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# Class Note for MATH 3321 at UH

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Date Created: 02/06/15

53 Solving Systems of Linear Equations In this section we will develop a systematic method for solving systems of linear equations We7ll begin with a simple case two equations in two unknowns azby a CI dy B Our objective is to solve this system of equations This means that we want to nd all pairs of numbers I y that satisfy both equations simultaneously As you probably know there is a variety of ways to do this Well illustrate an approach which we7ll extend to systems of m equations in n unknowns Example 1 Solve the system l l on 31 12y 2m 7 3y l l l N SOLUTION We multiply the rst equation by 13 divide by 3 This results in the system z4y 2 2173y 77 This system has the same solution set as the original system multiplying an equation by a nonzero number produces an equivalent equation Next we multiply the rst equation by 72 add it to the second equation and use the result as the second equation This produces the system z4y 2 lly 711 As we will show below this system also has the same solution set as the original systeml Finally we multiply the second equation by 7111 divide by ill to get I4y y 17 and this system has the same solution set as the original Notice the triangular form of our nal systeml The advantage of this system is that it is easy to solve From the second equation y ll Substituting y l in the rst equation gives 1 41 2 which implies z 72 The system has the unique solution I 72 y ll I The basic idea is this given a system of linear equations perform a sequence of operations to produce a new system which has the same solution set as the given system and which is easy to solve 162 Two systems of equations are equivalent if they have the same solution set The Elementary Operations ln Example 1 we performed a sequence of operations on a system to produce an equivalent system which was easy to solve The operations that produce equivalent systems are called elementary operations The elementary operations are 1 Multiply an equation by a nonzero number 2 Interchange two equations 3 Multiply an equation by a number and add it to another equationi It is obvious that the rst two operations preserve the solution set that is produce equivalent systemsi Using two equations in two unknowns welll justify that the third operation also preserves the solution set Exactly the same proof will work in the general case of m equations in n unknowns Given the system azby a czdy 9 Multiply the rst equation by k and add the result to the second equation Replace the second equation in the given system with this new equation We then have ax by a kazczkbydy ka which is the same as am by a b kaczkbdy ka i Suppose that 10 yo is a solution of system a To show that a and b are equivalent we need only show that 10 yo satis es the second equation in system b 1m czo kl dy0 kazo 010 kbyo dye kazo kbyo 010 dye Hare bye 010 dye ka Thus zoy0 is a solution of Now suppose that 10 yo is a solution of system In this case we only need to show that 163 zoy0 satis es the second equation in a We have 1m czo kl dy0 ka B kazo kbyo 010 dye ka B Hare bye are dye ka B kaczody0 ka 010 dye Thus zoy0 is a solution of a I The following examples illustrate the use of the elementary operations to transform a given system of linear equations into an equivalent system Which is in a triangular form from Which it is easy to determine the set of solutions To keep track of the steps involved we7ll use the notations kEi A Ei to mean multiply equation by the nonzero number kl Ei lt gt Ej to mean interchange equations 239 and j kEi Ej A Ej to mean multiply equation by k and add the result to equation j77 The interchange equations77 operation may seem silly to you but you7ll soon see its value Example 2 Solve the system z2y752 fl 73179y212 