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# Note 1 for MATH 290 at KU

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Chapter 1 Systems of Linear Equations 11 Intro to systems of linear equations Homework Textbook7 EX 137 157 417 477 497 517 657 73 page 11 Main points in this section 1 De nition of Linear system of equations and homogeneous sys tems 2 Row echelon form of a linear system and Gaussian elimination 5 Solving linear system of equations using Gaussian elimination 2 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS De nition 111 A linear equation in 72 unknown variables 17 7 has the form 11 lt122 ltlnn 5 Here 117127 7amb are real numbers We say b is the constant term and 12 is the coe icient of m For real numbers 317 73m if a131a282ltln8nb we say that 181728277n8n is a solution of this equation 1 An example of a linear equation in two unknowns is 2m 7y 5 A solution of this equation is m 7171 1 The equation has many more solutions The graph of this equation is a line 2 An example ofa linear equation in three unknowns is 2y7r2 7139 A solution of this equation is m 07 y 02 1 The equation has many more solutions The graph of this equation in 3space is a plane 3 See Textbook7 Example 17 page 2 for examples of linear and non linear equations De nition 112 By a System of Linear Equations in 72 variables 1727 7 we mean a collection of linear equations in these variables 11 INTRO TO SYSTEMS OF LINEAR EQUATIONS 3 A system of in linear equations in these n variables can be written as a111 112 113 air n b1 a211 122 123 a2nn 52 131 132 133 a3nn 53 am i am z am33 amr n brn where aij and bi are all real numbers Such a linear system is called a homogeneous linear system if b1b2bm0 1 A solution to such a system is a sequence of n numbers 317 7sn that is solution to all these in equations 2 In two variables7 the following is an example of a system of two equation 2w y 3 w 7 9y 78 Clearly7 w 17y 1 is the only solution to this system Geometrically solution given by precisely the point where the graphs two lines of these two equations rneet Also note that the system 2w y 3 2w y 7 does not have any solution Such a system would be called an inconsistent system Geometrically these two equations in the system represent two parallel lines they never meet CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS In three variables7 the following is an example of a system of two equation 2w y 22 3 w 7 9y 22 78 Clearly7 w 17y 172 0 is a solution to this system This system has many more solutions For example7 w 1171 02 7192 is also a solution of this system Geometrically solution given by precisely the points where the graphs two planes in 3 space of these two equations rneet Classi cation of linear systems Given a linear system in n variables7 precisely on the the following three is true a The system has exacly one solution consistent system b The system has in nitely many solutions consistent sys tem c The system has NO solution inconsistent system Two systems of linear equations are called equivalent7 if they have precisely the same set of solutions Following operations on a system produces an equivalent system a lnterchange two equations b Multiply an equation by a nonzero constant c Add a multiple of an equation to another one These three operations are sometimes known as basic or elemen tary operations 11 INTRO TO SYSTEMS OF LINEAR EQUATIONS 5 7 A linear system of the form 1 112 113 air n b1 2 lt1233 a2nn 52 z3ma3cb3 is said to be in rowechelon form In two variables L y this would sometime look like 1121 b1 1 52 In three variables 71172 this would sometime look like i my 1132 b1 ylt1232 52 Zb3 Theorem 113 1 Any system of linear equations is equivalent to a linear system in row echelon form 2 This can be achieved