizable Matrix 11.7.16 Minimal Polynomial 11.8 Summary 11.9 Key Words/Abbreviations 11.10 Learning Activity 11.11 Unit End Questions (MCQ and Descriptive) 11.12 References 11.0 Learning Objectives After studying this unit, you will be able to:  Describe the unitary matrix.  Discuss about the minimum polynomial symmetric matrix. 11.1 Introduction We will deal with problems related the Unitary matrix, Orthogonal matrix, Hermitian and Symmetric matrix. We will also come across concept of eigen values, eigen vectors and Cayley-Hamilton Theorem. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 135 11.2 Transpose of a Matrix The transpose of one matrix is another matrix that is obtained by using by using rows from the first matrix as columns in the second matrix. For example, it is easy to see that the transpose of matrix A is A'. Row 1 of matrix A becomes column 1 of A'; row 2 of A becomes column 2 of A',; and row 3 of A becomes column 3 of A'. 111 222  A = 333 444 111 333 555 555 666 A' = 222 444 666  Note that the order of a matrix is reversed after it has been transposed. Matrix A is a 2 x 3 matrix, but matrix A' is a 3 x 2 matrix. With respect to notation, this website uses a prime to indicate a transpose. Thus, the transpose of matrix B would be written as B'. 11.3 Orthogonal Matrix We know that a square matrix has an equal number of rows and columns. A square matrix with real numbers or elements is said to be an orthogonal matrix, if its transpose is equal to its inverse matrix or we can say, when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as orthogonal matrix. Suppose A is a square matrix with real elements and of n x n order and AT is the transpose of A.Then according to the definition, if, AT = A-1 is satisfied, then, A AT = I Where ‘I' is the identity matrix, A-1 is the inverse of matrix A and ‘n' denotes the number of rows and columns. For a orthogonal matrix if A.A = I, then matrix A is called orthogonal matrix. CU IDOL SELF LEARNING MATERIAL (SLM)

136 Mathematics Exercise 1: Check given matrix are orthogonal cos   sin  0 A = sin  cos  0  0 0 1   cos   sin  0 Solution: Given, A = sin  cos  0  0 0 1   cos  sin  0 A =  sin  cos  0  0 0 1  For orthogonal matrix, A.A = I cos   sin  0  cos  sin  0 cos  0  sin  cos  sin  0 1  0 0 0  0 1  cos2   sin 2   0 cos sin   sin  cos  0  0  0  = sin  cos   cos sin   0 sin 2   cos2   0 0  0  0 0  0  0 0  0  1  0 0 0  1 0 0 = 0 1 0 = I 0 0 1  Hence, A.A = I given matrix is orthogonal Exercise 2: Express the matrix A as the sum of a symmetric and a skew-symmetric matrix when  4 2  3 A=  1 3  6   5 0  7 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 4 137 Solution:   4 2  3   4 1  5 Given: A =  1 3  6 and A =  1 3 0    5 0  7  3  6  7   8 3  8  6  6 A + A =  3  8  6  4   0 1 2 A – A = 1 0  6  2 6 0  A = 1 (A + A) + 1 (A – A) 22  8 3  8  0 1 2 1   1  1 = 3 6  6 + 2  0  6 2  8 6 6 0  4  2    11.4 Minor and Co-factors of a Matrix The minor of a square matrix is the determinant obtained by deleting the rows and columns which intersect in that element. The co-factor of any element is it's minor with (–1)i+ j sign. 11.5 Adjoint of a Square Matrix The adjoint of a matrix A is the transpose of the matrix formed by the co-factor of matrix A. It is denoted by Adj A. CU IDOL SELF LEARNING MATERIAL (SLM)

138 Mathematics a11 a12 a13   Matrix of co-factor = a 21 a22 a 23  a 31 a32 a33   a11 a21 a31  Adj A = a 12 a22 a32  a13 a23 a33    11.6 Inverse of a Matrix If A be any non-singular square matrix and non-singular matrix B such that AB = I, then B is called Inverse of A such that denoted by A–1. Since, AB = A B = I = 0  A  0 and B  0 Hence, we can find A–1 by using Adjoint Method.  A–1 = Adj.A A Exercise 3: Find Inverse of the matrix by adjoint method.  1 2  2 A = 1 3 0  0 2 1    1 2  2 Solution: Given, A =  1 3 0  0 2 1   Step I: We have show A  0, then only A–1 exits. A = 1 (3 – 0) – 2 (–1 – 0) – 2(2 – 0) =3+2–4=5–4=1 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 139 A = 1  0, Hence, A–1 Exists. Step II: Co-factors of A 3 0 =3 a11 = +  2 1 a12 = – 1 0 = –(–1) = 1 0 1 a13 = + 1 3 =2 0 2 2  2 = – (2 – 4) = – (–2) = 2 a21 = –  2 1 a =+ 1 2 =1 22 0 1 1 2 = – (–2 – 0) = 2 a23 = – 0 2 a = + 2  2 = 0 – (–6) = 6 31 3 0 a = – 1  2 = – (0 –2) = 2 32 1 0 12 = 3+2=5 a33 = + 1 3 3 1 2 Co-factor of A = 2 1 2 6 2 5 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  140 3 2 6 Step III: Adjoint of A = 1 1 2 2 2 5  Adjoint of A = (Co-factor of A)   Adj.A 1 31 2 6 A 1 1 2 A–1 = = 2 5 2   3 2 6  A–1 = 1 1 2 2 2 5  Exercise 4: Find inverse of A, by using Adjoint Method. 