0 z6yillz l SOLUTION We7ll use the elementary operations to produce an equivalent system in a triangular77 forml z2y752 fl I2y752 fl 73179y212 E E EgtE E E 73y62 73 3 gt 71 gt I6y7112 7 12 2 12 2 4y762 7 2 z2y752 fl I2y752 fl 0st yigz 1 413473gt E yigz 1 W2 4y762 2 2 3 3 22 72 z 2y 7 52 fl A y 7 22 l 1233433 2 fl From the last equation 2 ill Substituting this value into the second equation gives y fl and substituting these two values into the rst equation gives 1 74 The system has the unique solution I 74 y fl 2 ill 164 Note the strategy use the rst equation to eliminate z from the second and third equations Then use the new second equation to eliminate y from the third equation This results in an equivalent system in which the third equation is easily solved for 2 Putting that value in the second equation gives y substituting the values for y and 2 in the rst equation gives 1 I We continue with examples that illustrate the solution method as well as the other two possibil ities for solution sets no solution in nitely many solutions Example 3 Solve the system 3174yiz 3 2173y2 l zi2y32 2 SOLUTION Smellyiz 3 zi2y32 2 2173y2 l 2173y2 l E1lt gtEg zi2y32 2 Smellyiz 3 Having z with coef cient 1 in the rst equation makes it much easier to eliminate z from the remaining equations we want to avoid fractions as long as we can in order to keep the calculations as simple as possible This is the value of the interchange operationl I72y32 2 zi2y32 2 2x73y2 Smellyiz l l l w 7 EE 7gtE 7 EE7gtE y 2 2 3 1 2 M 1 g g 2y7102 73 zi2y32 2 2E E yiSZ 73 2 3 3 02 3 Clearly the third equation in this system has no solution Therefore the system has no solution Since this system is equivalent to the original system the original system has no solution I Example 4 Solve the system zy732 l 2zy742 0 7312yiz 7 SOLUTION zy732 7 l zy732 7 l 21y742 0 2E E FEE E E inZZ 72 1E 7312y7z 7 1 2 1 3 5y7102 10 gt292 165 zy732 1 zy732 1 7 y722 yigz 5E E E 5y7102 10 21393 02 0 Since every real number satis es the third equation the system has in nitely many solutions Set 2 a a any real number Then from the second equation we get y 2 2n and from the rst equation I 1 7 y 3n 1 7 2 2a 3n 71 a Thus the solution set is z 71 a y 2 2n 2 a a any real number If we regard a as a parameter then we can say that the system has a oneparameter family of solutions I In our examples thus far our systems have been square777 the number of unknowns equals the number of equations As we7ll see this is the most interesting case However haVing the number of equations equal the number of unknowns is certainly not a requirement the same method can be used on a system of m linear equations in n unknowns Example 5 Solve the system 117212zg714 72 72115127134z4 1 311 7 712 413 7 4x4 74 Note We typically use subscript notation when the number of unknowns is greater than 3 SOLUTION 117212zg714 72 117212zg714 72 72 5 7 4 2 3 11 12 13 14 E12E2HE2gt3E1E2HE3 12 13 14 31177124zg74z4 74 712zg7z4 2 117212zg714 72 117212zg714 72 2 7 2 73 E2Egt Eg x2 3 x4 IQE Eg 2 3 114 1 213 514 g This system has in nitely many solutions set 14 a a any real number Then from the third equation 13 7 7 al Substituting into the second equation we7ll get 12 and then substituting into the rst equation we7ll get 111 The resulting solution set is 11 7 7 3a 12 7g 7 3a 13 7 7 a I4 a a any real number This system has a oneparameter family of solutions I Some terminology A system of linear equations is said to be consistent if it has at least one solution that is a system is consistent if it has either a unique solution of in nitely many solutions A system that has no solutions is inconsistent 166 This method of using the elementary operations to reduce a given system to an equivalent system in triangular form and then solving for the unknowns by working backwards from the last equation up to the rst equation is called Gaussian elimination with back substitution