by a sequence of application of the three basic operation described in 3 This process is known as Gaussian elimination Read Examples 5 9 page 6 Practice For exercise 31 56 page 1107 reduce the system to a row echelon form and solve 6 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Exercise 114 Ex 32 p 11 Reduce the following system and solve 475y3 Eqn71 7810y14 Eqn72 Add 2 times Enq l to Eqn 2 475y3 Eqn71 020 Eqn73 The Eqn 3 is absurd So7 the system has no solution The system is inconsistent Exercise 115 Ex 34 p 11 Reduce the following system and solve 9x74y5 Eqn71 1 1 x y0 Eqn72 Multiply Eqn 2 by 18 9x74y5 Eqn71 9x6y0 Eqn73 Subtract Eqn l from the Eqn 3 9m 7 4y 5 Eqn 7 1 10y 75 Eqn 7 4 Divide Eqn 4 by 10 9m 7 4y Eqn 7 1 y 7 Eqn 7 5 Divide Eqn l by 9 m 7 3y Eqn 7 6 Eqn75 11 INTRO TO SYSTEMS OF LINEAR EQUATIONS 7 This is the row echelon form Now substitute y 7 in Eqn 6 4 1 5 1 x77 7 or 9 2 9 3 So7 the solution is x i 1 i 71 7 3 y 7 239 Exercise 116 Ex 44 p 12 at 3 I 71 7 1T 2T 7 1 Eqn 7 1 2x17212 Eqn72 multiply Eqn 1 by 12 and simplify 3x1427 Eqn73 2x17212 Eqn72 2 A Add 73 times Eqn 3 to Eqn 2 31427 Eqn73 Tnmg Eqn 7 4 Multiply Eqn 4 by 1 5 31427 Eqn73 2 72 Eqn 7 5 Multiply Eqn 3 by 1 2 Eqn76 2 72 Eqn 7 5 So7 above is the row echelon form of the system Now substitute 2 72 in Eqn 6 and get 1 g g 5 So7 the systme is consistentand has unique solution 1 57 2 8 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Exercise 117 Ex 50 p 12 Deduce an equivalent row echelon form and solove the following system 5173x2233 Eqn71 2142737 Eqn72 1711x2433 Eqn73 First7 switch Eqn l and Eqn S 1711x2433 Eqn73 2142737 Eqn72 5173x2233 Eqn71 Subtract 2 times Eqn 3 from Eqn 2 and 5 times Eqn 3 from Eqn l 1711x2433 Eqn73 26m 7 9x3 1 Eqn 7 4 52m 7183 712 Eqn 7 5 Subtract 2 times Eqn 4 from Eqn E 1711x2433 Eqn73 26m 7 9x3 1 Eqn 7 4 0 714 Eqn 7 6 The system is inconsistent because Eqn 6 is absurd To obtain the row echelon form7 we diVid Eqn 4 by 26 1711x2433 Eqn73 2732 16 Eqn77 0714 Eqn76 Exercise 118 Ex 52 p 12 Deduce an equivalent row echelon form and solove the following system 1 43 41 7 22 3 21 7 22 7 73 Eqn71 Eqn72 Eqn73 11 INTRO TO SYSTEMS OF LINEAR EQUATIONS 9 Subtract 4 times Eqn 1 from Eqn 2 and suntract 2 times Eqn 1 from Equn 3 1 4x3 13 Eqn 71 72mg 715mg 745 Eqn 7 4 72mg 715mg 745 Eqn 7 5 Subtract Eqn 4 from Eqn S 1 4x3 13 Eqn 71 72mg 715mg 745 Eqn 7 4 0 0 Eqn 7 6 Multiply Eqn 4 by 5 and we get 1 4x3 13 Eqn 71 2 75 225 Eqn 7 7 0 0 Eqn 7 6 The above is the row echelon form of the system The system is con sistent Since the echelon form has actually two equations and number of variables is three7 the system has in nitely many solutions For any value 3 if7 we have 2 225 7 7575 and 1 13 7 4t So7 a parametric solution of this system is 1 13 7 4757 2 225 7 75757 3 75 Exercise 119 Ex 56 p 12 Deduce an equivalent row echelon form and solve the following system 1 34 4 Eqn 7 1 2x2 7mg 7M 0 Eqn 7 2 3x2 72 1 Eqn 7 3 2x1 7mg 43 5 Eqn 7 4 10 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Subtract 2 time Eqn l from Eqn 4 1 34 4 Eqn 7 1 2x2 73 74 0 Eqn 7 2 3x2 72 1 Eqn 7 3 7 43 76M 73 Eqn 7 5 Multiply Eqn 2 by 5 1 34 4 Eqn 7 1 2 75 75M 0 Eqn 7 6 3m 72M 1 Eqn 7 3 7 43 76M 73 Eqn 7 5 Subtract 3 times Eqn 6 from Eqn 3 and add Eqn 2 to Eqn S 1 34 4 Eqn 71 2 75 75M 0 Eqn 7 6 15m 75m 1 Eqn 7 7 35 765 73 Eqn78 11 INTRO TO SYSTEMS OF LINEAR EQUATIONS 11 Myltiply Eqn 7 by 1 34 4 Eqn 7 1 2 753 754 0 Eqn 7 6 3 74 Eqn 7 9 353 7654 73 Eqn 7 8 Subtract 35 times Eqn Q from Eqn B 1 34 4 Eqn 7 1 2 753 754 0 Eqn 7 6 3 74 Eqn 7 9 74 7 Eqn 7 10 Multiply Eqn 10 by 7 1 34 4 Eqn 7 1 2 753 754 0 Eqn 7 6 3 74 Eqn 7 9 4 1 Eqn 7 11 The above is a row echelon form of the system By back substitution 2 417 3 17 217 11 1 3 3 12 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS 12 Gaussian GaussJordan Elimination Homework Textbook7 Ex217 297 337 35by Gauss Jordan7 457 477 497 537 577 63 page 26 Main points in this section N De nition of matrices 2 Elementary row operation on a matrix 5 De nition of Row echelon form of matrix 4 Gaussian elimination and Gauss Jordan elimination 5 Solving sytems oflinear equations using Gaussian elimination and Gauss Jordan elimination 12 GAUSSIAN GAUSS JORDAN ELIMINATION 13 De nition 121 For two positive integers m7 n and m X nematrix is a rectangular array 111 112 113 39 39 39 am 121 122 123 39 39 39 a2n 131 132 133 39 39 39 Ian aml amZ am 39 39 39 04mm 1 the array has m rows aria 72 column 2 Here 127 is a real number7 to be called ijthientry This entry sits in the imirow jthicolumn The rst subscript 239 of 127 is called the row subscript and j is called the column subscript 3 It is possible to talk about matrices Whose entries 127 are not real numbers We can talk about matrices of any kind of objects We Will only talk about matrices With real entries and such matrices are also called real matrices 4 We say that the size of the matrix is m X n 5 A square matrix of order n is a matrix Whose number of rows and columns are same and equal to n 6 For a square matrix of order 717 the entries a117a2277am are called the main diaginal entries The most common use7 for this class7 of matrices is to represent system of liner equation Given a system of liner equations7 an asso ciated matrix to be called the augmented matrix contains all the information regarding the system 14 CHAPTER 1 De nition 122 Given a system of m linear equations SYSTEMS OF LINEAR EQUATIONS 111 112 113 a1nn 121 122 123 a2nn 52 131 132 133 a3nn am i am z am g amnn the augmented matrix of the system is de ned as 111 121 131 04ml and the coe icient matrix is de ned as 111 121 131 aml 1 Conversely7 given a m X n 1 matrix7 we can write down a system of linear m equations in n unknowns variables 2 Consider the linear system from exercise 114 112 122 132 amZ 112 122 132 amZ 113 123 133 04m 113 123 133 am 04171 04271 04371 04mm 04171 04271 04371 04mm 4x75y3 7810y14 b1 52 12 GAUSSIAN GAUSS JORDAN ELIMINATION 15 The augmented matrix of the system is 4 753 781014 and the coe icient matrix is 21 3 Consider the linear system from exercise 117 51732233 2142737 17112433 The augmented and the coe icient matrices of this system are 5 73 2 3 5 73 2 2 4 71 7 2 4 71 1 711 4 3 1 711 4 Recall that we deduced an equivalent system in row echelon form 17112433 Eqn73 273 Eqn77 0714 Eqn76 The augmented and the coe icient of this row echelon form is 16 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS 4 Consider the linear system from exercise 119 1 34 4 Eqn 7 1 2 7mg 7M 0 Eqn 7 2 3m 72M 1 Eqn 7 3 2 7mg 43 5 Eqn 7 4 The augmented and the coe icient matrices are 10 0 3 4 10 0 3 0 2 4141039 0 2 4141 0 3 0 721 0 3 0 72 2414 0 5 2414 0 Recall that we deduced an equivalent system in row echelon form 1 34 4 Eqn 7 1 2 75 75M 0 Eqn 7 6 3 74 Eqn 7 9 4 1 Eqn 7 11 The augmented and the coe icient matrices of this echelon form are given by 100 34 100 3 014545039 014545 127 1 0014EE 001 E 00011 0001 The above discussions and examples demonstrate that the three basic operations that we used to reduce a system of linear equations to a row echelon form7 can be translated to a version for matrices De nition 123 By an elementary row operation on a matrix we mean one of the following three 12 GAUSSIAN GAUSS JORDAN ELIMINATION 17 1 Interchange two rows 2 Multiply a row by a nonzero constant 3 Add a multiple of a row to another row Two matrices are said to be rowequivalent if one can be obtained from another by application of a sequence of elementary row operations Two row equivalent matrices7 correspond to two equivalent system