1 1 1  A = 1 2  3 2 1 3   1 1 1 Solution: Given: A = 1 2  3 2 1 3  Step I:  A = 1(6 – 3) – 1(3 + 6) + (–1 – 4) = 3 – 9 – 5 = –11 A = – 11  0. Hence, A–1 exists. Step II: Co-factor of A 2 3 a11 = + 1 =3 3 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 141 a = – 1 3 = –9 2 3 12 1 2 =–5 a13 = + 2 1 a21 = – 1 1 =–4 1 3 11 a22 = + 2 =1 3 a23 = – 1 1 =3 2 1 a31 = + 1 1 = –5 2 3 a32 = – 1 1 1 =4 3 11 a33 = 1 2 = 1  3  9  5 Co-factor of A =  4 1 3  5 4 1 Step III:   3  4  5 Adjoint of A =  9 1 4  5 3 1 Now, A–1 = Adj of A A  3  4  5 Hence, A–1 = 1  9 1 4 11  5 3 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  142 11.7 Eigen Values and Eigen Vectors 11.7.1 Vectors The row or column matrix elements will form a vector. Thus, any n numbers x1, x2….., xn written in a particular order constitute a vector x. 11.7.2 Vector Space The set of all n-vector of a field to be denoted by Vn(F) is called n-vector space over F. The elements of field F are called scalars. If x is a vector and K is an constant, then x implies multiplication of constant K to each element in x. e.g., x = 32and k = 3   5   32  6     Then, Kx = 3 =  9  5 15  11.7.3 Linear Combination of Vectors If a vector x1 is written in the form x1 = 2x2 + 3x3 + ….. + 1x1, then, the vector x1 is said to be linear combination of the vectors x2, x3, ….. x1. 11.7.4 Inner Product of Two Vectors Let p and q are the two vectors p = p1 i + p2 j + p3 k and q = q1 i + q2 j + q3 k . Now, p and q in matrix form, p = pp1  and q = qq1   2  2  p3  q3   Case of Real Vectors If p and q are real vectors, then inner product of these two vector is CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 143 p.q = p1q1 + p2q2 + p3q3 Also, p.q = q.p Case of Complex Vectors If p and q complex vectors, then the inner product of these vector is pq.   pq = p p p qq1  = p q + p q + p q 1 2 3  2  1 1 2 2 3 3 q3  Now,   qp = q q q pp1  = q p + q p + q p 1 2 3  2  1 1 2 2 3 3 p3   Note that pq  qp Exercise 5: Find the inner product of the given function a = [i + 1 –2 – i 3], b = [2 + 2i –1 + i 3i] Solution: a = [i + 1 –2 – i 3], a = [1– i –2 + i 3]  1 i  a = 2  i  3   b = [2 + 2i –1 + i 3i] , b = [2 – 2i –1 – i –3i] 2  2i –1 + i 3i] b = 1  i  3i   Inner product ab,   1 i    ab = 2  i [2 + 2i  3  CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  144 = (1 – i)(2 + 2i) + (–1 + i)(–2 + i) + 3(3i) = 2 + 2i – 2i – 2i2 + 2 – i – 2i + i2 + 9i  ab = 5 + 6i 2  2i Now, ba = 1  i[i+1 –2–i 3]  3i   = (i + 1)(2 – 2i) + (–2 – i)(–1 – i) + 3(–3i) = 5 – 6i  ab  ba 11.7.5 Concept of Eigen Values and Eigen Vectors Turn to matrices in particular. The eigen value idea not restricted to matrices, but is of wider application. The most general formulation is to write, Fy = y where, F is some sort of operator which is applied to y and  is a constant. The important point is that after operating on y with F, the result is proportional to y. The number  is called the eigen value and y is the eigen vector or eigen function. Characteristic Polynomial of A D(  ) = A   = a  n + a  n–1 + a  n–1 + a  n–2 + …..+ a is known as characteristic 01 2 2 n polynomial of matrix A. Characteristic Equation of A D()= A  I =0 i.e., a n + a n–1 + a n–2 + …..+ a = 0 is called characteristic of A. 01 2 n Obviously, the degree of the characteristic equation of a matrix is equal to the order of the matrix. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 145 Characteristic Roots or Eigen Values D()= A  I = ( – 1)( – 2)…( – n) = 0 where,  =  1,  2,  3, … n are called characteristic root's or eigen values or proper values or latent values of the matrix A. 11.7.6 Method of Finding Eigen Values  For a square matrix of order 2  2, say A = a11   a21 a12  is the characteristic equation of A is a 22   A  I = 0, i.e., a11   a12 = 0 a 21 a 22    2 – x 1 + A = 0 x1 = sum of minor's of order one along main diagonal A. x = a + a , Also A = a11 a12  = a a – a a  1 11 22 a 21 a  11 22 12 21  22 For a square matrix of order 3  3 say a11 a12 a13  A = a 21 a22 a 23  a 31 a32 a33   The characteristic equation of A is A   = 0 a11   a12 a13 i.e. a 21 a 22   a 23 = 0 a31 a32 a33    Important Step: 3 – x 2 + x  – A = 0 12 CU IDOL SELF LEARNING MATERIAL (SLM)