I Matrix Representation of Systems of Equations Look carefully at the examples we7ve donei Note that the operations we performed on the systems of equations had the effect of changing the coef cients of the unknowns and the numbers on the righthand side The unknowns themselves played no role in the calculations they were merely placeholders With this in mind we can save ourselves some time and effort if we simply write down the numbers in the order in which they appear and then manipulate the numbers using the elementary operationsi Consider Example 2 The system is z2y752 71 73179y212 0 z6y7122 1 Writing down the numbers in the order in which they appear we get the rectangular array 1 2 75 71 73 79 21 0 1 6 712 1 The vertical bar locates the 77 sign The rows represent the equations Each column to the left of the bar gives the coefficients of the corresponding unknown the rst column gives the coef cients of 1 etc the numbers to the right of the bar are the numbers on the right sides of the equations This rectangular array of numbers is called the augmented matrix for the system of equations it is a shorthand representation of the systemi In general a matrix is a rectangular array of objects arranged in rows and columns The objects are called the entries of the matrix A matrix with m rows and n columns is an m X n matrix The matrices that we will be considering in this chapter will have numbers as entriesi 1n the next chapter we will see matrices with functions as entriesi Right now we are concerned with the augmented matrix of a system of linear equations Reproducing Example 2 in terms of augmented matrices we have the sequence 1 2 75 71 1275 71 1275 71 73 79 21 73 A 0 1 72 1 O O l w on 1 6 712 1 0 4 76 2 0 4 76 2 1 2 75 71 1 2 75 71 A 0 1 72 1 A 0 1 72 1 0 0 2 72 0 0 1 71 167 The nal augmented matrix corresponds to the system I 2y 7 52 71 y 7 22 1 2 71 from which we can obtain the solutions I 74 y 71 2 71 as we did before Its time to look at systems of linear equations in general A system of m linear equations in n unknowns has the form a1111 1212 1313 ainrn bl a2111 2212 2313 a2n1n b2 a3111 3212 3313 asnrn 1 3 amiri am212 amsrs amnrn bm The augmented matrix for the system is the m X n 1 matrix an L112 an am b1 L121 L122 L123 a2n b2 L131 L132 ass asn b3 am am am amn bm The m X n matrix whose entries are the coef cients of the unknowns all 0412 0413 39 39 39 aln 0421 0422 0423 39 39 39 a2n 0431 0432 0433 39 39 39 a3n aml amZ ams 39 39 39 amn is called the matrix of coe cientsi Example 6 Given the system of equations zl2zg 7313 7414 2 211412753 7714 7 731176124 11134 1414 0 The augmented matrix for the system is the 3 X 5 matrix 1 2 73 74 2 2 4 75 77 7 73 76 11 14 0 168 and the matrix of coef cients is the 3 X 4 matrix 1 2 73 74 2 4 75 77 i I 73 76 11 14 Example 7 The matrix is the augmented matrix of a system of linear equations What is the system SOLUTION The system of equations is 2173y2 6 5y722 71 731 42 10 7z2y722 3 I In the method of Gaussian elimination we use the elementary operations to reduce the system to an equivalent system in triangular formi The elementary operations on the equations can be Viewed as operations on the rows of the augmented matrixi Rather than using the elementary operations on the equations7 well use corresponding operations on the rows of the augmented matrix The elementary Tow operations on a matrix are 1 Multiply a row by a nonzero number 2 lnterchange two rows 3 Multiply a row by a number and add it to another rowi Obviously these correspond exactly to the elementary operations on equations We7ll use the following notation to denote the row operations Ri A Ri means multiply row by the nonzero number kl Ri lt gt Rj means interchange rows 239 and j kRi Rj A Rj means multiply row by k and add the result to row Now we7ll redo Examples 3 and 4 using elementary row operations on the augmented matrixi Example 8 