Matrices Operations Linear Algebra Consider, for example, a system of equations such as x + 2y z + 4w = 0, 3x 4y + 2z 6w = 0, x 3y 2z + w = 0 The rectangular array 1 2 1 4 3 4 2 6 1 3 2 1 in which the numerical coefficients of the unknowns are displayed in the pattern of the equations, carries most of the essential information about the system Whether we call the unknowns x, y, z, and w or say, s, t, u, and v, is not very important Such a rectangular array is called a matrix Matrices may have numbers, functions, or other things in them Their important feature is that they provide a grid in which we can store objects in named locations (for example, row 1, column 2), and from which these objects can be recalled The following are some definitions of matrix: I An array of real numbers A = a 11 a 12 a 13 a 1n a 21 a 22 a 23 a 2n a m1 a m2 a m3 a mn is called an m n matrix with m rows and n columns (size or dimension) The a ij is referred to as the i, j-th element and denotes the element in the i-th row and j-th column If m = n then A is called a square matrix of order n If the matrix has one column or one row then it is called a column vector or a row vector respectively II In a square matrix A of order n, the diagonal containing the elements a 11, a 22,, a nn is called the principal or leading diagonal III A diagonal matrix is a square matrix that has its only non-zero elements along the leading diagonal A special case of a diagonal matrix is the unit or identity matrix I for which a 11 = a 22 = = a nn = 1 IV A zero or null matrix 0 is a matrix with every element zero V The transposed matrix A T is the matrix A with rows and columns interchanged, its i, j-th element being a ji VI A square matrix A is called a symmetric matrix if A T = A or a ij = a ji It is called skew symmetric if A T = A The following are the basic operations on matrices: In what follows the matrices A, B and C are assumed to have the i, j-th elements a ij, b ij and c ij respectively 1
I Equality The matrices A and B are equal, that is A = B, if they are of the same order m n and a ij = b ij, 1 i m, 1 j n II Multiplication by a scalar If λ is a scalar then the matrix λa has elements λa ij III Addition We can only add an m n matrix A to another m n matrix B and the elements of the sum A + B are a ij + b ij, 1 i m, 1 j n Properties of addition (i) commutative law: A + B = B + A (ii) associative law: (A + B) + C = A + (B + C) (iii) distributive law: λ (A + B) = λa + λb, where λ is a scalar IV Matrix multiplication If A is a m p matrix and B is a p n matrix, then we define the product C = AB as the m n matrix with elements c ij = p a ik b kj, i = 1, 2,,m; j = 1, 2,,n k=1 Properties of multiplication (i) The commutative law is not satisfied in general; that is, in general AB BA Order matters and we distinguish between AB and BA by the terminology: premultiplication of B by A to form AB and post-multiplication of B by A to form BA (ii) Associative law: A (BC) = (AB)C (iii) If λ is a scalar then (iv) Distributive law over addition: (λa)b = A (λb) = λ (AB) (A + B)C = AC + BC A (B + C) = AB + AC Note the importance of maintaining order of multiplication (v) If A is an m n matrix and if I m and I n are the identity matrices of order m and n respectively, then I m A = AI n = A V Properties of the transpose If A T is the transposed matrix of A then (i) (A + B) T = A T + B T (ii) ( A T) T = A 2
(iii) (AB) T = B T A T 1 Let A and B be 4 5 matrices and let C, D, and E be 5 2, 4 2, and 5 4 matrices, respectively Determine which of the following matrix expressions are defined For those that are defined, give the size of the resulting matrix BA AC + D (c) AE + B AB + B (e) E (A + B) (f) E (AC) 2 Find, where they exist, the matrices (i) A + B, (ii) A B, (iii) AB, (iv) BA, (v) (A + B) (A B) and (vi) A 2 B 2, if: 4 5 1 0 [ 2 6 A = 2 0 B = 0 7 1 2 3 A = B = 8 4 6 3 2 5 4 0 6 1 0 0 0 A = (c) A = [ 2 3 4 1 5 6 3 4 1 5 2 2 3 Consider the matrices Compute A = D = B = B = 3 0 1 2 1 1 1 2 3 1 5 2 1 0 1 3 2 4 [ 1 2 1 4 0 2 (e) A = [ 4 1 B = 0 2 E = B = 6 1 3 1 1 2 4 1 3 2 1 1 0 1 2 1 0 1 0 3 5 2 0 6 1 1 4 C = [ 1 4 2 3 1 5 AB DE (g) 3C D (j) A (BC) D + E (e) ED (h) (3E)D (k) (4B)C + 2B (c) D E (f) 7B (i) (AB)C (l) D + E 2 4 Find the transposes of 2 1 3 1 0 2 6 1 1 8 4 3 0 1 3 (c) [ 7 0 2 System of Linear Equations and Elementary Row Operations A system of equations that has no solution is said to be inconsistent If there is at least one solution, it is called consistent To illustrate the possibilities that can occur in solving systems of liner equations, consider a general system of two linear equations in the unknowns x and y: { l1 : a 1 x + b 1 y = c 1 (a 1,b 1 0) l 2 : a 2 x + b 2 y = c 2 (a 2,b 2 0) 3
