Chapter 2. Systems of Equations and Augmented Matrices. Creighton University

Transcription

1 Chapter Section - Systems of Equations and Augmented Matrices D.S. Malik Creighton University

2 Systems of Linear Equations Common ways to solve a system of equations: Eliminationi Substitution

3 Elimination Methods Consider the following system of equations: x + y = 7 () x +y = () Eliminate one of the variables, say x. Multiply the first equation by and then add the resulting equation with the second equation:

4 x + y = 7 () x + y = () x 4y y = 4 x + y = y = 4 add That is, y = 4 or y = 4. Next we substitute the value of y into one of equations and solve for x. Let us substitute y = 4 in the first equation. Thus,

5 x + y = 7 () x + y = () x + y = 7 x + 4 = 7, substitute y = 4 x + 8 = 7 x = 7 8 x =. Thus, the solution of the system of equations is (, 4).

6 x + y = 7 () x + y = () Note that, to find a solution of the given system of equations, the given system of equations is replaced with the following system of equations: x + y = 7 y = 4 That is, the second equation is replace with Eq + Eq.

7 Substitution Method In the substitution i method one of the equations is solved for one of the variables in terms of the remaining variables. The value of that variable is substituted in the remaining equations to eliminate that variable from those equations. Let us consider the following system of equations. x y = 4 x y = 7

8 x y = 4 x y = 7 To use the substitution method, solve one of these equations for x or y. Suppose that we solve the first equation for x as follows: x y = 4 x = y + 4. Next we use substitute this value of x into the second equation and solve for y. Thus,

9 x y = 7 x y = 4 x y = 7 x = y + 4 (y y + 4) y y = 7, substitute x = y y + 4 4y + 8 y = 7 y +8 = 7 y = 7 8 y =.

10 x y = 4 x y = 7 x = y + 4 y = Now we substitute the value of y into the equation x = y + 4, to get x = ( ) + 4 = + 4 =. The solution of the given system of equations is (, ).

11 Systems of Equations: Consistent or Inconsistent Consistent t Consistent and dependent (infinite solutions) Consistent and independent (unique solution) Inconsistent- No solutions

12 Consider the following systems of equations: () (a) x + y = x y = This has a unique solution: (, ) (b) x y = 4 4x y = 8 This has infinite solutions. Second equation is multiple of the first equation.

13 (c) x y = 5 x 6y = 4 Multiply pythe first equation with, to get =, which is not true. So this system of equations has no solutions.

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15 Matrices and Systems of Equations A system of equations can be represented by a matrix. Consider the following system of equations: x + y = 7 x + y = Suppose we write the coefficient of x first and then the coefficient of y, we can express the given system of equations as follows:

16 7 x + y = 7 x + y = The numbers in the first row correspond to the numbers in the first equation. The numbers in the second row correspond to the numbers in the second equation. The numbers in the first column are the coefficients of x. The numbers in the second column are the coefficients of y. The numbers in the third column are the constant terms on the right side of the equations. We have drawn a straight line to separate the coefficients of the variables and the constant terms of the equations. This matrix is called the augmented matrix of the system of equations.

17 Consider the following system of equations in two variables: a x + a y = b a x + a y = b, a, a, a, a, b, and b are real numbers. The matrix A = a a a a consists of the coefficients of the variables is called the coefficient matrix of the system of equations.

18 Now consider the matrices a x + a y = b a x + a y = b, X = x b = and B y b These are column matrices. The matrix X contains the variables. The matrix B contains the constant terms on the right side of the equations

19 Let us evaluate the following matrix multiplication: AX = a a x a x + a = a a y a x + a y y b = B b = That is AX = B. So the given system of equations can be expressed as a matrix equation.

20 Example: Consider the system of equations: x 5y = 8 4x + 6y = 7 Here 8 5 x = = = 7 8 and, B y x X A The matrix equation is: = y x y

21 Consider the following system of n equations in n variables, ibl x, x,, x n : a x + + = + a x L a x b n n a x + a x + L + a x = b n n M a x + a x + L + a x = n n nn n b n

22 Let b n n b b B x x X a a a a a a A L L = = = n b B X A M M M M M M and,, n n nn n n b x a a a L The matrix A is called the coefficient matrix of the system of equations. y q The equation AX = B is called the matrix equation of the system of equations equation of the system of equations.

23 The matrix a a L a b n a M a M a a n n L a b n M L M a nn M b n is called the augmented matrix of the system of equations.

24 Example: Consider the following system of equations: x y 4z = x y + 7z = 9 5x +y z = 6 Then, the coefficient matrix is: A = 4 7 5

25 x y 4z = x y + 7z = 9 5x +y z = 6 The augmented matrix is:

26 Example: Consider the following system of equations x + 7y z + w = x x +z +6w = 7 y + 5z w = x +4 4y z +8 8w =. This can be written as: x + 7y z + w = x + y + z + 6w = 7 x + y + 5z w = x + 4y z + 8w =.

