# Systems of Linear Equations

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1 A FIRST COURSE IN LINEAR ALGEBRA An Open Text by Ken Kuttler Systems of Linear Equations Lecture Notes by Karen Seyffarth Adapted by LYRYX SERVICE COURSE SOLUTION Attribution-NonCommercial-ShareAlike (CC BY-NC-SA) This license lets others remix, tweak, and build upon your work non-commercially, as long as they credit you and license their new creations under the identical terms. Systems of Linear Equations Page 1/51

2 Systems of Linear Equations Example A system of linear equations: x 1 2x 2 7x 3 = 1 x 1 + 3x 2 + 6x 3 = 0 variables: x 1, x 2, x 3. coefficients: 1x 1 2x 2 7x 3 = 1 1x 1 + 3x 2 + 6x 3 = 0 constant terms: x 1 2x 2 7x 3 = 1 x 1 + 3x 2 + 6x 3 = 0 Systems of Linear Equations Systems of Linear Equations Page 2/51

3 Example (continued) x 1 = 3, x 2 = 1, x 3 = 0 is a solution to the system x 1 2x 2 7x 3 = 1 x 1 + 3x 2 + 6x 3 = 0 as is x 1 = 6, x 2 = 0, x 3 = 1. However, x 1 = 1, x 2 = 0, x 3 = 0 is not a solution to the system. A solution to the system must be a solution to every equation in the system. The system above is consistent, meaning that the system has at least one solution. Systems of Linear Equations Systems of Linear Equations Page 3/51

4 Example (continued) x 1 + x 2 + x 3 = 0 x 1 + x 2 + x 3 = 8 is an example of an inconsistent system, meaning that it has no solutions. Why are there no solutions? Systems of Linear Equations Systems of Linear Equations Page 4/51

5 Graphical Solutions Example Consider the system of linear equations in two variables x + y = 3 y x = 5 A solution to this system is a pair (x, y) satisfying both equations. Since each equation corresponds to a line, a solution to the system corresponds to a point that lies on both lines, so the solutions to the system can be found by graphing the two lines and determining where they intersect. x + y = 3 y x = 5 ( 1, 4) Systems of Linear Equations Systems of Linear Equations Page 5/51

6 Given a system of two equations in two variables, graphed on the xy-coordinate plane, there are three possibilities, as illustrated below. intersect in one point consistent (unique solution) parallel but different inconsistent (no solutions) line are the same consistent (infinitely many solutions) Systems of Linear Equations Systems of Linear Equations Page 6/51

7 For a system of linear equations in two variables, exactly one of the following holds: 1 the system is inconsistent; 2 the system has a unique solution, i.e., exactly one solution; 3 the system has infinitely many solutions. (We will see in what follows that this generalizes to systems of linear equations in more than two variables.) Systems of Linear Equations Systems of Linear Equations Page 7/51

8 Example The system of linear equations in three variables that we saw earlier x 1 2x 2 7x 3 = 1 x 1 + 3x 2 + 6x 3 = 0, has solutions x 1 = 3 + 9s, x 2 = 1 + s, x 3 = s where s is any real number (written s R). Verify this by substituting the expressions for x 1, x 2, and x 3 into the two equations. s is called a parameter, and the expression x 1 = 3 + 9s, x 2 = 1 + s, x 3 = s, where s R is called the general solution in parametric form. Systems of Linear Equations Systems of Linear Equations Page 8/51

9 Problem Find all solutions to a system of m linear equations in n variables, i.e., solve a system of linear equations. Definition Two systems of linear equations are equivalent if they have exactly the same solutions. Example The two systems of linear equations 2x + y = 2 3x = 3 and x + y = 1 y = 0 are equivalent because both systems have the unique solution x = 1, y = 0. Systems of Linear Equations Systems of Linear Equations Page 9/51