The graphs of these equations are lines; call them l 1 and l 2 Since a point (x,y) lies on a line if and only if the numbers x and y satisfy the equation of the lines, the solutions of the system of equations will correspond to points of intersection of l 1 and l 2 There are three possibilities: I The lines l 1 and l 2 may intersect at only one point, in which case the system has exactly one solution or called unique solution II The lines l 1 and l 2 may coincide, in which case there are infinitely many points of intersection, and consequently infinitely many solutions to the system III The lines l 1 and l 2 may be parallel, in which case there is no intersection, and consequently no solution to the system y l 1 y l 1 and l 2 y l 2 l 1 x x x l 2 I II III Although we have considered only two equations with two unknowns here, we can generalize the results for any arbitrary systems, that is, every system of linear equations has either no solution, exactly one solution, or infinitely many solutions Consider the system of n equations in m unknowns: a 11 x 1 + a 12 x 2 + + a 1m x m = b 1, a 21 x 1 + a 22 x 2 + + a 2m x m = b 2, a n1 x 1 + a n2 x 2 + + a nm x m = b n Observe that the matrix equation a 11 a 12 a 1m a 21 a 22 a 2m a n1 a n2 a nm x 1 x 2 x m = becomes, after carrying out the product on the left, a 11 x 1 + a 12 x 2 + + a 1m x m a 21 x 1 + a 22 x 2 + + a 2m x m = a n1 x 1 + a n2 x 2 + + a nm x m By definition of equality of matrices (here, both sides are n 1 matrices), this is exactly the original system of equations Thus, the system can be written compactly as Ax = b, where A 4 b 1 b 2 b n b 1 b 2 b n
is the n m matrix of coefficients, called the coefficient matrix, having a ij as its i j entry, x 1 b 1 x 2 b 2 x = x m = [ T x 1 x 2 x m, and b = b n = [ T b 1 b 2 b n For the same system of equations, we define the augmented matrix as following: a 11 a 12 a 1m b 1 a 21 a 22 a 2m b 2 = [A b a n1 a n2 a nm b n The basic method for solving a system of linear equations is to replace the given system by a new system that has the same solution set, but is easier to solve This new system is generally obtained in a series of steps by applying the following three types of operations to systematically eliminate unknowns: I Interchange two equations II Multiply an equation through by a non-zero constant III Add a multiple of one equation to another Since the two rows (horizontal lines) in an augmented matrix correspond to the equations in the associated system, these three operations correspond to the following operations on the rows of the augmented matrix Denote r i be the i-th row in the augmented matrix and we have the following commands: I Interchange two rows, r i r j, where i j II Multiply a row through by a non-zero constant, tr i r i, where t 0 III Add a multiple of one row to another row, tr i + r j r j, where t 0 and i j These are called elementary row operations 1 Find the augmented matrix for each of the following systems of linear equations x 1 2x 2 = 0 x 1 + x 3 = 1 3x 1 + 4x 2 = 1 (c) 2x 2 x 3 + x 5 = 2 2x 1 x 2 = 3 2x 3 + x 4 = 3 { x1 + x 3 = 1 x 1 + 2x 2 x 3 = 3 2 Find a system of linear equations corresponding to each of the following augmented matrices 5
1 0 1 2 2 1 1 3 0 1 2 4 1 0 0 0 1 0 1 1 1 (c) [ 1 2 3 4 5 5 4 3 2 1 Gaussian Elimination A matrix that is in reduced row-echelon form must have the following properties I If a row does not consist entirely of zeros, then the first non-zero number in the row is a 1 (We call this a leading 1) II If there are any rows that consist entirely of zero, then they are grouped together at the bottom of the matrix III In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row IV Each column that contains a leading 1 has zeros everywhere else A matrix having properties I, II, and III is said to be in row-echelon form If, by a sequence of elementary row operations, the augmented matrix for a system of linear equations is put in reduced row-echelon form, then the solution set for the system can be obtained by inspection or, at worst, after a few simple steps Suppose that a system of linear equations is being reduced