1. Systems of Linear Equations

Transcription

1 CHAPTER Systems of Linear Equations 1. Systems of Linear Equations Solution of an equation with one variable: is a solution of x + 6 = 16 since () + 6 = 16 is a true statement. Solution of an equation with two variables: (, 7) is a solution of x y = since () 7 = is a true statment. We might also say that x = and y = 7 is a solution. Also, (, ) is a solution of x = (thought of as x + 0y = ) since = is a true statement. Definition. A system of equations in two variables is a collection of two or more linear equations with two variables. The solution to the system of equations, if it exists, is the point(s) of intersection of the graphs of the equations. In other terms, a solution is an ordered pair that satisfies each of the equations in the system. Note. This definition is easily generalized to more variables. 9

2 0. SYSTEMS OF LINEAR EQUATIONS Graphical Method. It is hard to be accurate here. Example. (1) x y = () x + y = The solution appears to be x = 4, y = 1 or (4, 1), but we need to check to be sure, because maybe it is (.98, 1.0). There is no visual way to tell. 4 (1) = = The algebra convinces us that we have the correct solution. These equations are consistent (and independent), i.e., they have a single solution.

3 Example. 1. SYSTEMS OF LINEAR EQUATIONS 1 (1) x + y = 4 () x + 4y = 8 The graphs do not intersect since the lines are parallel. Thus there is no solution. These equations are inconsistent.

4 . SYSTEMS OF LINEAR EQUATIONS Example. (1) x + 4y = 8 () x + y = 4 Both equations simplify to y = 1 x +, so they have the same graph. These equations are consistent and dependent, i.e., there are infinitely many solutions. You can say the solution is all points on the line x + 4y = 8 or all points on the line x + y = 4.

5 1. SYSTEMS OF LINEAR EQUATIONS Example. Solve using the calculator: (1) 6x y = 10 () x + y = 4 We need to solve these equations for y: (1) y = 6x 10 =) y = x () y = x + 4 Find the approximate solution by using the Technology Tip on page. We get x =.444, y = The exact solution is x = 7 11, y = Substitution Method. This method is appropriate for cases where a variable has a coe cient of ±1. Solve for that variable in that equation, and then substitute into the other equation. Example. x y = 7 x + y = 1 x 7 = y x + (x 7) = 1 x + 9x 1 = 1 11x = x = y = () 7 y = 1

6 4. SYSTEMS OF LINEAR EQUATIONS Elimination (by Addition). Example. 4x + y = 6 ( 11) To give opposite x 11y = 7 ( ) coe cients for y. 44x + y = 86 9x y = 1 x = 6 x = Substitute this value into any previous equation with two variables to get y: 4() + y = y = 6 y = 6 y = Example. x + y = ( ( )) 4x + 10y = 4x 10y = 6 4x + 10y = 0 = 8 When a problem reduces to a false numerical equation, there is no solution, like the second graphical example. When a problem reduces to a true numerical equation, the two lines coincide, like the third graphical example.

7 Example. 1. SYSTEMS OF LINEAR EQUATIONS (1) y =.x + 1 () y = 4x + 18 () y = x + 6 We need to find points common to all three lines. We first pick any two lines to see if there is a solution. If not, the problem has no solution. We will choose equations (1) and () and use substitution..x + 1 = 4x + 18 x + 4 = 16x x = 68 x = 4 Now substitute this value for x into one of the two equations we chose. We substitute into equation (). y = 4(4) + 18 y = y = We have now found a solution for (1) and (). We now substitute this into () to see if we get a true statement. = (TRUE) Since we get a true statement, (4, ) is the solution of this system of three equations. If we had gotten a false statement from (), such as if () was y = x + 10, then there is no solution.

8 6. SYSTEMS OF LINEAR EQUATIONS Problem (page 7 #8). t = # of years since the school year. F (t) = # of Female athletes in year t. F (t) = 0.104t million M(t) = # of Male athletes in year t. M(t) = t +. million According to the model, when will the number of female athletes surpass the number of male athletes? Solve Use substitution: y = 0.104t y = t t = t t = 1. t = = 4.6 Thus the number of female athletes would surpass the number of male athletes in the 0-6 school year if the model continued to hold.

9 . USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 7. Using Matrices to Solve Linear Systems of Equation Definition. A = is a matrix, an array of numbers in rows 4 1 and columns. The plural of matrix is matrices. The 1 numbers are the entries of A. A is a 4 matrix since it has rows and 4 columns. An n n matrix (same number of rows as columns) is a square matrix. B = is a square matrix The 4 1 matrix 6 7 4is a column matrix. 4 The 1 matrix is a row matrix. The entries of a 4 matrix A may be represented as A = 4 a 11 a 1 a 1 a 14 a 1 a a a 4 a 1 a a a 4 a means the entry in row, column.