of equations Now we de ne the matrix version of row echelon form De nition 124 A matrix is said to be in rowechelon form7 if it has the following properties 1 All rows conssiting entierly of zeros occur at the bottom of the matrix 2 For each non zero row7 rst nonzero entry is 1 called the leading 1 3 For each successive nonzero rows7 the leading 1 in the higher row is farther to the left than the leading 1 in the lower row A matrix in row echelon form is said to be in reduced rowechelon form7 if every column that has a leading 1 has zeros in evey position above and below the leading 1 Theorem 125 A matrix is row equivalent to a matrix is row echelon form We gave see below de nition 122 above the augmented matrix of the system in exercise 119 and that of the equivalent system in row echelon form 18 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Read Textbook7 Example 47 p 16 for examples of matrices in row echelon form De nition 126 The method Gaussian elimination with beck substitution of solving a system os equation is described as follows 1 Write the augmented matrix of the system 2 Use the elemetary row operations tto reduce the augmented ma trix to a matrix in row echelon form 3 Write the linear system corresponding to the row echeclon matrix and solve by back substitution Exercise 127 Ex 11988 GEES We Will use the method of Gaus sian elimination With beck substitution to solve exercise 1197 using analogous steps Recall the system 1 34 4 Eqn 7 1 2x2 7mg 7M 0 Eqn 7 2 3x2 72 1 Eqn 7 3 2x1 7mg 43 5 Eqn 7 4 The augmented matrix is NOOH N3 l H ll NH WHOVP Subtract 2 times row 1 from row 4 1 0 0 3 4 0 2 71 71 0 0 3 0 72 1 0 71 4 76 73 12 GAUSSIAN GAUSS JORDAN ELIMINATION 19 Multiply row 2 by 5 1 0 0 3 4 0 1 75 75 0 0 3 0 72 1 0 71 4 i6 73 Subtract 3 times row 2 from row 3 and add row 2 to Eqn 4 1 0 0 3 4 0 1 75 75 0 0 0 15 75 1 0 0 35 765 73 Myltiply row 3 by g 1 0 0 3 4 0 1 75 75 0 0 0 1 4 g 0 0 35 765 73 Subtract 35 times row 3 from row 4 10034 0145450 12 0014EE 1616 00044 Multiply row 4 by 7 OOOH OOHO H1 0quot ll 0quot HWINOHgt 20 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS The above is a matrix in row echelon form row equivalent to the aug mented matrix Now the system of linear equations corresponding this row echelon matrix is 1 34 4 2 753 754 0 3 i 4 4 1 By back substitution 2 4 17 3 g Read Textbook7 Example 57 67 p 19 for other such use ofthis method De nition 128 A matrix in row echelon form is said to be in Gauss Jordan form7 if all the entries above leading entries are zero The method of Gaussian elimination With back substitution to solve system of linear equations can be re ned by rst further reducing the augmented matrix to a Gauss Jordan form and work With the sytem corresponding to it This method is called GaussJordan elimina tion method of solVing linear sytems Consider exercise 1277 the matrix in the row echelon form7 equiv alent to the augmented matrix7 is 100 34 0175750 0017 00011 12 GAUSSIAN GAUSS JORDAN ELIMINATION 21 All the entries above the leading 1 in row 2 is zero So7 we try to achieve the same above the leading 1 in row 3 Add 5 times row 3 to row 2 10034 21 0107EE 12 0017EE 00011 Now we want to get zeros above the leading 1 in row 4 Subtract 3 times the row 4 from row 1 add times the row 4 from row 2 add times the row 4 from row 3 0 0 0 0 1 0 0 1 HHHH 0 1 0 0 COOH This matrix is in Gauss Jordan form The system of linear equation corresponding to this one is 1 1 2 1 3 1 4 1 So7 the solution to the system is 417 317 217 11 Read Textbook7 Example 77 8 p 22 23 for more on Gauss Jordan elimination Remark If you feel comfortable working with matrices7 it is best to reduce a system to Gauss Jordan7 instead of only to row echelon form 22 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Exercise 