to row-echelon form, except the last row without a leading 1, with augmented matrix as follows, this method of solving system of equations is called the Gaussian elimination : Considering the following cases: 1 e f g 0 1 c d 0 0 a b I If a is zero, but b is non-zero, then the last row of the system implies 0x 1 + 0x 2 + 0x 3 = 0 = b 0, which is impossible, so the system is inconsistent, which means there is no solution for the system II If a is non-zero, then the system of reduced equations becomes x 1 + ex 2 + fx 3 = g (1) x 2 + cx 3 = d (2) ax 3 = b (3) From (3), we know that x 3 = b/a If we substitute x 3 = b/a into (2), then x 2 can be solved Similarly, x 1 can also be solved by substituting the known x 2 and x 3 The way for solving the reduced system of equations is called the backward substitution So, the system is consistent and has unique solution 6
III If both a and b are zeros, then the system becomes { x1 + ex 2 + fx 3 = g (1) x 2 + cx 3 = d (2) Let t = x 3, then x 2 = d ct and hence x 1 can be solved in terms of t t is a constant or called a free variable If an n equations in m unknowns system, where n < m, has p entirely zeros rows in the reduced augmented matrix, then there are m n+p number of free variables in the solution set So, the system is consistent and has infinity many solutions If a system is being reduced to reduced row-echelon form, again except the last row without a leading 1, with augmented matrix as follows, this is called the Gauss-Jordan method: 1 0 0 p 0 1 0 q 0 0 a b We can obtain the values of x 1 and x 2 directly from the first and second rows of the reduced augmented matrix, x 1 = p and x 2 = q The system is consistent or inconsistent is totally depending on the value of a 1 Which of the following matrices are in row-echelon form? 1 2 3 1 1 0 0 0 0 (c) 0 1 0 0 0 1 0 0 0 [ 1 7 5 5 0 1 3 2 1 3 0 2 0 1 0 2 2 0 0 0 0 0 1 0 0 0 0 0 (e) (f) 2 3 4 0 1 2 0 0 3 0 0 0 0 0 0 0 0 0 2 Which of the following matrices are in reduced row-echelon form? 1 0 0 0 0 0 0 0 1 1 1 0 (c) 0 1 0 0 0 0 (e) 0 1 0 1 0 0 0 0 0 1 2 0 3 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 3 Given a system of linear equations 2x 1 4x 2 + x 3 = 9 4x 1 + 2x 2 3x 3 = 17 x 1 x 2 + 5x 3 = 11 The augment matrix is given as: 2 4 1 9 4 2 3 17 1 1 5 11 (f) 1 0 0 5 0 0 1 3 0 1 0 4 [ 1 0 3 1 0 1 2 4 Perform the following sequences of elementary row operations one after the other: 7
i r 1 r 3 ; ii 2r 1 + r 3 r 3 ; iii 4r 1 + r 2 r 2 ; iv r 2 r 3 ; v 3r 2 + r 3 r 3 ; vi 1 50 r 3 r 3 ; vii 1 2 r 2 r 2 Solve the system by applying the backward-substitution process to the result of the last operation 4 In each part suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row-echelon form Solve the system, if it is consistent 1 2 4 2 0 1 2 1 0 0 1 2 1 0 4 7 10 0 1 3 4 2 0 0 1 1 2 (c) 1 5 4 0 7 5 0 0 1 1 7 3 0 0 0 1 4 2 0 0 0 0 0 0 1 2 2 2 0 1 3 3 0 0 0 1 5 In each part suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given reduced row-echelon form Solve the system, if it is consistent 1 0 0 4 0 1 0 3 0 0 1 2 1 0 0 3 2 0 1 0 1 4 0 0 1 1 2 (c) 1 5 0 0 5 1 0 0 1 0 3 1 0 0 0 1 4 2 0 0 0 0 0 0 1 2 0 0 0 0 1 0 0 0 0 1 6 Solve each of the following systems by (i) Gaussian elimination (ii) Gauss-Jordan method x 1 + x 2 + 2x 3 = 8 3x 1 + 2x 2 x 3 = 15 x 1 2x 2 + 3x 3 = 1 5x (e) 1 + 3x 2 + 2x 3 = 0 3x 1 7x 2 + 4x 3 = 10 3x 1 + x 2 + 3x 3 = 11 2x 1 + 2x 2 + 2x 3 = 0 11x 1 + 7x 2 = 30 2x 1 + 5x 2 + 2x 3 = 0 4x 1 8x 2 = 12 7x 1 + 7x 2 + x 3 = 0 (f) 3x 1 6x 2 = 9 x 1 x 2 + 2x 3 x 4 = 1 2x 1 + 4x 2 = 6 2x (c) 1 + x 2 2x 3 2x 4 = 2 { x 1 + 2x 2 4x 3 + x 4 = 1 5x1 + 2x (g) 2 + 6x 3 = 0 3x 1 3x 4 = 3 2x 1 + x 2 + 3x 3 = 0 2x 1 3x 2 = 2 x 1 2x 2 + x 3 4x 4 = 1 2x 1 + x 2 = 1 (h) x 1 + 3x 2 + 7x 3 + 2x 4 = 2 3x 1 + 2x 2 = 1 x 1 12x 2 11x 3 16x 4 = 5 7 For which value(s) of the constant k does the following system of linear equations have (i) no solution (ii) exactly one solution (iii) infinitely many solutions? 8