Example 3 Solve the system 3174yiz 3 2173y2 1 172y32 2 169 SOLUTION The augmented matrix for the system of equations is 3 74 71 3 2 73 1 1 1 72 3 2 Mimicking Example 37 we have 3 74 71 3 1 72 3 2 2 73 1 1 R 2 73 1 1 2B R R831 R R 172 3 2 mg 374 713 l2gt212gt2 1 72 3 2 1 72 3 2 0 1 75 73 mi E 0 1 75 73 0 2 710 73 2 3 3 0 0 0 3 The last augmented matrix represents the system of equations zi2y32 2 Ozy75z 73 Oz0y02 3 As we saw in Example 37 the third equation in this system has no solution which implies that the original system has no solution I Example 9 Example 4 Solve the system zy732 1 2zy742 0 7312yiz 7 SOLUTION The augmented matrix for the system is 1 1 73 1 2 1 74 0 73 2 71 7 Performing the row operations to reduce the augmented matrix to triangular form7 we have 1 1 73 1 1 1 73 1 2 1 74 0 A 0 71 2 72 7H 73 2 71 7 2R1R2 gtR23R1R3 gtR3 0 5 710 1R2gtR2 1 1 73 1 1 1 73 1 0 1 72 2 is 0 1 72 2 0 5 710 10 MW 0 0 0 0 170 The last augmented matrix represents the system of equations zy732 1 Ozy72z Oz0y02 0 As we saw in Example 4 this system has in nitely many solutions given by z 71 a y 2 2a 2 a a any real number I Example 10 Solve the system of equations 1173122zg7142z5 2 311 79127zg7z4315 7 21176127zg4z4 7515 7 SOLUTION The augmented matrix is 1 73 2 71 2 2 3 79 7 71 3 7 2 76 7 4 75 7 Perform elementary row operations to reduce this matrix to triangular form 17327122 17327122 37977137 SRRVZRRR 0012731 2 76 7 4 75 7 M1272 11373 0 0 3 6 79 3 1 73 2 71 2 2 3R7 R 0 0 1 2 73 1 2 27gt g 0 0 0 0 0 0 The system of equations corresponding to this augmented matrix is 1173122zg7142z5 2 011012I32z4 7315 1 011012013Oz4 015 0 Since all values of the unknowns satisfy the third equation the system can be written equivalently as 1173122zg7142z5 2 zg2z47315 1 From the second equation 13 1 7 214 3151 Substituting into the rst equation we get 11 2312 72zgz4 7 2x5 312514 7 8151 If we set 12 a 14 I I5 cl Then the solution set can be written as 11 3a 512 7 Sc 12 a 13 17 2b Sc 14 I I5 c a b c arbitrary real numbers 171 The system has a threeparameter family of solutions I ROW Echelon Form and Rank The Gaussian elimination procedure applied to the augmented matrix of a system of linear equations consists of row operations on the matrix which convert it to a triangular formi Look at the examples above This triangular form is called the rowechelon form of the matrix A matrix is in rowechelon form if 1 Rows consisting entirely of zeros are at the bottom of the matrix 2 The rst nonzero entry in a nonzero row is a 1 This is called the leading 11 3 1f row i and row i1 are nonzero rows then the leading 1 in row i 1 is to the right of the leading 1 in row ii This implies that all the entries below a leading 1 are zero NOTE Because the leading 17s in the row echelon form of a matrix move to the right as you move down the matrix the number of leading 17s cannot exceed the number of columns in the matrix To put this another way the number of nonzero rows in the row echelon form of a matrix is always less than or equal the number of columns in the matrix Stated in simple terms the procedure for solving a system of m linear equations in n unknowns is 1 Write down the augmented matrix for the systemi 2 Use the elementary row operations to reduce the matrix to rowechelon formi 3 Write down the system of equations corresponding to the rowechelon form matrix 4 Write down the solutions beginning with the last equation and working upwardsi It should be clear from the examples we7ve worked that it is the nonzero rows in the rowechelon form of the augmented matrix that determine the solution set of the system of equations The zero rows if any represent redundant equations in the original system the nonzero rows represent the independent equations in the systemi If an m X n matrix A is reduced to row echelon form then the number of nonzero rows in its row echelon form is called the rank of Al