10 Elementary Operations We solve a system of linear equations by using Elementary Operations to transform the system into an equivalent but simpler system from which the solution can be easily obtained. Three types of Elementary Operations Type I: Interchange two equations. Type II: Multiply an equation by a nonzero number. Type III: Add a multiple of one equation to a different equation. Systems of Linear Equations Elementary Operations Page 10/51

11 Elementary Operations Example Consider the system of linear equations 3x 1 2x 2 7x 3 = 1 x 1 + 3x 2 + 6x 3 = 1 2x 1 x 3 = 3 Interchange first two equations (Type I elementary operation): x 1 + 3x 2 + 6x 3 = 1 3x 1 2x 2 7x 3 = 1 2x 1 x 3 = 3 Multiply first equation by 2 (Type II elementary operation): 6x 1 + 4x x 3 = 2 x 1 + 3x 2 + 6x 3 = 1 2x 1 x 3 = 3 Add 3 time the second equation to the first equation (Type III elementary operation): 7x x 3 = 2 x 1 + 3x 2 + 6x 3 = 1 2x 1 x 3 = 3 Systems of Linear Equations Elementary Operations Page 11/51

12 Theorem (Elementary Operations and Solutions) If an elementary operation is performed on a system of linear equations, the resulting system of linear equations is equivalent to the original system. (As a consequence, performing a sequence of elementary operations on a system of linear equations results in an equivalent system of linear equations.) Systems of Linear Equations Elementary Operations Page 12/51

13 Solving a System with Elementary Operations Problem Use Elementary Operations to solve the system 2x + y = 4 x 3y = 1 Solution Add ( 2) times the second equation to the first equation. 2x + y + ( 2)x ( 2)(3)y = 4 + ( 2)1 x 3y = 1 The result is an equivalent system 7y = 2 x 3y = 1 Systems of Linear Equations Elementary Operations Page 13/51

14 Solution (continued) The first equation of the system, can be rearranged to give us 7y = 2 y = 2 7. Substituting y = 2 7 into second equation: ( ) 2 x 3y = x 3 = 1, 7 and simplifying, gives us x = = Therefore, the solution is x = 13/7, y = 2/7. The method illustrated in this example is called back substitution. Systems of Linear Equations Elementary Operations Page 14/51

15 The Augmented Matrix Represent a system of linear equations with its augmented matrix. Example The system of linear equations x 1 2x 2 7x 3 = 1 x 1 + 3x 2 + 6x 3 = 0 is represented by the augmented matrix [ (A matrix is a rectangular array of numbers.) Note. Two other matrices associated with a system of linear equations are the coefficient matrix and the constant matrix. [ ] [ ] , ] Systems of Linear Equations Gaussian Elimination Page 15/51

16 Elementary Row Operations For convenience, instead of performing elementary operations on a system of linear equations, perform corresponding elementary row operations on the corresponding augmented matrix. Type I: Interchange two rows. Example Interchange rows 1 and Systems of Linear Equations Gaussian Elimination Page 16/51

17 Elementary Row Operations Type II: Multiply a row by a nonzero number. Example Multiply row 4 by Systems of Linear Equations Gaussian Elimination Page 17/51

18 Elementary Row Operations Type III: Add a multiple of one row to a different row. Example Add 2 times row 4 to row Systems of Linear Equations Gaussian Elimination Page 18/51

19 Definition Two matrices A and B are row equivalent (or simply equivalent) if one can be obtained from the other by a sequence of elementary row operations. Systems of Linear Equations Gaussian Elimination Page 19/51

20 Row-Echelon Matrix All rows consisting entirely of zeros are at the bottom. The first nonzero entry in each nonzero row is a 1 (called the leading 1 for that row). Example Each leading 1 is to the right of all leading 1 s in rows above it. where can be any number Systems of Linear Equations Gaussian Elimination Page 20/51

21 Reduced Row-Echelon Matrix Example Row-echelon matrix. Each leading 1 is the only nonzero entry in its column. where can be any number Systems of Linear Equations Gaussian Elimination Page 21/51