{ x1 x 2 = 3 2x 1 2x 2 = k x 1 + 2x 2 3x 3 = 4 3x 1 x 2 + 5x 3 = 2 4x 1 + x 2 + (k 2 14)x 3 = k + 2 8 Consider the system of equations x 1 + x 2 + 2x 3 = a x 1 + x 3 = b 2x 1 + x 2 + 3x 3 = c Show that in order for this system to be consistent, a, b, and c must satisfy c = a + b 9 Solve the system x 1 + 2x 2 + x 3 = b 1 x 1 x 2 + x 3 = b 2 x 1 + x 2 = b 3 when b 1 = 1, b 2 = 3, b 3 = 4 b 1 = 5, b 2 = 0, b 3 = 0 (c) b 1 = 1, b 2 = 1, b 3 = 3 b 1 = 1 2, b 2 = 3, b 3 = 1 7 10 What conditions must the b s satisfy in order for the given system to be consistent? x 1 x 2 + 3x 3 = b 1 3x 1 3x 2 + 9x 3 = b 2 2x 1 + 2x 2 6x 3 = b 3 11 Let a 0 b 2 a a 4 4 0 a 2 b 2x 1 + 3x 2 x 3 + x 4 = b 1 x 1 + 5x 2 + x 3 2x 4 = b 2 x 1 + 2x 2 + 2x 3 3x 4 = b 3 3x 1 + x 2 3x 3 + 4x 4 = b 4 be the augmented matrix for a linear system For what values of a and b does the system have a unique solution a one-parameter solution (c) a two-parameter solution no solution 12 Find a matrix K such that AKB = C given that 1 4 [ A = 2 3 2 0 0, B = 0 1 1 1 2 Systems of Homogeneous Equations, C = 8 6 6 6 1 1 4 0 0 A system of linear equations is said to be homogeneous if all the constant terms are zero, that is, the system has the form a 11 x 1 + a 12 x 2 + + a 1m x m = 0, a 21 x 1 + a 22 x 2 + + a 2m x m = 0, a n1 x 1 + a n2 x 2 + + a nm x m = 0 9
Every homogeneous system of linear equations is consistent, since x 1 = 0, x 2 = 0,, x m = 0 is always a solution This solution is called the trivial solution; if there are other solutions, they are called non-trivial solutions Since a homogeneous system of linear equations must be consistent, there is either one solution or infinitely many solutions Since one of these solutions is the trivial solution, we can make the following statement For a homogeneous system of linear equations, exactly one of the following is true I The system has only the trivial solution, which is the unique solution II The system has infinitely many solutions in addition to the trivial solution A homogeneous system of linear equations with more unknowns than equations always has infinitely many solutions 1 Without solving the systems, determine which of the following homogeneous systems have non-trivial solutions x 1 + 2x 2 + 3x 3 = 0 x 1 + 3x 2 + 5x 3 + x 4 = 0 x 2 + 4x 3 = 0 (c) 4x 1 7x 2 3x 3 x 4 = 0 5x 3 = 0 3x 1 + 2x 2 + 7x 3 + 8x 4 = 0 { x1 + x 2 = 0 2x 1 + 2x 2 = 0 2 Solve the given homogeneous system of linear equations 2x 1 + x 2 + 3x 3 = 0 2x 1 4x 2 + x 3 + x 4 = 0 x 1 + 2x 2 = 0 x 1 5x 2 + 2x 3 = 0 x 2 + x 3 = 0 (c) 2x 2 2x 3 x 4 = 0 x 1 + 3x 2 + x 4 = 0 x 1 2x 2 x 3 + x 4 = 0 { { 3x1 + x 2 + x 3 + x 4 = 0 x1 + 6x 2 2x 3 = 0 5x 1 x 2 + x 3 x 4 = 0 2x 1 4x 2 + x 3 = 0 3 For which value(s) of λ does the following system of equations have non-trivial solutions? { (λ 3)x1 + x 2 = 0 Non-singular Matrix x 1 + (λ 3) x 2 = 0 Let A be a square n n matrix We shall say that A is invertible or non-singular if there exists an n n matrix B such that AB = BA = I n Such a matrix B is uniquely determined by A and matrix B will be called the inverse of A and will be denoted by A 1 Thus AA 1 = I and A 1 A = I Consider the 2 2 matrix A = [ a b c d 10,
if ad bc 0, then A 1 = [ 1 ad bc d b c a since AA 1 = I 2 and A 1 A = I 2 (verify) In the following section we shall show how to find inverse of invertible matrices whose size are greater than 2 2 If A and B are non-singular matrices of the same size, then I AB is non-singular; II AB 1 = B 1 A 1 An n n matrix is called an elementary matrix if it can be obtained from the n n identity matrix I n by performing a single elementary row operation When a matrix A is multiplied on the left by an elementary matrix E the effect is to perform an elementary row operation on A This is the content of the following theorem, which we state without proof If the elementary matrix E results from performing a certain row operation on I m and if A is an m n matrix, then the product EA is the matrix that results when this same row operation is performed on A If an elementary row operation is applied to an identity matrix I to produce an elementary matrix E, then there is a second row operation which, when applied to E, produces I back again The various possibilities are list in the following table: Elementary row operation Row operation on I that produces E Row operation on E that reproduces I I r i r j Interchange rows i and j Interchange rows i and j II tr i r i Multiply row i by t 0 Multiply row i by 1/t III tr i + r j r j Add t times row i to row j Add t times row i to row j The operations on the right side of the table are called the inverse operations If A is an n n matrix, then the following statements are equivalent, that is, are all true or all false I A is invertible II Ax = 0 has only trivial solution III A is row equivalent to I n, that is, A can be reduced to I n by a finite sequence of elementary row operations Every elementary matrix is invertible, and the inverse is also an elementary matrix From statement III above, we can apply a finite sequence of elementary row operations to reduce A to I n, which implies we can find elementary matrices E 1, E 2,, E k such that E k E 2 E 1 A = I n As E 1, E 2,, E k are invertible Multiplying both sides of equation on the left successively by E 1 k,, E 1 2, E 1 1 we obtain (E k E 2 E 1 ) 1 E k E 2 E 1 A = (E k E 2 E 1 ) 1 I n = ( ) E 1 1 E 1 2 E 1 k Ek E 2 E 1 A = E 1 1 E 1 2 E 1 k = (I I I)A = A = E 1 1 E 1 2 E 1 k, 11
