Elimination and back substitution

Transcription

1 Roberto s Notes on Linear Algebra Chapter 3: Linear systems and matrices Section 2 Elimination and back substitution What you need to know already: What a (linear) system is. What it means to solve such a system. What you can learn here: An elementary, but inefficient way to solve any linear system. As it is usual in mathematics, after we identify an interesting problem, in this case how to find the solutions set of a linear system, we want to develop a method for solving such problem. And in this section we shall review one such method. Is this the method we saw in high school? Yes, at least for those who have seen it and although you may have seen it by a different name or with no name at all. Here is how it works. Strategy for the method of Elimination and Back substitution The solution set of a linear system of the form: a x a x... a x c a x a x... a x c... a x a x... a x c n n n n 2 m1 1 m2 2 mn n m can be obtained by using the following process: 1. Solve one equation for any one variable, say x i, in terms of the others. You may pick any equation and any variable for which this process is easiest. 2. Eliminate the variable x i from the other equations by substituting the expression found for it in the previous step. 3. Solve another equation for another variable as in step Eliminate this second variable from the remaining equations by substituting its expression as in step Continue in this way until all equations have been used to solve for one variable, Linear Algebra Chapter 3: Linear systems and matrices Section 2: Elimination and back substitution Page 1

2 6. Substitute back the value or expression of the last variable in the other equations to obtain a value or expression for the previous variables. 7. Continue back substituting until the values for all variables have been obtained. The order of the equations may be changed at any stage, since it does not affect the solutions. Any duplicate equation may be eliminated, as it adds no useful information. This is complicated! It looked a lot simpler in high school! It is a lot simpler, in fact it is one of those methods that are easier done than said, especially when one uses small systems, so here are some examples to refresh your memory. Example: 3x2y 1 To solve the system we first solve the second equation for x: x 3y 5 3x2y 1 x 5 3y Then we substitute this value of x in the first equation: 3 5 3y 2y 1 x 53y Now the first equation only contains y, so we solve it for y: 14 11y 14 y 11 x 53y x 53y Finally, we use this value of y to find the value of x: 14 y 14 y x 5 3 x Therefore, the only possible solution of this system is: Example: To solve the system since its coefficient there is 1: 3x 5y 4z 0 3x 2y 4z 0 6x y 8z 0 3x 5y 4z 0 3x 2y 4z 0 y 8z 6x we solve the third equation for y, Then we substitute this value of y in the other equations: 3x 58z 6x 4z 0 27x 36z 0 3x 28z 6x 4z 0 9x 12z 0 y 8z 6x y 8z 6x 3x4z 0 3x 4z 0 y 8z 6x Now we eliminate one of the first two equations and solve the remaining one for x: Linear Algebra Chapter 3: Linear systems and matrices Section 2: Elimination and back substitution Page 2

3 4 3x4z 0 x z 3 y 8z 6x y 8z 6x Finally, we use this expression for x to find one for y: 4 x z 4 3 x z 3 4 y 8z 6 z y 0 3 Therefore, any solution will have y 0 and x and z related as conclusion, the solution set is the set of vectors of the form: 4 4 z 0 z z Example: To solve the system 3x 5y 4z 1 3x 2y 4z 2 6x 7 y 8z 3 y, since its coefficient there is the smallest one: x 4 3 z. In we solve the second equation for 3x 5y 4z 1 3x 5y 4z 1 2y 4z 3x 2 y 2z 1.5x 1 6x 7 y 8z 3 6x 7 y 8z 3 Then we substitute this value of y in the other equations: 3x 52z 1.5x 1 4z 1 4.5x 6z 6 y 2z 1.5x 1 y 2z 1.5x 1 6x 72z 1.5x 1 8z 3 4.5x 6z 10 But now the first and third equations have the same left side, but different and constant right side: there is no way of picking values for x and z so that both equations can be true! This means that this system has NO solutions. I can see that this method can be quite long and tedious. And it is very easy to make mistakes when doing it by hand, given all the variables, fractions and other numbers that one must move from one place to the other. There is one trick that can simplify the computations in many situations. I will show you now what the trick is and then we ll take full advantage of it in the next section and in the one after that and in many more settings. Are you saying that I should pay attention to this trick? You should pay attention to anything stated here, but YES! You should pay particular attention to this trick because it is going to be a central tool in what we ll do later. Technical fact Every step in the method of elimination and back substitution generates a new linear system with the same set of solutions as the previous one. In particular, at each step, the equations that make up the system are all linear. Well, that seems obvious! As any good trick does once you explain it! But it is a good trick that will also produce additional fruits and that s what the next sections of this chapter are about. But even now, notice what interesting consequence this simple fact has. Linear Algebra Chapter 3: Linear systems and matrices Section 2: Elimination and back substitution Page 3

4 Proof Technical fact A linear system can only have no solutions, one solution or infinitely many solutions. As we apply the method of elimination and back substitution, we change the original linear system to other linear systems with the same set of solutions. At the end of the process we can end up only with one of these possibilities: One of the equation presents an inconsistency, such as 1=0, in which case there is no solution Each equation states that one variable equals a value and there are no inconsistencies. This means that there is only one possible combination of values, hence only one solution. There are more remaining equations than variables, so that one variable is not required to take a specific value. Since there are infinitely many possible values for that variable, there are, correspondingly, infinitely many solutions for the system. If you look back at the examples of this section, or forward to the Learning questions, you will see that this is indeed the case. Summary We can solve any linear system by systematically isolating variables and eliminating variables. A linear system may have a single solution, infinitely many or none at all! Common errors to avoid The method of elimination and back substitution is a basic tool that is very popular at beginner s level. Although it is valid and effective, it is not efficient and you should look forward to learning better methods. Linear Algebra Chapter 3: Linear systems and matrices Section 2: Elimination and back substitution Page 4

5 Learning questions for Section LA 3-2 Review questions: 1. Describe how the method of elimination and back-substitution works. 2. Explain why a linear system cannot have three solutions only. Memory questions: 1. How many solutions can a linear system have? 2. What is eliminated in the method of elimination and back-substitution? Solve the systems presented in questions 1-5 by using elimination and back substitution. Computation questions: x 3y z 25 x 2y 4z 25 3x y 2z 2 3x y 2z 1 3x y 2z 5 x y z x 2z 1 3x y 4z 7 6x y z 0 3x y z 4 x 2y 3z 2 4x y 2z 6 5. x y 2z w 1 2x y 2z 2w 2 x y 4z w 2 4x 6z 4w 3 Linear Algebra Chapter 3: Linear systems and matrices Section 2: Elimination and back substitution Page 5

6 Theory questions: 1. Can the solution set of a linear system consist of 3 solutions? 2. Is it possible for a non-linear system to have more than one solution, but not infinitely many? Application questions: 1. An internet media outlet has obtained $30 million in revenue for the sale or rent of a certain movie. The movie could be rented for $6 or purchased for $15 and was acquired by 3,531,800 customers. Determine how many customers purchased the movie and how many rented it by constructing a suitable linear system and solving it with the method of elimination and back substitution. Templated questions: 1. Construct a system consisting of a no more than 5 equations and involving no more than 4 variables and solve it by using the method of elimination and back substitution. What questions do you have for your instructor? Linear Algebra Chapter 3: Linear systems and matrices Section 2: Elimination and back substitution Page 6