22 Example Suppose that the following matrix is the augmented matrix of a system of linear equations. We see from this matrix that the system of linear equations has four equations and seven variables Note that the matrix is a row-echelon matrix. Each column of the matrix corresponds to a variable, and the leading variables are the variables that correspond to columns containing leading ones (in this case, columns 1, 3, 4, and 7). The remaining variables (corresponding to columns 2, 5 and 6) are called non-leading variables. We will use elementary row operations to transform a matrix to row-echelon or reduced row-echelon form. Systems of Linear Equations Gaussian Elimination Page 22/51

23 Solving Systems of Linear Equations Solving a system of linear equations means finding all solutions to the system. Method I: Gauss-Jordan Elimination 1 Use elementary row operations to transform the augmented matrix to an equivalent (not equal) reduced row-echelon matrix. The procedure for doing this is called the Gaussian Algorithm, or the Reduced Row-Echelon Form Algorithm. 2 If a row of the form [ ] occurs, then there is no solution to the system of equations. 3 Otherwise assign parameters to the non-leading variables (if any), and solve for the leading variables in terms of the parameters. Systems of Linear Equations Gaussian Elimination Page 23/51

24 Gauss-Jordan Elimination Problem Solve the system 2x + y + 3z = 1 2y z + x = 0 9z + x 4y = 2 Solution Systems of Linear Equations Gaussian Elimination Page 24/51

25 Solution (continued) Given the reduced row-echelon matrix x and y are leading variables; z is a non-leading variable and so assign a parameter to z. Thus the solution to the original system is given by x = s y = s z = s where s R. Systems of Linear Equations Gaussian Elimination Page 25/51

26 Solving Systems of Linear Equations Method II: Gaussian Elimination with Back-Substitution 1 Use elementary row operations to transform the augmented matrix to an equivalent row-echelon matrix. 2 The solutions (if they exist) can be determined using back-substitution. Systems of Linear Equations Gaussian Elimination Page 26/51

27 Gaussian Elimination with Back Substitution Problem Solve the system 2x + y + 3z = 1 2y z + x = 0 9z + x 4y = 2 Solution Systems of Linear Equations Gaussian Elimination Page 27/51

28 Solution (continued) This row-echelon matrix corresponds to the system x + 2y z = 0 y 5 3 z = 1 3, so y = z, x = 2y + z and thus x = 2( z) + z = z y = z Setting z = s, where s R, gives us (as before): x = s y = s z = s Systems of Linear Equations Gaussian Elimination Page 28/51

29 Problem Solve the system x + y + 2z = 1 y + 2x + 3z = 0 z 2y = 2 Solution The unique solution is x = 5 3, y = 4 3, z = 2 3. Check your answer! Systems of Linear Equations Gaussian Elimination Page 29/

30 Problem Solve the system 3x 1 9x 2 + x 3 = 9 2x 1 + 6x 2 x 3 = 6 x 1 + 3x 2 x 3 = 2 Solution The last row of the final matrix corresponds to the equation 0x 1 + 0x 2 + 0x 3 = 1 which is impossible! Therefore, this system is inconsistent, i.e., it has no solutions Systems of Linear Equations Gaussian Elimination Page 30/51

31 General Patterns for Systems of Linear Equations Problem Find all values of a, b and c (or conditions on a, b and c) so that the system 2x + 3y + az = b y + 2z = c x + 3y 2z = 1 has (i) a unique solution, (ii) no solutions, and (iii) infinitely many solutions. In (i) and (iii), find the solution(s). Solution 2 3 a b c c 2 3 a b Systems of Linear Equations Gaussian Elimination Page 31/51

32 Solution (continued) c c 2 3 a b 0 3 a + 4 b c c c 0 3 a + 4 b a 2 b 2 3c Case 1. a 2 0, i.e., a 2. In this case, ( ) c c 4 b 2 3c ( a 2) c c + 2 b 2 3c b 2 3c a a b 2 3c a 2 Systems of Linear Equations Gaussian Elimination Page 32/51