so, A = E 1 1 E 1 2 E 1 k and A 1 = E k E 2 E 1 We wish to reduce A to the identity matrix by row operations and simultaneously apply those operations to I to produce A 1 This can be accomplished by adjoining the identity matrix to the right of A, that is [A I, and applying row operations to both sides until the left side is reduced to I The final matrix will then have the form [I A 1 If A is an n n matrix, then the following statements are equivalent I A is invertible II Ax = 0 has only the trivial solution III A is row equivalent to I n IV Ax = b is consistent for every n 1 matrix b 1 Use the formula to compute the inverse of the following matrices [ [ [ 3 1 2 3 2 0 A = B = C = 5 2 4 4 0 3 2 Let A be an invertible matrix whose inverse is [ 3 4 5 6 Find the matrix A 3 Let A be an invertible matrix, and suppose that the inverse of 7A is [ 1 2 4 7 Find the matrix A 4 Let A be the matrix [ 1 0 2 3 Compute A 3, A 3, and A 2 2A + I 5 Assume that A is a square matrix which satisfies A 2 3A+I = 0 Show that A 1 = 3I A 6 Let A be a square matrix Show that (I A) 1 = I + A + A 2 + A 3 if A 4 = 0 Show that (I A) 1 = I + A + A 2 + + A n if A n+1 = 0 7 Which of the following are elementary matrices? [ [ 2 0 1 0 0 1 3 1 (c) [ 2 0 0 2 12
(e) 0 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 1 (f) 1 0 0 0 1 3 0 0 1 (g) 1 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 8 Find matrices which will perform each of the following operations on a 3 3 matrix add three times row 3 to row 1 halve column 2 (c) subtract row 3 from row 2 9 Perform the following row operations on A = 3 1 0 2 1 4 3 5 5 by multiplying A on the left by a suitable elementary matrix Check your answer in each case by performing the row operation directly on A Interchange the first and third row Multiply the second row by 1/3 (c) Add twice the second row to the first row 10 Determine the row operation that will restore the given elementary matrix to an identity matrix [ 1 0 0 0 1 1 0 0 0 5 1 0 1 0 (c) 0 8 0 0 1 0 0 0 0 1 0 0 0 0 1 11 Consider the matrices 1 2 3 A = 4 5 6 B = 7 8 9 7 8 9 4 5 6 1 2 3 C = Find elementary matrices E 1, E 2, E 3, and E 4 such that 1 2 3 4 5 6 9 12 15 E 1 A = B E 2 B = A (c) E 3 A = C E 4 C = A 12 Consider the matrix [ 1 0 3 4 Find elementary matrices E 1 and E 2 such that E 2 E 1 A = I Write A 1 as a product of two elementary matrices (c) Write A as a product of two elementary matrices 13
13 Express the matrix A = 1 3 3 8 2 5 1 8 0 1 7 8 in the form A = EFR, where E and F are elementary matrices, and R is in row-echelon form 14 Let A be the matrix 1 1 0 0 1 1 1 0 1 Determine if A is invertible, and if so, find its inverse 15 Find the inverse of the given matrix by row operations if the matrix is invertible [ 1 2 3 1 5 1 0 1 3 5 (e) 2 4 1 (h) 1 1 1 4 2 9 0 1 0 (c) [ 2 3 3 5 [ Determinant 8 6 4 3 3 4 1 1 0 3 2 5 4 (f) (g) 1 0 1 0 1 1 1 1 0 2 6 6 2 7 6 2 7 7 (i) (j) 1 0 0 0 1 2 0 0 1 2 4 0 1 2 4 8 5 11 7 3 2 1 4 5 3 2 8 7 0 0 0 0 The order of a determinant is the number of rows (or columns) it has The value of the determinant of A 2 2, where [ a11 a A 2 2 = 12, is a 21 a 22 a 11 a 12 a 21 a 22 = a 11a 22 a 12 a 21 We emphasis the determinants is a number in contrast to the corresponding square matrix which is an array For another 3 3 matrix A = A 3 3 = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 the determinant of this matrix A is denoted by deta or A We wish to study this and higher order determinants a little more fully Formally, the value of deta is given by a deta =a 22 a 23 11 a 32 a 33 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 =a 11 a 22 a 33 a 11 a 23 a 32 a 12 a 21 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 13 a 22 a 31,, 14
