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Chapter – 1 Matrices and Its Applications 1.1 Introduction Matrix: A system of mn numbers arranged in a rectangular formation along m rows and n columns and bounded by the brackets [] is called an m by n matrix, written as m n . A matrix is a single entity with a number of components rather than collection of numbers. Unlike a determinant, a matrix cannot reduce to a single number and the question of finding value of matrix never arises. Interchange of rows and columns does not alter the determinant but gives an entirely different matrix. 1.2 Elementary Transformations(or operations) The following operations on a matrix are called elementary transformations: 1. Interchange of any two rows or columns. 2. Multiplication of any row (column) by a non zero number. 3. The addition of k times the multiple of the elements of any row(column) to the corresponding elements of any other row(column) Elementary Transformations do not change either the order or rank of a matrix. Equivalent Matrices: Two matrices A and B are said to be Equivalent if one can be obtained from the other by a sequence of elementary transformations. Two equivalent matrices have the same order & rank and is written as A ~ B. Elementary Matrices: Matrix obtained from a unit matrix I by subjecting it to one of the E-operations is called an elementary matrix. 1 0 0 Example1: Let I = 0 1 0 0 0 1 1 0 0 Operating R23 on I we get the elementary matrix 0 0 1 . It is denoted by E23. 0 1 0 Theorems 1. Any E-Row transformation on the product of two matrices is equivalent to the same E-Row operation on the pre-factor i.e R(AB) = R(A).B 2. Any E-Column transformation on the product of two matrices is equivalent to the same E-Column operation on the post-factor i.e C(AB) = A.C(B) 3. Every E- Row operation on a matrix is equivalent to pre-multiplication by the corresponding Elementary Matrix i.e. R(A) = R(E).A 4. Every E- Column operation on a matrix is equivalent to post-multiplication by the corresponding Elementary Matrix i.e. C(A) = A.C(E) 1 1.3 Inverse of a Matrix by E-Operations(Gauss-Jordan method) The Elementary row transformations which reduce a square matrix A to the unit matrix, when applied to the unit matrix, give the inverse matrix A-1. To find the inverse of A by E-Row operations, we write A and I side by side and the same operations are performed on both. As soon as A is reduced to I, I will get reduce to A-1. 1 1 3 Example 2: Find the inverse of the matrix A = 1 3 3 2 4 4 Solution: Writing the given matrix side by side with the unit matrix I3, we get 1 1 31 0 0 1 [A:I3] = 3 3 0 1 0 2 4 4 0 0 1 Operating R2 –R1, R3 + 2R1 1 1 3 1 0 0 ~ 0 2 6 1 1 0 0 2 2 2 0 1 Operating R2/2, R3/2 1 1 3 1 0 0 0 1 3 1 / 2 1 / 2 0 ~ 0 1 11 0 1 / 2 Operating R3 + R2 1 1 3 1 0 0 ~ 0 1 3 1 / 2 1 / 2 0 0 0 2 1 / 2 1 / 2 1 / 2 Operating R3/-2 1 1 3 1 0 0 0 1 3 1 / 2 1 / 2 ~ 0 0 0 1 1 / 4 1 / 4 1 / 4 Operating R1 - 3 R3, R2 + 3 R3 2 1 1 0 7/4 3/ 4 3/ 4 ~ 0 1 0 5 / 4 1 / 4 3 / 4 0 0 1 1/ 4 1/ 4 1/ 4 Operating R1-R2 1 0 0 3 1 3/ 2 0 1 ~ 0 5 / 4 1 / 4 3 / 4 0 0 1 1/ 4 1/ 4 1/ 4 3 1 3/ 2 -1 Therefore A = 5 / 4 1 / 4 3 / 4 1 / 4 1 / 4 1 / 4 1.4 Rank of a Matrix Let A be m n matrix. It has square sub matrices of different orders. The determinants of these square sub- matrices are called minors of A. If all minors of order (r + 1) are zero but there is at least one non-zero minor of order r, then r is called rank of A.Symbolically rank of A is written as ( A) . 1 2 3 For example: The rank of A = 2 3 4 is 2 3 5 7 1 2 Since 1 0 while A 0 . 