129 Ex 30 p 26 Solve the following using Gaussian elemination or Gauss Jordan elemination 21 7mg 33 24 Eqn 7 1 2 7mg 14 Eqn 7 2 71 75mg 6 Eqn 7 3 The augmented matrix is 2 71 3 24 0 2 71 14 7 75 0 6 Divide rst row by 2 and divide second row by 2 1 0 7 1 7 g 12 0 1 7 7 0 g 722 1 778 Add times second row to the third row 1 3 1 75 5 12 1 0 1 75 45 135 0 0 1 7 Multiply third row by 744 5 1 3 1 75 5 12 1 0 175 7 0 1 6 12 GAUSSIAN GAUSS JORDAN ELIMINATION 23 The above matrix is in row echelon form So7 we can use back substi tution and solve the system The system corresponding to this matrix is 1 752 3 1 2 753 3 7 By back substitution 1 3 1 6 7 610 127 6 108 3 7 2 2 7 1 2 2 Alternately we could reduce the row echelon matrix 12 H NIH MILO 1 0 1 7 7 0 0 6 to a Gauss Jordan form We will do this To do this add time the second row to the rst 1 0 125 155 1 0 1 75 7 0 0 1 6 Subtract 125 times third rwo from the rst 1 0 0 1 7 0 0 HNIH o 8 7 6 Now add 5 time the third row to the second 000 1 0 0 0 1 0 1 0 0 1 G 24 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS This matrix is in Gauss Jordan form The system of liner equations corresponding to this matrix is 1 8 2 3 6 This gives the solution of our system Exercise 1210 Ex 32 p 26 Solve the following using Gaussian elemination or Gauss Jordan elemination 21 33 3 Eqn 7 1 4m 73m 73 5 Eqn 7 2 8m 79m 153 10 Eqn 7 3 The augmented matrix is 2 0 3 3 4 73 7 5 8 79 15 10 We Will reduce this matrix to row echelon form Subtract 2 times rst row from second row and subtract 4times rst row from 3rd row 2 0 3 3 0 73 1 71 0 79 3 72 Subtract 3 times the second row from third 2 0 3 3 0 73 1 71 0 0 0 1 12 GAUSSIAN GAUSS JORDAN ELIMINATION 25 Divide rst row by 2 and second row by 3 33 11 017E 0001 The matrix is in rwo echelon form The sytern corresponding to thsi equation is 1 3 2 i 3 The last equation is absurd So7 the sytern is inconsistent Exercise 1211 Ex 34 p 26 Solve the following using Gaussian elernination or Gauss Jordan elernination 2y 2 8 73 76y 732 721 The augrnented matrix is 1 2 1 8 73 76 73 721 Add 3 times rst row to the second row 1218 0003 The above matrix is in row echelon form The corresponding system of linear equations is 2y 2 8 0 3 The last equation is absurd So7 the system is inconsistent 26 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Exercise 1212 Ex 44 p26 Solve the homogeneous linear sys tem corresponding to the coei cient matrix 1000 0110 Since it is a homogeneous sytems have all the constants zero and they system is 1 0 22 23 0 The system is already in row echelon form So7 by back substitution 2 7237 1 0 With 23 t a paramentric solution is 1 0722 775723 75 Note that this is a four variable problem and unknowns are 17 2 7 37 4 The variable 24 does not appear in these equations So for any 24 3 for any 37 for each solutions above So7 nal parametric solution is 1 0722 775723 7524 3 Where 3775 are paramenters Exercise 1213 Ex 50 p27 Consider the system of linear equa tions 2 y 0 Eqn 7 1 y 2 0 Eqn 7 2 2 2 0 Eqn 7 3 a2 by c2 0 Eqn 7 4 Find the values of 17 b7 0 such that the system has a a unique solution7 b no solution c an in nite number of solution 12 GAUSSIAN GAUSS JORDAN ELIMINATION 27 Solution The augmented matrix of the equation 0 1 1 Q HOH O OHH 0000 C Subtract 1 times rst row from third and a times rst row from fourth 0 1 1 1 COOH 1 H 0000 Add second row to third OOOH 0 WHO 0000 1 1 0 0 1 1 0 0 1 0 bid 0 GOOD Divide third row by 2 Add a 7 b second row to fourth 11 0 0 01 1 0 0 0 1 0 0 0 caib 0 28 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Subtract c a 7 b tirnes third row from fourth i OHHO GOOD 1 1 0 0 COOH The matrix is in row echelon form The corresponding liner system is 2 y y 2 0 2 0 0 0 The system is consistent for all values of 