which is called expanding along the top row Notice that each element in the top row is multiplied by the determinant of the elements left when the column and row through that element are struck out For example, looking at element a 12 the determinant of elements left over is a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 a 21 a 23 a 31 a 33 These determinants left are called minors and denote the minor for the element a ij by M ij Next we define the cofactor, denoted by C ij for the element a ij by multiply the minor and a sign as follow: C ij = ( 1) i+j M ij The determinant of an n n matrix A can be computed by multiplying the entries in any row (or column) by their cofactors and adding the resulting products, that is, for each 1 i n and 1 j n, deta =a i1 C i1 + a i2 C i2 + + a in C in (cofactor expansion along the i-th row) =a 1j C 1j + a 2j C 2j + + a nj C nj (cofactor expansion along the j-th column) A square matrix is called upper triangular if all the entries below the main diagonal are zeros Similarly, a square matrix is called lower triangular if all the entries above the main diagonal are zeros A matrix that is either upper or lower triangular is called triangular If T is an n n triangular matrix, T = a 11 a 12 a 13 a 14 0 a 22 a 23 a 24 0 0 a 33 a 34 0 0 0 a 44 then dett is the product of the entries on the main diagonal, that is, dett = a 11 a 22 a 33 a 44 Properties of Determinants I If two rows (or two columns) are interchanged, then deta deta, that is, the value of the determinant is multiplied by 1 II If the rows and columns are interchanged (transposed) the value is not changed, that is, deta T = deta III If the two rows or columns are identical, deta = 0, IV If a row or column has a common factor we proceed thus: a 11 a 12 a 13 ka 21 ka 22 ka 23 a 31 a 32 a 33 = k a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = k deta 15
V a 11 + ka 12 + la 13 a 12 + na 13 a 13 a 21 + ka 22 + la 23 a 22 + na 23 a 23 a 31 + ka 32 + la 33 a 32 + na 33 a 33 = deta This means that we can add multiples of corresponding elements of one column to another (Similarly for rows) VI a 11 + α a 12 a 13 a 21 + β a 22 a 23 a 31 + γ a 32 a 33 = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 + α a 12 a 13 β a 22 a 23 γ a 32 a 33 VII If A and B are square matrices of the same size, then detab = detadetb VIII Since AA 1 = I and detab = detadetb, then detadeta 1 = deti = 1 = deta 1 = 1 deta XI Since we can express a non-singular matrix A into product of elementary matrices and an upper triangular matrix U, with the property of detab = detadetb, we can find deta by deta = det ( E 1 1 E 1 2 E 1 k U) = dete 1 1 dete 1 2 dete 1 k U = Similarly, deta 1 = det ( E k E 2 E 1 U 1) = dete 1 dete 2 dete k detu detu dete 1 dete 2 dete k Suppose B = E i A, then the properties of the determinant of matrix B after applied an elementary row operation on matrix A is as following: Elementary row operation dete i dete 1 i detb I r i r j 1 1 deta II tr i r i t 1/t t deta III tr i + r j r j 1 1 deta **************** For AMA226 only *********************** Suppose a non-singular matrix A can be reduced to an upper triangular matrix U by the elementary row operations σ 1, σ 2,, σ k, that is, A σ 1 B σ 2 C σ 3 σ k U When a square matrix is being reduced by an elementary row operation, let say apply σ 1 on A to obtain B, the determinant of deta and detb might change Once the matrix is being reduced by σ 1, detb is equal to deta times a factor The following table shows the relationships between elementary row operation and the factor: Elementary row operation factor from A to B inverse factor from B back to A detb I r i r j 1 1 deta II tr i r i t 1/t t deta III tr i + r j r j 1 1 deta 16
Let µ 1, µ 2,, µ k be the factors corresponding to the elementary row operations σ 1, σ 2,, σ k Then detb =µ 1 deta detc =µ 2 detb = µ 2 µ 1 deta detu =µ k µ 2 µ 1 deta Therefore, deta = detu µ 1 µ 2 µ k or deta 1 = µ 1µ 2 µ k detu Let A be an n n matrix and C ij be the cofactor of the entry a ij, then we define the adjoint matrix of A as T C 11 C 12 C 1n C 21 C 22 C 2n adja = C n1 C n2 C nn If A is an invertible matrix, non-singular, ie, deta 0, then A 1 = 1 deta adja The system Ax = b has a unique solution if any only if A is non-singular In this case, the unique solution is x = A 1 b If Ax = b is a system of n linear equations in n unknowns such that deta 0, then the system has a unique solution This solution is x 1 = deta 1 deta, x 2 = deta 2 deta,, x n = deta n deta, where A j is the matrix obtained by replacing the entries in the j-th column of A by the entries in the matrix b = [b 1 b 2 b n T This formula is known as the Cramer s rule 1 Evaluate the determinant by definition 1 2 8 2 1 1 3 (e) 3 4 6 6 4 1 7 2 3 2 1 0 3 (c) 1 7 (f) 4 0 1 2 8 6 8 3 1 4 3 1 1 2 7 3 5 1 (g) 2 0 6 3 4 1 2 5 4 3 8 1 0 2 4 2 Evaluate the following by inspection (h) (i) 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 5 0 0 0 0 0 4 0 0 0 0 0 0 2 0 0 0 3 0 0 0 0 0 0 1 5 0 0 0 0 17