10 8. SYSTEMS OF LINEAR EQUATIONS Particular System General System x + 8y = 1 a 11 x + a 1 y = k 1 x y = 6 a 1 x + a y = k a 1 means nd equation, coe cient of the 1st variable. k 1 means constant or RHS (right-hand side) of 1st equation. Coe cient matrices: 8 Constant or RHS matrices: 1 6 a11 a 1 a 1 a k1 Augmented matrices: We will often skip the vertical bar. k a11 a 1 k 1 a 1 a k Since an augmented matrix clearly represents a system of linear equations, we can work with these matrices to solve the equations. Equivalent Systems of Equations The following operations on a system of equations result in a system of equations with the same solution as the original system: (1) Interchange (change the position) of two equations. () Multiply an equation by a nonzero number. () Add a nonzero multiple of one equation to a nonzero multiple of another equation and replace either equation with the result.

11 . USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 9 Related to this, we have the following. Row Operations For any augmented matrix of a system of equations, the following row operations yield an augmented matrix of an equivalent system of equations: (1) Interchange (change the position) of two rows. () Multiply a row by a nonzero number. () Add a nonzero multiple of one row to a nonzero multiple of another row and replace either row with the result. Our goal : A reduced matrix, a matrix in reduced row echelon form (RREF). Definition. An augmented matrix A is in RREF if: (1) The leading entry (first nonzero entry) in each row is a 1. () The leading entry in each row is the only nonzero entry in its corresponding column. () The leading entry of each row is to the right of the leading entry in the row above it. (4) All rows of zeros are at the bottom of the matrix. Problem (page 89 #1 0 even). (1) RREF (14) not RREF The leading entry in row does not have 0 s in the rest of the column the is the problem. (16) RREF (18) RREF (0) not RREF the leading entry in row 1 is not 1

12 40. SYSTEMS OF LINEAR EQUATIONS In general: With two equations, our goal is 1 0 where 0 1 can be any number of columns; With three equations, our goal is where can be any number of columns; With four equations, our goal is where can be any number of columns; 7 and so on. We need to make 1 s and 0 s. To make 1 s, we swap rows (row operation 1) or multiply a row by a nonzero number (row operation ). To make 0 s, we use (high school) elimination (row operation ).

13 Example.. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 41 x 4y = 8 x + y = Our goal is Making a 0 R 1 + R! R Making a 1 (1/17)R! R R + R 1! R 1 Making a (1/)R 1! R 1 Making a 1 x = 0 y =

14 4. SYSTEMS OF LINEAR EQUATIONS Example. x y + z = x + y z = 7 x y + z = Our goal is R $ R 1 Making a 1 1 R 1 $ R Making 0 s R 1 + R! R R 1 + R! R

15 . USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION R 1 + R! R 1 Making 0 s R + R! R Making 1 s R! R R + R 1! R 1 7R + R! R Making 0 s (1/10)R 1! R 1 (1/10)R! R Making 1 s x = y = 0 z = 1

16 44. SYSTEMS OF LINEAR EQUATIONS Example. x + y + z = 1 x y z = x + y z = Our goal is R $ R 1 R 1 $ R making 1 s Making 0 s R 1 + R! R R 1 + R! R Making 1 s 0 1 (1/)R! R 0 9 1

17 . USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION R + R 1! R 1 Making 0 s R + R! R How do we find a solution in a case like this where we have lost a row? Finding Solutions from Reduced Matrices: Example. (1) x y z Every column except the last is associated with a variable. x = y = 0 z = This is the kind of case with a single solution. () Look to the previous example: x z = y + z = Solve each equation for the leftmost variable, assigning parameters (such as t or s) to variables that are not associated with leading entries, i.e. are not leftmost variables in any equation.

18 46. SYSTEMS OF LINEAR EQUATIONS x = z + y = z = t Then, for the general solution, replace each variable to the right of the = sign by its parameter. z x = t + y = z = t Cases with parameters have infinitely many solutions. To get some specific solutions, replace the parameters by various real numbers. () t t = 0 t = 1 t = t = x = x = x = 7 x = y = y = y = 1 y = 14 z = 0 z = 1 z = z = Our goal was x + y = 6 z = 4

19 . USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 47 x = 6 y y = t z = 4 x = 6 t y = t z = 4 (4) Our goal was The second row represents 0x + 0y + 0z = 1, which is impossible. As soon as you encounter a row where the only nonzero number is the last number, you conclude there is no solution. () w x y z Our goal was

20 48. SYSTEMS OF LINEAR EQUATIONS w x + z = 4 y + z = 1 w = 4 + x z x = s y = 1 z z = t The general solution is w = 4 + s t x = s y = 1 t z = t Two particular solutions are s = 0, t = 1 s = 1, t = 0 w = 1 w = 6 x = 0 x = 1 y = y = 1 z = 1 z = 0 Problem (page 90 #6). 9x 6y = 0 4x + y = x z = 6