27 x + 7y z + w = x + y + z + 6w = 7 x + y + 5z w = x + 4y z + 8w =. The coefficient matrix is: The augmented matrix is: is: 7 7 = 5 6 A

28 Example: Suppose that the augmented matrix of a system of equations in variables x, y, and z is: The system of equations corresponding to this augmented matrix is: x 7y + 4z = x + z = 5 x + y 9z = 8.

29 In general, to solve a system of equations, we will write the augmented matrix of the system of equations and then use row operations to transform the augmented matrix in a form, called reduced row-echelon form. We will determine the solutions, if any of the system of equations from the reduced row-echelon form. Next, we discuss the reduced row-echelon from of a matrix and the row operations performed on an augmented matrix.

30 educed ow-echelon Form of a Matrix Consider the following matrices: 4, 4,, 5 5

31 In each row, the first nonzero entry, if any, is, called the leading. If two successive rows contain a leading, then the leading in the lower row is (farther) to the right of the leading in the upper row. The rows consisting of all the zero entries are grouped at the bottom. If a column contains a leading, then all other entries in that column are. All of these matrices are said to be in reduced row-echelon form.

32 Definition: A matrix is said to be in reduced row- echelon form if the following conditions are satisfied: i. If a row contains at least one nonzero entry, then the first nonzero entry of that row is. This is called the leading. ii. If two successive rows contain a leading, then the leading of the lower row is (farther) to the right of the leading of the upper row. iii.the rows consisting of all the zero entries are grouped at the bottom. iv. If a column contains a leading, then all other entries in that column are.

33 Suppose that a matrix that satisfies (i), (ii), and (iii) of previous definition and (iv ) Suppose that a column contains a leading, and it is in row, say i. Then all entries in this column below the ith row, if any, are. Then the matrix is said to be in row-echelon hl form. It follows that if a matrix is in reduced row- echelon form, then the matrix is in row-echelon form.

34 Example: The following matrices are in reduced rowechelon form echelon form.,, 5 4

35 Example: The following matrices are not in reduced rowechelon form , 4 4 In the first matrix, the leading in the fourth row is to the left of the leading one in the third row. In the second matrix, the third row consists of zeros and the fourth row has a nonzero entry. So the third row should be towards the bottom.

36 Gauss-Jordan Elimination Method Obtain the augmented matrix of the system of equations. Use elementary row operations, defined next, to find the reduced row-echelon form of the augmented matrix.

37 Let A = [a ij ] m n be an m n matrix. For each i, i =,,, m, let i denote the ith row of A. The elementary row operations we typically perform on A matrix are: T(): Interchange any two rows of the matrix, i.e., interchange i and j, written i j, where i =,,, m and j =,,, m. T(): eplace a row with a nonzero multiple of that row. That is, multiply l each element of a row i with a nonzero number, written as a i i, where a is a nonzero number and i =,,, m. T(): eplace a row with the sum of that row and a multiple of another row, written a i + j j, where a is a nonzero number and i =,,, m and j =,,, m.

38 Note the following: T() corresponds to interchanging two equations. T() corresponds to multiplying both sides of an equation by a nonzero number. T() corresponds to replacing an equation by the sum of that equation and a multiple of another equation.

39 Example: Consider the following matrix: Apply to get:

40 Example: Consider the following matrix: Apply the row operations (/), i.e., di id h l t f th fi t b t t 5 divide each element of the first row by, to get: 5

41 Example: Consider the following augmented matrix: 4 8 Let us apply the row operations ( ) +. We multiply each element of with, add the corresponding elements of and, and finally replace with the new elements. That is, 4 6 ( ) ( ) +

42 The resulting matrix is: 5 4 Note that in this matrix, the elements of rows and are unaffected. Also, if you write the system of equations corresponding to this matrix, then in the second equation, the coefficient i of x is, i.e., the variables x is eliminated.

43 5 4 Next, if you apply the row operation () + to the previous matrix, then the resulting matrix is: If you write the system of equations corresponding to If you write the system of equations corresponding to this matrix, then in the second and the third equations, the coefficient of x is, i.e., the variables x is eliminated. So, the last two rows correspond to the system of equations in two variables.

44 In order to find the reduced row-echelon form of a matrix we make the leading entry, if any, in each row. Consider the following matrix: In this matrix, in the first row, the first nonzero entry is and all other elements of the first row are multiples. So in the first row, we can make the leading entry by dividing all the elements by, i.e., by applying the row operation (/).

45 Now consider the following matrix: In this matrix, the first nonzero entry in the first row is. However, not all the elements of the first row are multiples of. So if we divide the elements of row by, then some of the elements will be fractions. Sometimes we have no choice and we must work with fractions. However, sometimes, we can make the leading entry in a row and avoid idfractions in that trow by using rows below that row. For example, in the previous matrix, we can apply the row operation and make the leading entry in first row.

46 In general, to find the reduced row-echelon form we start with the first row and make the leading entry in this row a. Then using the leading entry, we make all other entries in the column containing the leading entry. We continue this process. To make the leading entry in a nonzero row a, we always use that row and the rows below, if any, not the rows above, if any.