33 Solution (continued) ( ) c 4 b 2 3c ( a 2) c + 2 b 2 3c a b 2 3c a 2 (i) When a 2, the unique solution is ( ) ( b 2 3c b 2 3c x = 1 + 3c 4, y = c + 2 a 2 a 2 z = b 2 3c. a 2 ), Systems of Linear Equations Gaussian Elimination Page 33/51

34 Solution (continued) Case 2. If a = 2, then the augmented matrix becomes c c c c 0 0 a 2 b 2 3c b 2 3c From this we see that the system has no solutions when b 2 3c 0. (ii) When a = 2 and b 3c 2, the system has no solutions. Systems of Linear Equations Gaussian Elimination Page 34/51

35 Solution (continued) Finally when a = 2 and b 3c = 2, the augmented matrix becomes c c c c b 2 3c and the system has infinitely many solutions. (iii) When a = 2 and b 3c = 2, the system has infinitely many solutions, given by x = 1 + 3c 4s y = c + 2s z = s where s R. Systems of Linear Equations Gaussian Elimination Page 35/51

36 Uniqueness of the Reduced Row-Echelon Form Theorem Systems of linear equations that correspond to row equivalent augmented matrices have exactly the same solutions. Theorem Every matrix A is row equivalent to a unique reduced row-echelon matrix. Systems of Linear Equations Uniqueness Page 36/51

37 Homogeneous Systems of Equations Definition A homogeneous linear equation is one whose constant term is equal to zero. A system of linear equations is called homogeneous if each equation in the system is homogeneous. A homogeneous system has the form a 11 x 1 + a 12 x a 1n x n = 0 a 21 x 1 + a 22 x a 2n x n = 0. a m1 x 1 + a m2 x a mn x n = 0 where a ij are scalars and x i are variables, 1 i m, 1 j n. Notice that x 1 = 0, x 2 = 0,, x n = 0 is always a solution to a homogeneous system of equations. We call this the trivial solution. We are interested in finding, if possible, nontrivial solutions (ones with at least one variable not equal to zero) to homogeneous systems. Systems of Linear Equations Rank and Homogeneous Systems Page 37/51

38 Homogeneous Equations Example Solve the system x 1 + x 2 x 3 + 3x 4 = 0 x 1 + 4x 2 + 5x 3 2x 4 = 0 x 1 + 6x 2 + 3x 3 + 4x 4 = The system has infinitely many solutions, and the general solution is x 1 = 9 5 s 14 5 t x 2 = 4 5 s 1 5 t x 3 = s x 4 = t or x 1 x 2 x 3 x 4 = 9 5 s 14 5 t 4 5 s 1 5 t s t, where s, t R. Systems of Linear Equations Rank and Homogeneous Systems Page 38/51

39 Definition If X 1, X 2,..., X p are columns with the same number of entries, and if a 1, a 2,... a p R (are scalars) then a 1 X 1 + a 2 X a p X p is a linear combination of columns X 1, X 2,..., X p. Example (continued) In the previous example, x 1 x 2 x 3 x 4 = 9 5 s 14 5 t 4 5 s 1 5 t s t = 9 5 s 14 5 t 4 5 s s t 0 0 t = s t Systems of Linear Equations Rank and Homogeneous Systems Page 39/51

40 Example (continued) This gives us where X 1 = x 1 x 2 x 3 x = s and X 2 = + t = sx 1 + tx 2, The columns X 1 and X 2 are called basic solutions to the original homogeneous system. The general solution to a homogeneous system can be expressed as a linear combination of basic solutions. Systems of Linear Equations Rank and Homogeneous Systems Page 40/51

41 Example (continued) Notice that x 1 x 2 x 3 x 4 = s t = s 5 = r t 5 + q = r(5x 1 ) + q(5x 2 ) where r, q R. Systems of Linear Equations Rank and Homogeneous Systems Page 41/51

42 Example (continued) The columns 5X 1 = and 5X 2 = to the original homogeneous system are also basic solutions In general, any nonzero multiple of a basic solution (to a homogeneous system of linear equations) is also a basic solution. Systems of Linear Equations Rank and Homogeneous Systems Page 42/51