1 b7 07 and by back substi tution the sytern has unique solution 2 y 2 0 13 APPLICATION OF LINEAR SYSTEMS 29 13 Application of Linear systems Homework Textbook7 EX 17 47 137 217 23 page 38 As the section heading suggests7 we do a few applications of linear systems We do the following applications 1 Fitting polynimials 2 Network arilysis 5 Kircho s Laws for electrical networks 30 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS System of linear equations is much easier to handle than nonlinear systems I do not mean for this class only7 I mean for expert math ematicians and scientists ln fact7 it is really very dif cult to handle nonlinear systems That is why7 there is a wide range of applications of linear systems 131 Polynomial curve tting Recall two facts 1 there is exactly one line y c m2 that passes through two given points 2 there is exactly one parabola y 122 n0 that passes through three given points barring some exceptions These facts generalizes to 72 number of points in the plane and polyno mials p of degree n 7 17 so that these points will pass through the graph of y We describle it as follows Suppose a collection of data is represented by n points 17y17 27y27 39 39 39 7 n7yn39 lfthe micoordinates 1227 7 are distinct7 then there is a UNIQUE polynomial PW lt10 11 lt122 an71n71 of degree n 7 1 or less so that the graph of y pm passes through these points Given n such points7 to determine p we need to nd the coef cents 107117 71711 Since the points 221715 pass through the graph of y p7 we have yi More explicitly7 2 71 ac lt111 ltl21 39 quot ltln71f y1 2 71 a0 a1x2 M222 minim yz 2 71 ac lt113 ltl23 39 quot an71x f ya 2 H71 7 ac lt11n ltl2n 39 quot ltln71n 7 yn This is a linear system of n equations7 with n unknowns variables 10 a1 a2 an1 It is known that7 under our condition that 21227 7 are distinct7 the system has a unique solution 13 APPLICATION OF LINEAR SYSTEMS 31 The augmented matrix of this linear system is 2 H71 1 1 1 1 yl 2 H71 1 2 2 2 yg 2 H71 1 3 3 3 yg 1 mm mi 24 yn and the coe icients matrix is 2 H71 1 1 1 1 1 2 n7 1 2 2 2 1 2 n7 1 3 3 3 2 H71 1 7 7 7 The coe icients matrix is called Vandermonde rnatrix in 17 2 7m Read Textbook7 Example 1 47 p 29 32 Exercise 131 Ex 3 p 38Changed Determine the polynomial function of degree 2 that passes through the points 2437674710 Let pm a In m2 Since these points pass through the graph of y pm a In m2 we have a b2 022 4 a 2b 40 4 a b3 032 6 or a 3b 90 6 a b4 042 10 a 4b 160 10 The augmented matrix of this system is 32 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Now we reduce the matrix to the row echelon form To do this subtract row 1 from row 2 and row 3 1 2 4 4 0 1 5 2 0 2 12 6 Now7 subtract 2 times row 2 from row 3 1 2 4 0 1 5 2 0 0 2 Divide the last row by 2 1 2 4 0 2 0 0 1 The matrix is in row echelon form They linear system correspond ing to this matrix is 1 21 40 4 b 50 2 c 1 By back substitution 7 b275737 a47466 So pm a in 0x2 6 7 3m 2 Use your Tl to graph it Exercise 132 Ex 13 p 38Changed Some US census popula tion data is given in the following table Year 1980 1990 2000 population y 227 249 281 Here population is given in millions 13 APPLICATION OF LINEAR SYSTEMS 33 1 Fit a second degree polynomial passing through these points 2 Use it to predict popolation in year 2010 abd 2020 Solution Let t be the variable time and set if 0 for the year 1980 The table reduces to t 0 10 20 y 227 249 281 Let pt a bt ct2 be the polynomial that ts this data Since the data points pass through the graph of y pt a bt 0752 we have a b0 002 227 a 227 a b10 0102 249 or 1 101 1000 249 a b20 6202 281 1 201 4000 281 The augrnented matrix is 1 0 0 227 1 10 100 249 1 20 400 281 Now use Tl 84 or you can hand reduce to reduce the matrix to Gauss Jordan form So a 2271 