It is obvious that the rank of a matrix is less than or equal to the number of rows in the matrix Also as we noted above the number of nonzero rows in the row echelon form is always less than or equal to the number of columns Therefore the rank of a matrix is also less than or equal to the number of columns in the matrix The last nonzero row of the augmented matrix usually determines the nature of the solution set of the systemi Case 1 1f the last nonzero row has the form 000 01b b7 0 then the row corresponds to the equation 011012013quot01nb 120 172 Which has no solutionsi Therefore the system has no solutionsi See Example 8 Case 2 1f the last nonzero row has the form 000 1 6 lb Where the 1 is in the 16 k lt n column and the 7s represent numbers in the k1St through the nth columns then the row corresponds to the equation 011012quot3901k71Ik1k139quotIn 12 Which has in nitely many solutions Therefore the system has in nitely many solutions See Examples 9 And 10 Case 3 1f the last nonzero row has the form 000 01112 then the row corresponds to the equation 01101201301n71zn 12 Which has the unique solution In 12 The system itself either Will have a unique solution or in nitely many solutions depending upon the solutions determined by the rows above Note that in Case 1 the rank of the augmented matrix is greater by 1 than the rank of the matrix of coef cients 1n Cases 2 and 3 the rank of the augmented matrix equals the rank of the matrix of coef cients Thus we have 1 1f rank of the augmented matrix rank of matrix of coef cients77 the system has no solutions the system is inconsistent 2 1f rank of the augmented matrix rank of matrix of coef cients77 the system either has a unique solution or in nitely many solutions the system is consistent Exercises 53 Each of the following matrices is the row echelon form of the augmented matrix of a system of linear equations Give the ranks of the matrix of coef cients and the augmented matrix and nd all solutions of the systemi 1 72 0 71 1i 0 1 1 2 0 1 71 1 72 1 71 2 0 1 1 2 0 0 0 0 1 1 2 3i 1 72 0 0 0 173 10 11 12 13 14 15 16 H 1 N O 71 CO OH OH LCM O H gt707 N 1 H 0 0 1 71 0 0 0 1 73 1 0 2 0 3 6 0 1 5 0 4 7 0 0 0 1 9 73 Solve the following systems of equations I72y78 2173y711 zz3 2y722741 y7225 z2y7321 215y78241 318y71327 zyz6 z2y429 21y6211 z2y72271 317y227 513y7422 7zy7272 31y210 4z2y3214 11212 7313 7414 2 211 412 7 513 7 7x4 7 7311 7 612 1113 1414 0 3116zg73143 1131271374z4712 11712I32148 2113128 174 17 18 19 20 21 22 23 24 25 1121221351411 21141221381414 1131241381419 117127132 1121273134142 2115127213141 51112127713614 7 112127137140 11212144 71172122134145 211 7 412 1613 71414 10 711 512 71713 1914 72 11 7 312 1113 71114 4 311 7 412 1813 71314 17 11 7 12 213 7 211 7 212 213 7 414 12 71112 713214 74 7311 12 7 813 71014 729 211 7 512 313 7 414 215 4 311 7 712 213 7 514 415 9 511 71012 7 513 7 414 715 22 Determine the values of k so that the system of equations has a unique solution7 no solutions7 in nitely many solutions 1y7z1 213yk23 1ky322 What condition must be placed on a 127 c so that the system 12y732a 216y7112b 172y7zc has at least one solution Consider two systems of linear equations having augmented matrices Al b1 and A l 122 Where the matrix of coef cients of both systems is the same 3 X 3 matrix A a Is it possible for Al 121 to have a unique solution and Al 122 to have in nitely many solutions Explain b Is it possible for Al 121 to have a unique solution and Al 122 to have no solution Explain c Is it possible for Al 121 to have in nitely many solutions and Al 122 to have no solutions Explain 175 26 Suppose that you wanted to solve the systems of equations z2y4za z3y2zb 2z3y112c for a 71 3 0 b 2 2 72 c 3 4 1 respectively Show that you can solve all three systems simultaneously by row reducing the matrix 1 2 4 71 3 0 1 3 2 2 2 72 2 3 11 1 176

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