2 3 As rank is the largest order of a non-vanishing minor of the matrix. In the example above minor of order 3 is zero and there exists one minor of order 2 which is non-zero. From the definition of rank of a matrix it is clear that: 1. Rank of a null matrix is zero. 2. Rank of a non null matrix 1 3. If A is a non singular n n matrix then ( A) n 4. If A is m n matrix, then ( A) minimum of m and n. This method usually involves a lot of computational work since we have to evaluate several determinants. Normal Form If A is an m n matrix and by series of elementary (row or column or both) transformations, it can be put into one of the following forms (called normal forms): 3 I r 0 I r , 0 0 , I r 0 , I r where Ir is the unit matrix of order r. 0 Since the rank of a matrix is not changed as a result of elementary transformations, it follows that ( A) = r. [ rth order minor I r 1 0 ] Cor.1 Since each Elementary transformation can be affected by pre-multiplication or post-multiplication with suitable Elementary matrix and each Elementary matrix is non- singular, therefore we have the result: Corresponding to every matrix A of rank r non singular matrices P and Q such that PAQ is in the normal form. Method to find matrices P and Q 1. Write A = IAI 2. Reduce matrix on LHS to normal form by E-Row/E-Column transformations. 3. Every E-row transformation on A must be accompanied by the same transformation on the pre factor on RHS. 4. Every E-column transformation on A must be accompanied by the same transformation on the post factor on RHS. Remarks: 1. Matrices P and Q are not unique. 2. A-1 = QP. Example 3: Reduce the following matrix to normal form and hence find its rank 8 1 3 6 A= 0 3 2 2 8 1 3 4 Operate R3 + R1 8 1 3 6 ~ 0 3 2 2 0 0 0 10 Operate R4/10 8 1 3 6 ~ 0 3 2 2 0 0 0 1 Operate C1/8 1 1 3 6 ~ 0 3 2 2 0 0 0 1 Operate R2 – 2R3, R1 – 6R3 4 1 1 3 0 ~ 0 3 2 0 0 0 0 1 Operate C3 – 3C1 1 1 0 0 ~ 0 3 2 0 0 0 0 1 Operate C2-C3 1 1 0 0 ~ 0 1 2 0 0 0 0 1 Operate C3/2 1 1 0 0 ~ 0 1 1 0 0 0 0 1 Operate C2 – C3 1 1 0 0 ~ 0 0 1 0 0 0 0 1 Operate C1 – C3 0 1 0 0 ~ 0 0 1 0 0 0 0 1 = 0 I 3 Therefore ( A) =3 Example 4: Find Non-Singular matrices P & Q such that PAQ is in the normal form for 1 1 1 the matrix A= 1 1 1 3 1 1 Solution: We write A = IAI 1 1 1 1 0 0 1 0 0 i.e. 1 1 1 = 0 1 0 A 0 1 0 3 1 1 0 0 1 0 0 1 We shall affect every elementary row (column) transformation of the product by subjecting the pre factor (post factor) of A to the same. Operate C2 + C1, C3 + C1 5 1 0 0 1 0 0 1 1 1 ~ 1 2 2 = 0 1 0 A 0 1 0 3 4 4 0 0 1 0 0 1 Operate R2 – R1, R3-3R1 1 0 0 1 0 0 1 1 1 ~ 0 2 2 = 1 1 0 A 0 1 0 0 4 4 3 0 1 0 0 1 Operate R3-2R2 1 0 0 1 0 0 1 1 1 ~ 0 2 2 = 1 1 0 A 0 1 0 0 0 0 1 2 1 0 0 1 Operate C3-C2 1 0 0 1 0 0 1 1 0 ~ 0 2 0 = 1 1 0 A 0 1 1 0 0 0 1 2 1 0 0 1 Operate C3/2 1 0 0 1 0 0 1 1 0 ~0 1 0 = 1 1 0 A 0 1 / 2 1 / 2 0 0 0 1 2 1 0 0 1 I 2 : 0 1 0 0 1 1 0 ~ =PAQ where P= 1 1 0 and Q = 0 1 / 2 1 / 2 0 : 0 1 2 1 0 0 1 1.5 Solution of a System of Linear Equations Consider the system of equations a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3 (3 equations in 3 unknowns). In the matrix notation these equations can be written as a 1 x b1 y c1 z d 1 a 1 b1 c1 x d 1 a x b y c z = d or a b c y = d 2 2 2 2 2 2 2 2 a1x b3 y c3z d 3 a1 b3 c3 z d 3 or AX = B a 1 b1 c1 where A= a 2 b 2 c 2 is called the coefficient matrix. a1 b3 c3 6 x X= y is the column matrix of unknowns z d1 B= d 2 is the matrix of constants. d3 If matrix B = 0 i.e. AX = 0, then system of equations is called a system of Homogeneous linear equations. If B 0 i.e. AX = B, then the system of equations is called Non- Homogeneous linear equations. A system of equations having no solution is called an inconsistent system of equations. A