MA 3280 Lecture 05 - Generalized Echelon Form and Free Variables. Friday, January 31, 2014.

Transcription

1 MA 3280 Lecture 05 - Generalized Echelon Form and Free Variables Friday, January 31, Objectives: Generalize echelon form, and introduce free variables. Material from Section 3.5 starting on page 64 of the Schaum s Outline book. Consider this example from the book. (1) 2x 1 + 6x 2 x 3 + 4x 4 2x 5 = 15 x 3 + 2x 4 + 2x 5 = 5 3x 4 9x 5 = 6 We will call the first unknown/variable in each of these equations (i.e. the first unknown/variable with a non-zero coefficient) a leading unknown/variable. The book uses the term unknown, and we ll use the terms unknown and variable interchangeably. I ll probably start saying variable most of the time. In particular, x 1 is the leading variable for the first equation, x 3 is the leading variable for the second equation, and x 4 is the leading variable for the third. We can also say that x 1, x 3, and x 4 are pivot variables. It won t be a big deal to me, but there is a slight distinction between these terms. Leading refers to a particular equation, and pivot refers to the system as a whole. The book describes this system of equations as being in echelon form, and we will too. A system of equations is in echelon form if (1) No equation is degenerate (i.e., of the form 0 = c with c 0). (2) Each leading variable lies to the right of the leading variables above it. (3) (optional) Each leading variable has coefficient 1. For a system in echelon form, the variables that are not pivot/leading variables are called free variables. This definitely looks different from the examples we ve looked at so far. There are more variables than equations. We can see what s going on by trying to do some back-substitution, and solving for our pivot variables. In the last equation, we can solve for x 4 to get (2) x 4 = 2 + 3x 5. Substituting this into the second equation, we get (3) x 3 + 2(2 + 3x 5 ) + 2x 5 = 5, and so (4) x 3 = 1 8x 5. And finally, we can substitute these values into the first equation to get (5) 2x 1 + 6x 2 (1 8x 5 ) + 4(2 + 3x 5 ) 2x 5 = 15, and then solving for x 1 to get (6) x 1 = 4 3x 2 9x 5. We have all of our pivot variables solved in terms of our free variables. It s common to think of the free variables as parameters, and give them different names, like x 2 = a and x 5 = b. Then (7) x 1 x 2 x 3 x 4 x 5 = 4 3a 9b = a = 1 8b = 2 + 3b = b In this formulation, x 2 and x 5 are free in that they are free to take any value, and the pivot variables depend on them. So no matter what values a and b take, the 5-tuple (8) (4 3a 9b, a, 1 8b, 2 + 3b, b) 1

2 MA 3280 Lecture 05 - Generalized Echelon Form and Free Variables 2 is a solution to the system of equations. For example, pick any two numbers for a and b, like a = 2 and b = 3, and you ll have a solution to the system of equations, (9) (4 3(2) 9( 3), (2), 1 8( 3), 2 + 3( 3), ( 3)) = (25, 2, 25, 7, 3). All of the solutions to this system will be a plane in 5-dimensional space. Basic Principle 1. Given a system of linear equations, if you can get it into echelon form, then that system will have solutions. The dimension of the solution space will correspond to the number of free variables. A 0-dimensional solution space is a single point, a 1-dimensional solution space is a line, a 2-dimensional solution space is a plane, and in general, an n-dimensional solution space is an n-dimensional hyper-plane (we ll talk about this more later). Example 1. Let s look at a few really simple examples. Consider the system (10) 4x 1 + 6x 2 = 12. You can see that the second equation is just a multiple of the first, so there s really only one equation here. Let s get this to echelon form. We can avoid fractions by not worrying about getting 1 s on the leading variables, so let s do that. In that case, our first operation is 2 E 1 + E 2, which gives us (11) 0x 1 + 0x 2 = 0. The second equation is 0 = 0, which is OK, and we re in echelon form. We have that x 1 is a pivot variable, and x 2 is free. Solving the first equation for x 1 gives us (12) x 1 = x 2. We have one free variable, so our solution set will have dimension 1, a line. The points on this line will take the form ( (13) 3 32 ) a, a We can graph the line by finding two points. Two easy values for a are a = 0 and a = 2, which correspond to (3, 0) and (0, 2). Of course, all we re doing is graphing the equation, which has x 1 -intercept 3 and x 2 -intercept 2.

3 MA 3280 Lecture 05 - Generalized Echelon Form and Free Variables 3 Example 2. We can change this last system slightly by making the two lines parallel. (14) Trying the same operation, 2 E 1 + E 2, gives us (15) 4x 1 + 6x 2 = 10. 0x 1 + 0x 2 = 2. The second equation is 0 = 2, which is degenerate. This system has no solutions. The solution set is empty, which is less than 0-dimensional in our view. Example 3. Here s one more example. Consider the system (16) Let s do this one using matrix notation. (17) 2x 1 + x 2 + 5x 3 = 8 4x 1 + 2x x 3 = 22 2x 1 + x 2 + 6x 3 = Doing the operations 2 R 1 + R 2 and 1 R 1 + R 3 gives us (18) We can then do 1 3 R 2 + R 3 to get (19) Now, a row of 0 s is OK. We just don t want 0 s and something non-zero in the last column, which would mean, no solutions. We can see from the last matrix that x 1 and x 3 are pivot variables, and x 2 is free. The second equation is (20) 3x 3 = 6 which simplifies to (21) x 3 = 2. Substituting this into the first equation gives us (22) 2x 1 + x 2 + 5(2) = 8, and solving for x 1 gives us (23) x 1 = 1 x 2 2. Our solutions, therefore, take the form ( (24) 1 a ) 2, a, 2. These points form a line in x 1 x 2 x 3 -space.. Find three solutions to this last system of equations. Quiz 05

4 MA 3280 Lecture 05 - Generalized Echelon Form and Free Variables 4 Homework Consider the system of equations that corresponds to the following matrix. (25) Consider the system of equations that corresponds to the following matrix. [ ] (26) Consider the system of equations that corresponds to the following matrix. [ ] (27)

5 MA 3280 Lecture 05 - Generalized Echelon Form and Free Variables 5 4. Consider the system of equations that corresponds to the following matrix. (28) Quiz 05: There are infinitely many answers. For a = 0, 1, 2, you d get ( 1, 0, 2), ( 3, 1, 2), and ( 2, 2, 2). 2 HW: 1) a) or b) x 1, x 3, and x 4. c) x 2. d) (2 2a, a, 4, 3) e) A line. 2) a) Already in echelon form. b) x 1 and x 3. c) x 2 and x 4. d) ( 7 a + 3b, a, 4 2b, b). e) A plane. 3) a) Already in echelon form. b) x 1 and x 2. c) x 3 and x 4. d) ( 4 a b, 4, a, b). e) A plane. 4) a) b) x 1. c) x 2, x 3, x 4. d) (1 a b c, a, b, c). e) A 3-D hyper-plane.