47 Example: Consider the following system of equations: x + y z = x + y + z = x + y + z = 4 The augmented matrix for this system of equations is: 4

48 Next, we apply row operations to find the reduced row-echelon from of this matrix. 4 4

49 4 4 ) ( () ) ( ( )

50 ) ( ) ( ( /) 7 5 ( /) 7

51 7 5 5) ( + 7) ( (5) + + The system of equations corresponding to this augmented matrix is: x =, y =, z =., y, The solution of the given system of equations is ( ) is (,, ).

52 Example: Consider the following system of equations: x y + z = 4 x y + z = 4x 5y +8 8z = The augmented matrix corresponding to this system of equations is: Next, we find the reduced row-echelon form of this augmented matrix.

53 ) ( 4 4) ( ) ( (/) 4 6 6

54 ) ( () ) ( This matrix is in reduced row-echelon form. The system of equations corresponding to The system of equations corresponding to this augmented matrix is: x + 7z = y + z =

55 x + 7z = y + z = We have two equations and three variables. The system of equations has at least one solution. This is the case when the system of equations has infinite solutions. We assume one of the variables as a free variable, i.e., which can assume any real number as its value. Let us assume z as the free variable and write z = r, where r is any real number. Next, we solve the remaining equations in terms of r.

56 We have: x = 7z = 7r y = z = r x + 7z = y + z = Hence, solutions of the given system of equations are of the form ( 7r, r, r), where r is any real number. If r =, then the solution is ( 7r, r, r) = ( 7,, ) = (,,). If r =, then the solution is ( 7r, r, r) = ( 7,,) = ( 7,, ) = ( 7, 4, ). Hence, two particular solutions are: (,, ) and ( 7, 4, ).

57 To find the solutions of a consistent and dependent system of equations, we assume some of the variables as the free variables and then solve the remaining variables in terms of the free variables. We can assume any of the variables as the free variable. For example, in the previous, we assumed z as the free variable. However, we could have assumed x or y as the free variable. To be consistent, as well as keeping the reduced row-echelon form in mind, we will assume the variables occurring towards the end of the equation as the free variables.

58 Example: Suppose that the reduced row-echelon form of a system of equations in variables x, y, and z is: 4 The system of equations corresponding to this augmented matrix is: x y + 4z = This system of equations has infinite solutions. We have one equation and three variables. So we will have = free variables. We assume that y and z are the free variables.

59 x y + 4z = Let y = r and z = s, where r and s are any real numbers. Then x = + y 4z = + r 4s. The solutions of the given system of equations are of the form ( +r 4s, r, s), where r and s are any real numbers.

60 Example: Suppose that the reduced row-echelon form of a system of equations in variables x, y, and z is: 5 The system of equations corresponding to this augmented matrix is: x + y = 5 z =

61 x + y = 5 z = This system of equations has infinite solutions. We have two equations and three variables. So we will have = free variable. Note that z has a specific value, which is. We assume that y is a free variable. Let y = r, where r is any real number. Then x = 5 y = 5 r. The solutions of the given system of equations are of the form (5 r, r, ), where r is any real number.

62 Example: Suppose that the reduced row-echelon form of a system of equations in variables x, y, and z is: 5 The system of equations corresponding to this augmented matrix is: y = 5 z =

63 y = 5 z = This system of equations has infinite solutions. We have two equations and three variables. So we will have = free variable. Note that both y and z have specific values. In this case, we assume that x is a free variable. Let x = r, where r is any real number. The solutions of the given system of equations are of the form (r, 5, ), where r is any real number.

64 Example: Suppose that the reduced row-echelon form of a system of equations in variables x, y, z, and w is: 5 4 The system of equations corresponding to this augmented matrix is: x + y + 5w = z + w = 4

65 x + y + 5w = z + w = 4 This system of equations has infinite solutions. We have two equations and four variables. So we will have 4 = free variables. The second equations is in terms of z and w. So we assume w as one of the free variables. The first equation contains the variables x, y and w. We have already assumed that w is a free variable. We assume that y is also a free variable.

66 x + y + 5w = z + w = 4 Let y = r and w = s, where r and s are any real numbers. Then x = y 5w = r 5s z = 4 w = 4 s. The solutions of the given system of equations are of the form ( r 5s, r, 4 s, s), where r and s are any real numbers.

67 Example: Consider the following system of equations x + y 4z = x y z = 5x 9y + z = The augmented matrix corresponding to this system of equations is: 4 5 9

68 To determine solutions of the given system of equations, we find the reduced row-echelon form of this augmented g matrix ( ) 4 9 5

69 ) ( ) ( ) ( 8 6 8/ 4 (/) 8 6 8/ 8 6

70 () + ( 6) + 5 8/ 9 4 8/ 6 8 This matrix is in reduced row-echelon form. The system of equations corresponding to this matrix is: x 5z =, y z = 8/, = 9 From the last equation, we have = 9, which is false. Hence, the given system of equations has no solutions.

71 When obtaining the reduced row-echelon form of the augmented matrix, if you come across a matrix in which all entries, except the last entry, of a row are zero, you can stop obtaining the reduced rowechelon form. Because in this case, the given system of equations is inconsistent. That is, it has no solutions