system of equations having one or more solution is called a consistent system of equations For a system of non-homogeneous linear equations AX = B (i) if A B A , the system is consistent. (ii) If A B A = number of unknowns, the system has unique solution. (iii) If A B A (= r) < number of unknowns(n), the system has an infinite number of solutions. Number of parameters = n – r. The matrix A B in which the elements of A and B are written side by side is called the augmented matrix. For a system of homogeneous linear equations AX = 0 Homogeneous system is always consistent. (i) If A = number of unknowns, the system has only the trivial solution. (ii) If A < number of unknowns, the system has infinite number of non- trivial solution. Gauss Elimination Method Find the ranks of matrix A and matrix A B by reducing A to the triangular form by elementary row transformations. Example 5: Test for consistency and solve 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5. 5 3 7 x 4 Solution: We have 3 26 2 y = 9 7 2 10 z 5 Or A X = B 5 3 7 4 A B = 3 26 29 7 2 105 Reducing matrix A to upper triangle form by elementary row operations Operate R2 5R2 – 3R1, R3 5R3 - 7R1 7 5 3 7 4 0 1 21 11 33 ~ 0 11 1 3 Operate R3 3R3 + R2 5 3 7 4 0 1 21 11 33 ~ (i) 0 0 0 0 From above it is clear that A = 2 and A B = 2. Hence the equations are consistent. Number of unknowns (n) = 3 Rank of matrix(r) = 2 So r < n, System of equations has infinite solutions. Number of parameters = n – r =3–2=1 From (i) we have 5x + 3y + 7z = 4 121y – 11z = 33 Let z = k (parameter) then y = (33+11k)/121 = (3 + k)/11 And x = (7 – 16k)/11 Example 6: Solve the equations x + 2y + 3z = 0, 3x + 4y + 4z = 0, 7x + 10y + 12z = 0. 1 2 3 x 0 Solution: We have 3 4 4 y 0 7 10 12 z 0 Or A X = 0 (System of homogeneous equations) Now we find the rank of matrix A by reducing it to upper triangular form by Elementary row transformations 1 2 3 3 4 4 7 10 12 Operate R2 – 3R1, R3 – 7R1 1 2 3 ~ 0 2 5 0 4 9 Operate R3 – 2R2 1 2 3 ~ 0 2 5 0 0 1 8 From above it is clear that A = 3 and number of unknowns = 3 So r = n, the equations have only trivial solution: x = 0, y = 0, z = 0. 1.6 Vectors Any ordered n-tuple of numbers is called an n-vector. By an ordered n-tuple, we mean a set consisting of n numbers in which the place of each number is fixed. If x1, x2, …., xn be any n numbers then the ordered n-tuple X=( x1, x2, …., xn) is called an n-vector. Linear Dependence and Linear Independence of vectors A set of r n-vectors X1, X2, …., Xr is said to be linearly dependent if there exist r scalars k1,k2,…kr not all zero, such that k1X1 + k2X2 + …. + krXr = 0 i.e. atleast one member of the set can be expressed as a linear combination of the remaining vectors. A set of r n-vectors X1, X2, …., Xr is said to be linearly independent if every relation of the type k1X1 + k2X2 + ….+ krXr = 0 implies k1=k2=…=kr = 0. Example 7:Are the vectors x1 = (1,3,4,2), x2 = (3, -5,2, 2) and x3 = (2, -1, 3, 2) linearly dependent ? If so express one of these as a linear combination of the others. Solution: Consider the relation k1x1 + k2x2 + k3x3 = 0 i.e. k1 (1, 3,4 ,2) + k2 (3, -5,2, 2) + k3(2, -1, 3, 2) = 0 is equivalent to k1+3k2+2k3 = 0, 3k1-5k2 -k3 = 0, 4k1+2k2 +3k3 = 0, 2k1+2k2 +2k3 = 0 1 3 2 3 5 1 k1 0 or k 2 0 4 2 3 k 3 0 2 2 2 or AK = 0 Reducing matrix A to simpler form with the help of elementary row operations to calculate rank of matrix A Operate R2 – 3R1, R3 – 4R1, R4 – 2R1 1 3 2 0 14 7 ~ 0 10 5 0 4 2 Operate R2/-7, R3/-5, R4/-2 1 3 2 0 2 1 ~ 0 2 1 0 2 1 Operate R3 – R2, R4 – R2 9 1 3 2 0 2 1 ~ (i) 0 0 0 0 0 0 Rank of matrix A is 2. The system of equations is homogeneous and number of unknowns is 3 So n > r (3 > 2), Hence system of equations have infinite non-trivial solution. i.e. not all k1, k2,k3 are zero. Vectors k1, k2, k3 are linearly dependent. From matrix (i) we have k1 + 3k2 + 2k3 = 0 (ii) 2k2 + k3 = 0 (iii) From eq. (iii) k3 = -2k2 And from eq. (ii) k1 = k3/2 or 2k1 = k3 2k1 2k 2 k 3 k k k or 1 2 3 1 1 2 k1 = 1, k2 = -1 and k3 = 2. Putting values of k1, k2, k3 we have x1 – x2 + 2x3 = 0 or x1 + 2x3 = x2 (relation between the vectors). 