2 40 17 0 1 11 0 0 3 1 0 0 0 9 1 0 0 12 7 8 0 4 5 7 2 (c) 1 2 3 3 7 6 1 2 3 3 1 2 6 2 4 1 7 3 3 Evaluate the determinants of the given matrices by reducing the matrix to row-echelon form 1 1 2 3 7 2 4 8 1 1 2 2 2 0 0 3 2 7 2 1 2 7 0 1 5 (g) 1 1 0 1 2 2 2 2 1 1 0 3 3 3 3 6 9 3 1 2 1 1 4 2 3 (e) 1 0 1 0 1 1 0 3 3 1 3 2 1 1 3 0 1 2 2 1 1 3 1 3 5 2 1 3 1 2 7 0 4 2 1 2 0 (c) 3 5 1 (f) 1 0 1 1 (h) 0 0 1 0 1 0 2 1 0 0 0 2 1 1 4 3 2 0 1 2 3 0 0 0 1 1 4 Find all values of λ for which deta = 0 [ λ 1 2 A = 1 λ 4 5 Assume a b c d e f g h i d e f g h i a b c a b c 2d 2e 2f g h i = 5 Find A = (c) λ 6 0 0 0 λ 1 0 4 λ 4 a + d b + e c + f d e f g h i a b c d 3a e 3b f 3c 2g 2h 2i 6 Use row reduction to show that 1 1 1 a b c = (b a) (c a) (c b) a 2 b 2 c 2 1 a b a 1 b = (a 1) (b 1) (a + b + 1) a b 1 (c) 1 a bc 1 b ca 1 c ab = (a b) (b c) (c a) 7 Assume deta = 5, where Find A = 18 a b c d e f g h i
det (3A) det (2A 1 ) (c) det (2A) 1 a g d b h e c i f 8 Determine which of the following matrices are invertible 1 0 0 2 1 4 7 2 1 3 6 7 1 1 2 (c) 7 2 1 0 8 1 3 1 6 3 6 6 9 For which value(s) of k does A fail to be invertible? [ k 3 2 A = 2 k 2 B = 1 2 4 3 1 6 k 3 2 0 7 5 0 1 1 0 3 2 10 Let A = 1 6 3 2 7 1 3 1 4 Find all the minors Find all the cofactors (c) Evaluate deta by a cofactor expansion along i the first row ii the first column Find adj A (e) Find A 1 iii the second row iv the second column v the third row vi the third column 11 Evaluate deta by a cofactor expansion along a row or column of your choice 0 6 0 4 4 0 4 A = 8 6 8 (c) A = 1 1 0 1 3 2 2 3 0 3 1 6 14 3 6 A = 1 3 7 2 0 8 1 3 4 A = 4 3 1 9 2 0 3 2 4 2 0 3 4 6 4 1 1 2 2 2 0 0 3 3 3 12 A = 1 3 1 1 2 5 2 2 1 3 8 9 1 3 2 2 Evaluate A 1 using the adjoint method 19
Evaluate A 1 using the product of elementary matrices 13 For the following systems, find A 1 for the augmented matrix [A b Hence, solve the systems by x = A 1 b { x1 + 2x 2 = 7 x 1 + x 2 + x 3 = 5 2x 1 + 5x 2 = 3 (e) x 1 + x 2 4x 3 = 10 { 3x1 6x 2 = 8 4x 1 + x 2 + x 3 = 0 2x 1 + 5x 2 = 1 3x 1 + x 2 + 7x 3 + 9x 4 = 4 x 1 + 2x 2 + 2x 3 = 1 x (f) 1 + x 2 + 4x 3 + 4x 4 = 7 (c) x 1 + 3x 2 + x 3 = 4 x 1 2x 3 3x 4 = 0 x 1 + 3x 2 + 2x 3 = 3 2x 1 x 2 4x 3 6x 4 = 6 2x 1 + x 2 + x 3 = 7 3x 1 + 2x 2 + x 3 = 3 x 2 + x 3 = 5 14 Solve the systems by Cramer s rule, where it applies { 3x1 4x 2 = 5 x 1 3x 2 + x 3 = 4 2x 1 + x 2 = 4 2x 1 x 2 = 2 4x 1 3x 3 = 0 4x 1 + 5x 2 = 2 2x 1 x 2 + x 3 = 8 11x 1 + x 2 + 2x 3 = 3 (e) 4x 1 + 3x 2 + x 3 = 7 x 1 + 5x 2 + 2x 3 = 1 6x 1 + 2x 2 + 2x 3 = 15 2x 1 x 2 + x 3 4x 4 = 32 x 1 + x 2 2x 3 = 1 7x (f) 1 + 2x 2 + 9x 3 x 4 = 14 (c) 2x 1 x 2 + x 3 = 2 3x 1 x 2 + x 3 + x 4 = 11 x 1 2x 2 4x 3 = 4 x 1 + x 2 4x 3 2x 4 = 4 15 Use Cramer s rule to solve x 3 without solving x 1, x 2, and x 4 4x 1 + x 2 + x 3 + x 4 = 6 3x 1 + 7x 2 x 3 + x 4 = 1 7x 1 + 3x 2 5x 3 + 8x 4 = 3 x 1 + x 2 + x 3 + 2x 4 = 3 Linear Dependence A column vectors with n entries, a n 1 matrix, can be denoted by R n Similarly, for a row vectors with n entries, a 1 n matrix, can be denoted by R n For any two vectors u = [u 1 u 2 u n T and v = [v 1 v 2 v n T in R n, they are called equal if The sum u + v is defined by u 1 = v 1, u 2 = v 2,, u n = v n u + v = [u 1 + v 1 u 2 + v 2 u n + v n T and if k is any scalar, the scalar multiple ku is defined by ku = [ku 1 ku 2 ku n T 20
The operations of addition and scalar multiplication in this definition are called the standard operations on R n We define the zero vector in R n to be the vector 0 = [0 0 0 T A vector w is called a linear combination of the vectors v 1, v 2,, v r if it can be expressed in the form w = k 1 v 1 + k 2 v 2 + + k r v r, where k 1, k 2,, k r are scalars If S = {v 1,v 2,,v n } is a set of vectors in R n and k 1, k 2,, k r are scalar, then the vector equation k 1 v 1 + k 2 v 2 + + k r v r = 0 has at least one solution, namely k 1 = 0, k 2 = 0,, k r = 0 If this is the only solution, then S is called a linearly independent set or v 1, v 2,, v n are linearly independent If there are other solutions, then S is called a linearly dependent set or v 1, v 2,, v n are linearly dependent 1 Let u = [1 2 3 T, v = [2 3 1 T, and w = [3 2 1 T Find the components of u w 7v + 3w (c) w + v 3 (u 7v) (e) 3v 8w (f) 2v (u + w) 2 Explain why the following are linearly dependent sets of vectors (solve this problem by inspection) {[ [ } {[ [ [ } 1 3 2 5 6,,, 2 6 3 8 1 3 Which of the following sets of vectors in R 3 are linearly dependent? 2 3 2 6 1 1, 6, 10 (c) 0, 1 4 2 4 1 4 3 2 4 1 0 5 1, 1, 0 3, 1, 6, 1 5 3 3 4 3 4 Which of the following sets of vectors in R 4 are linearly dependent? 1 0 0 3 4 2 2 1, 2 2, 2 3, 0 3 4 8, 2 4, 2 0 1 6 0 0 6 0 0 2, 7 2 1 6 3 3 0 21