1770 005 and y pt 227 17t05t2 This answers part For part 27 for year 20107 wehave t 30 and prediceted population is p30 227 17 gtk 30 05 gtk 302 323 mi 34 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Similarly for year 2020 wehave t 40 and prediceted population is p30 227 17 gtk 40 05 gtk 402 375 mi 132 Network Analysis A network consists of junctions and branches Following is an example of netwrok V av 13 X It is a connected diagram which I could not draw Look at Textbook p Such network systems are used to model diverse situations including in economics tra ic telephone signal and electrical engineer ing Such models assumes the total ow into a junction is equal to total ow out of the junction Accordingly above network is represented by m y 13 2 Read Textbook Example 5 7 p 33 Exercise 133 Ex 22 p 39 The ow of tra ic in vehicles per hour through a network of streets is shown in the following gure 300 9101 150 ml A l W TXW 1 Solve this system for 12345 2 Find the tra ic ow when 2 200 and 3 50 13 APPLICATION OF LINEAR SYSTEMS 35 3 Find the tra ic ow when 2 150 and 3 0 Solution From junction A7 we get 1 2 300 From junction B7 we get 131504 OR 1374150 From junction Y7 we get 220035 OR 273757200 From junction Z7 we get 4 5 350 We will write the system in a better way 1 2 300 1 3 in 150 2 in 7 7200 4 5 350 To solve thsi linear system7 we write the augmented matrix 1 1 0 0 0 300 1 0 1 71 0 150 0 1 71 0 71 7200 0 0 0 1 1 350 We will reduce this matrix to row echelon form Subtract row 1 from row 2 1 0 0 0 300 71 1 71 0 7150 71 0 71 7200 0 0 1 1 350 OOOH H 36 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS Add second row to third 1 1 0 0 0 300 0 1 1 1 0 150 0 0 1 1 350 0 0 1 1 350 Add third roe to fourth 1 1 0 0 0 300 0 1 1 1 0 150 0 0 0 1 1 350 0 0 0 0 0 Multiply second row by 1 and third row by 1 1 1 0 0 0 300 0 1 1 1 0 150 0 0 0 1 1 350 000000 The matrix is in row echelon form The corresponding linear system is given by 1 2 300 2 3 4 150 4 5 350 0 0 Pararnetrically7 With 2 if7 3 37 we have 1 300757 2 if7 3 3 4 150t57 5 3504 150t3 This answers For 2 t X2 2003 3 50 So7 1 1007 2 2007 3 507 4 07 5 300 For 3 75 X2 15073 3 0 So7 1 2 3 07 4 07 5 13 APPLICATION OF LINEAR SYSTEMS 37 133 Kirchhoff s Laws Sirnilarly7 system of Linear equations is also applicable in electrical network Analysis of electrical network is guided by two properties known as Kirchhoff 5 Laws 1 All the current owing into a junction rnust ow out of it 2 The sum of the products IR I is current and R is resistance around a closed path is equal to the total voltage A battery is denoted by lk or 4 and the resistance is denoted by Exercise 134 Ex 25 p 40 Consider the electrical circuit I1 3 1 mll 12 m R23 mlt R31 lb The circuit should be connected I could not draw a better one See Textbook Ex 25 p 40 for the actual circuit Use Kirchho Law to determine I17 I27 I3 Solution Apply 1 of Kirchho Law to junction Jl7 we have 1 13 2 1 Applying the same to J2 wil give the same equation So7 we will not write it 38 CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS NOW apply 2 of Kirchho Law R111 R2I2 3 411 3 R212 R313 1 312 13 1 The the network system is given by 1 712 13 0 1 The augmented matrix is 71 oqgtgta HOH 0 3 3 3 1 NOW7 we reduce this matrix to row echelon form To dothis7 rst sub tract 4 time rst reo from second 1 i1 1 0 0 7 i4 3 0 3 1 1 Divide row two by 7 1 i1 1 0 4 3 0 1 7 7 0 3 1 1 Subtract 3 times rWO two from row three 13 APPLICATION OF LINEAR SYSTEMS 39 Divide row three by Now7 we further reduce it to Gauss Jordan form To do this7 add second row to rst 9 10 0 E 4 3 01 7 2 0 0 1 7E Now7 add tirne third roe to second 9 10 0 E 7 010 E 0 0171 The corresponding linear system s given by7 ii 1 19 l

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