1.7 Characteristic Equation If A is a matrix of order n, we can form the matrix A - I, where is a scalar and I is the unit matrix of order n. The determinant of this matrix equated to zero, i.e., A - I = 0 is called the characteristic equation of A. On expanding the determinant, the characteristic equation can be written as a polynomial equation of degree n in . The roots of this equation are called characteristic roots or latent values or Eigen values of A. Eigen vectors Consider the linear transformation Y = AX which transforms the column vector X into the column vector Y. In practice, we are required to find those vectors X which transform into scalar multiples of themselves. Let X be such a vector which transforms into X by the transformation Y = AX. Then Y = X AX = X ( A I ) X 0 These equations will have a non trivial solution only if the co-efficient matrix A - I is singular i.e. if A - I = 0 This is the characteristic equation of the matrix A and has n roots which are Eigen values of A. Corresponding to each root; the homogeneous system has a non-zero solution 10 x1 x 2 X = . which is called an Eigen-vector or Latent vector. .. xn The Eigen vector corresponding to an Eigen value is not unique. Because if X is a solution of ( A I ) X 0 , then so is kX, where k is an arbitrary constant. 1 1 3 Example 8:Find the Eigen Values and Eigen Vectors of the matrix 1 5 1 3 1 1 1 1 3 Solution: The characteristic equation is A I 1 5 1 3 1 1 i.e. 3 72 36 0 Since = -2 satisfies it, we can write this equation as ( 2) (2 9 18) 0 or ( 2) ( 3) ( 6) = 0 Thus the eigen values of A are = -2, 3, 6. If x, y, z be the components of an eigen vector corresponding to the eigen value , 1 1 3 x We have A I 1 5 1 y = 0 3 1 1 z Putting = -2, we have 3x + y + 3z = 0, x + 7y + z = 0, 3x + y + 3z = 0 The first and third equations being same, we have from the first two x y z x y z or 20 0 20 1 0 1 1 Hence the Eigen vector is 0 . Also every non- zero multiple of this vector is an eigen 1 vector to = -2. Similarly the eigen vectors corresponding to = 3 and = 6 are the arbitary non-zero 1 1 multiples of the vectors 1 and 2 which are obtained from(i). 1 1 1 1 1 Hence the three eigen vectors may be taken as 0 , 1 and 2 . 1 1 1 11 1.8 Cayley-Haimilton Theorem Every square matrix satisfies its own characteristic equation. 1 4 Example 9: Verify Cayley- Hamilton theorem for the matrix A = and find its 2 3 5 4 3 2 inverse. Also express A - 4A – 7A + 11A – A – 10I as the linear polynomial in A. Solution: The characteristic equation of A is 1 4 = 0 or 2 4 5 0 (i) 2 3 By Cayley- Hamilton theorem, A must satisfy its characteristic equation (i), so that A2 – 4A – 5I = 0 (ii) 1 4 1 4 1 4 1 0 Now A2 – 4A – 5I = - 4 - 2 3 2 3 2 3 0 1 9 16 4 16 5 0 0 0 = 8 17 8 12 0 5 0 0 This verifies the theorem. Multiplying (ii) by A-1, we get A – 4I – 5A-1 = 0 1 1 1 4 1 0 1 3 4 or A-1 = ( A 4 I ) 4 5 5 2 3 0 1 5 2 1 5 4 3 Now dividing the polynomial - 4 – 7 + 11 2 – – 10 by the polynomial 2 - 4 - 5 we obtain 5- 4 4 – 7 3 + 11 2 – – 10 = ( 2 - 4 - 5) ( 3 - 2 + 3) + + 5 = +5 Hence A5- 4A4 – 7A3 + 11A2 – A – 10I = A + 5 which is a linear polynomial in A. 12

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