(c) 4 4 0 0, 0 0 6 6, 5 0 5 5 3 0 4 1, 6 2 1 2, 1 3 5 1, 3 7 8 3 5 Express the given vector w as a linear combination of the given vector v 1 and v 2 w = [1 0 T, v 1 = [1 1 T, v 2 = [0 1 T w = [4 3 T, v 1 = [2 1 T, v 2 = [ 1 0 T 6 Express the given vector w as a linear combination of the given vector v 1, v 2, and v 3 w = [1 0 0 T, v 1 = [1 1 1 T, v 2 = [ 1 1 0 T, v 3 = [1 0 1 T w = [1 1 1 T, v 1 = [0 1 1 T, v 2 = [1 1 0 T, v 3 = [1 0 2 T Eigenvalues and Eigenvectors If A is an n n matrix, then a non-zero vector x in R n is called an eigenvector of A if Ax is a scalar multiple of x, that is, Ax = λx for some scalar λ The scalar λ is called an eigenvalue of A and x is said to be an eigenvector corresponding to λ To find the eigenvalues of an n n matrix A we rewrite Ax = λx as or equivalently Ax = λix (A λi)x = 0 For λ to be an eigenvalue, there must be a non-zero solution of this equation However, this homogeneous system has a non-zero solution (non-trivial solution) if and only if det (A λi) = 0 This is called the characteristic equation of A; the scalars satisfying this equation are the eigenvalues of A When expanded, the determinant det (A λi) is a polynomial in λ called the characteristic polynomial of A If A is an n n matrix, the the following are equivalent I λ is an eigenvalue of A II The system of equations (A λi) x = 0 has non-trivial solutions III There is a non-zero vector x in R n such that Ax = λx IV λ is a real solution of the characteristic equation det (A λi) = 0 When the eigenvalues are solved, then we can solve the eigenvectors The eigenvectors of A corresponding to an eigenvalue λ are the non-zero vectors that satisfy Ax = λx Equivalently the entries in the eigenvectors corresponding to λ are the non-trivial solution of (A λi)x = 0 1 Find (i) the characteristic polynomials, (ii) eigenvalues, and (iii) eigenvectors of the following matrices: 22
[ 3 0 8 1 [ 10 9 4 2 [ 0 3 (c) 4 0 [ 2 7 1 2 [ 0 0 (e) 0 0 (f) (g) (h) 4 0 1 2 1 0 2 0 1 2 0 1 6 2 0 19 5 4 1 0 1 1 3 0 4 13 1 (i) (j) (k) 5 0 1 1 1 0 7 1 0 5 6 2 0 1 8 1 0 2 0 0 2 0 1 0 1 0 0 1 2 0 0 0 0 1 Diagonalization A square matrix A is called diagonalizable if there is an invertible matrix P such that P 1 AP is diagonal; the matrix P is said to diagonalize A If A is an n n matrix, then the following are equivalent I A is diagonalizable II A has n linearly independent eigenvectors For a diagonalizable matrix A, there is an invertible matrix p 11 p 12 p 1n p 21 p 22 p 2n P = = [p 1 p 2 p n p n1 p n2 p nn such that P 1 AP is diagonal, say P 1 AP = D, where λ 1 0 0 0 λ 2 0 D = 0 0 λ n and p 1, p 2,, p n denote the column vectors of P Since P is invertible, its column vectors are all non-zero, hence, λ 1, λ 2,, λ n are eigenvalues of A, and p 1, p 2,, p n are corresponding eigenvectors Diagonal matrices have many desirable properties For example, the k-th power, where k is a positive integer, of a diagonal matrix e 1 0 0 0 e 2 0 E = 0 0 e n is E k = e k 1 0 0 0 e k 2 0 0 0 e k n 23
As P 1 AP = D, then P ( P 1 AP ) P 1 = ( PP 1) A ( PP 1) = PDP 1 = A Hence, A k = ( PDP 1) k ( ) ( = PDP 1 PDP 1) (PDP 1) =PD ( P 1 P ) D ( P 1 P ) D DP 1 = PDD DP 1 =PD k P 1 1 Show that the following matrices are not diagonalizable [ [ 2 0 2 3 3 0 0 1 2 1 1 (c) 0 2 0 0 1 2 1 0 1 1 3 0 4 13 1 2 Find a matrix P that diagonalizes A, and determine P 1 AP [ 14 12 1 0 0 A = 20 17 (c) A = 0 1 1 0 1 1 [ 2 0 2 1 0 A = 0 3 0 A = 6 1 0 0 3 3 Determine if A is diagonalizable If so, find a matrix P that diagonalizes A, and determine P 1 AP 19 9 6 0 0 0 A = 25 11 9 A = 0 0 0 17 9 4 3 0 1 A = (c) A = 1 4 2 3 4 0 3 1 3 5 0 0 1 5 0 0 1 5 (e) A = (f) A = 2 0 0 0 0 2 0 0 0 0 3 0 0 0 1 3 2 0 0 0 0 2 5 5 0 0 3 0 0 0 0 3 4 Evaluate the following matrices [ 18 1 0 1 5 [ 3 2 1 4 16 (c) 1 1 2 1 2 1 0 1 1 15 24