21 USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION R $ R 1 Making 1 s R 1 $ R Making 0 s 4R 1 + R! R 9R 1 + R! R Making 1 s R R! R a di erent way Making 0 s 6R + R! R

22 SYSTEMS OF LINEAR EQUATIONS Making 1 s ( 1/69)R! R R 1 + R! R 1 1R + R! R making 0 s Using the TI: x = y = z = 4 (1) Enter the matrix above using the Technology Tip on page 87. () Find the reduced version of the matrix by using the Technology Tip on page 88. Example. 4x y + z = 6x + y z = 10x y + 9z = 4

23 USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION (1/4)R 1! R Making 1 s Making 0 s R 1 + R! R We can stop here and say no solution since is impossible. Problem (page 91 #4). x 4y + z = 6 x y + z = x + 0y + 0z = 11 Our goal is

24 SYSTEMS OF LINEAR EQUATIONS (1/)R 1! R 1 Making 1 s Making 0 s R 1 + R! R R 1 + R! R 1 Making 0 s (1/)R 1! R 1 (1/)R! R Making 1 s x y z = 1 4 z = 14 x = z 1 y = 4 z 14 z = t

25 . LINEAR SYSTEM APPLICATIONS x = t 1 y = 4 t 14 z = t. Linear System Applications Problem (page 99 #). (a) Go to STAT/EDIT and enter t in L1, M in L, and C in L. Turn on Plot1 as usual. Turn on Plot with L for Ylist and + for Mark. Set a WINDOW as [ 1, 0] [ 1, 1]. Go to Y= and clear any functions.. Hit GRAPH to see t on the x-axis and M and C on the y-axis, M with and C with +. Hit STAT/CALC/4:LinReg(ax+b)/VARS/Y-VARS/1:Function/ 1:Y1/ENTER. Hit STAT/CALC/4:LinReg(ax+b)/nd 1/,/nd /,/VARS/Y-VARS/ 1:Function/:Y/ENTER. Now hit GRAPH to see the two regression lines. (b) By rounding, M(t) =.18t and C(t) =.t +.1. We solve: M(t) = C(t).18t =.t +.1.t = 8.7 t = Or, if using the TI, assuming the recent graph window is showing, hit nd Calc/:intersect/ENTER/ENTER/ENTER.

26 4. SYSTEMS OF LINEAR EQUATIONS The output of the two models will be equal about two months into (c) As per capita margarine consumption decreased, mozzarella cheese consumption has increased at almost twice the rate and has surpassed margarine consumption in Problem (page 10 #). x = # of Corollas bought. y = # of Camry s bought. z = # of Prius bought. x + y + z = 0 x = y x + y + z = 0 x y = ! (RREF ) x + 6 z = y z = x = 6 z

27 . LINEAR SYSTEM APPLICATIONS y = z = t 1 6 z x = y = z = t 6 t 1 6 t We need x, y, and z to be whole numbers greater than or equal to 0 (and less than or equal to 0). Thus t is limited to multiples of 6. t x y z These are all possible values. Problem (page 104 #8). x = # of chicken bags. y = # of lamb bags. z = # of turkey bags. 8x + 8y + 8z = 10 (amount).(8x) +.8(8y) +.6(8z) =.7(10).0(8x) +.0(8y) +.0(8z) =.0(10) (protein) (fiber)

28 6. SYSTEMS OF LINEAR EQUATIONS Both the first and third equations reduce to x + y + z = 1, so this is really a two equation system. The second equation simplifies to x +.4y +.08z = ! (RREF ) x + z = y + 1 z = 10 x = y = 10 z = t x = y = 10 z = t z 1 z t 1 t Now x, y, and z must be integers greater than or equal to 0, so z must be a multiple of. Also, Possible solutions are: cost = 7.99(x + y) z.

29 . LINEAR SYSTEM APPLICATIONS 7 x y z cost Problem (page 106 #40). Variables are given in the diagram. Note that cars entering each intersection = cars leaving each intersection (hopefully). Rewriting, (N + E) x = x 1 (E + S) x 1 = x + 10 (S + W ) x + 0 = x + 40 (W + N) x = x x 1 x 4 = 400 x 1 x = 10 x x = 10 x x 4 = ! (RREF ) >< x 1 x 4 = 400 x x 4 = 0 >: x x 4 = 10! 8 x 1 = x >< x = x x = x >: x 4 = t! 8 x 1 = t >< x = t + 0 x = t + 10 >: x 4 = t

30 8. SYSTEMS OF LINEAR EQUATIONS t = 100 t = 00 x 1 = 00 x 1 = 900 x = 0 x = 70 x = 0 x = 60 x 4 = 100 x 4 = 00