43 Problem Find all values of a for which the system x + y = 0 ay + z = 0 x + y + az = 0 has nontrivial solutions, and determine the solutions. Solution Non-trivial solutions occur only when a = 0, and the solutions when a = 0 are given by x y z = s 1 1 0, s R. Systems of Linear Equations Rank and Homogeneous Systems Page 43/51

44 Rank Definition The rank of a matrix A, denoted rank A, is the number of leading 1 s in any row-echelon matrix obtained from A by performing elementary row operations. Problem Find the rank of A = [ a b ]. Solution [ a b ] [ a b 5 ] [ b + 2a 5 a If b + 2a = 0 and 5 a = 0, i.e., a = 5 and b = 10, then rank A = 1. Otherwise, rank A = 2. ] Systems of Linear Equations Rank and Homogeneous Systems Page 44/51

45 What does the rank of an augmented matrix tell us? Suppose A is the augmented matrix of a consistent system of m linear equations in n variables, and rank A = r. 1 m } {{ } n } {{ } r leading 1 s Then the set of solutions to the system has n r parameters, so if r < n, there is at least one parameter, and the system has infinitely many solutions; if r = n, there are no parameters, and the system has a unique solution. Systems of Linear Equations Rank and Homogeneous Systems Page 45/51

46 Solutions to a System of Linear Equations For any system of linear equations, exactly one of the following holds: 1 the system is inconsistent; 2 the system has a unique solution, i.e., exactly one solution; 3 the system has infinitely many solutions. Systems of Linear Equations Rank and Homogeneous Systems Page 46/51

47 Problem Solve the system 3x 1 + 6x 2 4x 3 9x 4 + 3x 5 = 1 x 1 + 2x 2 2x 3 4x 4 3x 5 = 3 x 1 2x 2 + 2x 3 + 2x 4 5x 5 = 1 x 1 2x 2 + x 3 + 3x 4 x 5 = 1 Solution Begin by putting the augmented matrix in reduced row-echelon form The system has 5 variables, and the rank of the augmented matrix is 3. Since the system is consistent, the set of solutions has 5 3 = 2 parameters. Systems of Linear Equations Rank and Homogeneous Systems Page 47/51

48 Solution (continued) From the reduced row-echelon matrix we obtain the general solution x 1 = 9 + 2r + 13s x 2 = r x 3 = 2 x 4 = 2 4s x 5 = s r, s R The solution has two parameters (r and s) as we expected. Systems of Linear Equations Rank and Homogeneous Systems Page 48/51

49 Review Problem Problem The following is the reduced row-echelon form of the augmented matrix of a system of linear equations How many equations does the system have? 5 How many variables does the system have? 5 If the variables are labeled x 1, x 2, x 3, x 4, x 5, which variables are are the - leading variables? x 1, x 2, x 4 - non-leading variables? x 3, x 5 Systems of Linear Equations Review Problem Page 49/51

50 Problem (continued) What is the rank of this matrix? 3 How many parameters (if any) do we need for the general solution? 2 What is the system of equations corresponding to this matrix? x 1 3x 3 + 6x 5 = 8 x 2 + x 3 + 2x 5 = 0 x 4 + 5x 5 = 5 Is this system homogeneous or inhomogeneous? Inhomogeneous Systems of Linear Equations Review Problem Page 50/51

51 Problem (continued) x 1 3x 3 + 6x 5 = 8 x 2 + x 3 + 2x 5 = 0 x 4 + 5x 5 = 5 What is the general solution? Assign paramaters s, t R to the non-leading variables x 3 and x 5, respectively. x 1 = 8 + 3s 6t x 2 = s 2t x 3 = s x 4 = 5 5t x 5 = t The general solution may also be written as x 1 x 2 x 3 x 4 x 5 = s t , for s, t R. Systems of Linear Equations Review Problem Page 51/51

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