Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:40 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Monday, January 10 Chapter 1.1 and 1.2 Page 11, Problems 11,12,13 Main Idea: Look at the coefficients of the equations. Key Words: Simultaneous Linear Equations, Gauss Jordan Elimination Reduced Row Echelon Form, Row Canonical Form Goal: Learn to reduce matrices to Row Canonical Form. We need a name for an object. Draw a rectangle. Start somewhere on the top edge and make a "stair step" path down and to the right. 1. All of your step heights have to be one step high. 2. The horizontal segments can be a long as you want. 3. You stop at the right hand edge of the rectangle. Example: ________________________________________ | |___ | | |_ | | |________________| | | | | | | | | | | | | | | |______________________________________| The matrix is in Row Canonical Form if: (1) You can find the stairs. (2) The first entry after the vertical segment is a 1 called a stairstep one. (3) There are all zeros below the stairs. (4) There are zeros above the stairstep ones. Find the stairs in this matrix. | 1 2 0 4 5 0 7 8 9 | | 0 0 1 3 4 0 8 9 1 | | 0 0 0 0 0 1 3 2 1 | | 0 0 0 0 0 0 0 0 0 | Why is this matrix not in Row Canonical Form | 1 2 0 4 5 0 7 8 9 | | 0 1 3 4 0 8 9 1 0 | | 0 0 0 0 0 1 3 2 1 | | 0 0 0 0 0 0 0 0 0 | Why is this matrix not in Row Canonical Form | 1 0 0 0 5 0 7 8 9 | | 0 0 0 0 0 8 9 1 0 | | 0 0 1 0 0 1 3 2 1 | | 0 0 0 1 0 0 0 0 0 | Theorem: The Row Canonical Form can be used to solve systems of linear equations. Solve 2x + 8y + 4z = 2 2x + 5y + z = 5 4x +10y - z = 1 We need only keep track of the coefficients, but we have to keep them straight. We put the unknowns above the columns x y z RHS 2 8 4 2 2 5 1 5 4 10 -1 1 Now each equation can be converted from the "matrix representation". Furthermore, the row canonical form gives the answer. x y z RHS Answer: | 1 0 0 11 | x = 11 | 0 1 0 -4 | y = -4 | 0 0 1 3 | z = 3 11 -4 3 Check 2x + 8y + 4z = 2 2x + 5y + z = 5 4x +10y - z = 1 22 -32 +12 = 2 22 -20 + 3 = 5 It checks. 44 -40 -3 = 1 With every procedure we give a way to check. If your answer is right you will receive full credit whether or not you check. If your answer is wrong, you will get up 80% partial credit if you have done the following. (1) Showed the correct method in solving the problem. (2) Checked the problem and wrote that you noticed that the answer was wrong and said how you knew it was wrong. If you do not check, you will receive much less credit, 50% or less. The correct method is measured by things like, (1) You set up the augmented matrix correctly. (2) You stopped with a row canonical form. It might not be the correct row canonical form. (3) Your answer agrees with your row canonical form. Essentially, I have to be convinced that you could go back and get the correct answer if you had more time. Write these equations as a matrix and reduce to Row Canonical Form using the Elementary Row operations: (a) Multiply row by nonzero number. (b) Add a multiple of one row to another. (c) Interchange two rows. 2 x + 3 y + 8 z + 11 w + 4 u = 20 x + 2 y + 5 z + 7 w = 8 2 x + 3 y + 8 z + 11 w + u = 14 2 x + 3 y + 8 z + 11 w + 4 u = 20 Solution: x y z=a w=b u RHS 1____ 0 + 1 + 1 0 | 0 0 |_1____ 2______3___ 0 | 4 0 + 0 0 0 |_1____|__2 0 0 0 0 0 | 0 Reduce the following matrix to Row Canonical Form x y z w u RHS 0 0 1 -1 -1 4 2 4 2 4 2 4 2 4 3 3 3 4 3 6 6 3 6 6 - Solution: x y=a z w=b u 1 2 0 3 0 2 0 0 1 -1 0 2 0 0 0 0 1 -2 0 0 0 0 0 0 Solution to Row Canonical Form for class problem 1. 2 x + 3 y + 8 z + 11 w + 4 u = 20 x + 2 y + 5 z + 7 w = 8 2 x + 3 y + 8 z + 11 w + u = 14 2 x + 3 y + 8 z + 11 w + 4 u = 20 x y z w u RHS 2 + 3 + 8 + 11 + 4 | 20 1 + 2 + 5 + 7 0 | 8 2 + 3 + 8 + 11 + 1 | 14 2 + 3 + 8 + 11 + 4 | 20 x y z w u RHS 1 + 2 + 5 + 7 0 | 8 2 + 3 + 8 + 11 + 4 | 20 2 + 3 + 8 + 11 + 1 | 14 2 + 3 + 8 + 11 + 4 | 20 x y z w u RHS 1 + 2 + 5 + 7 0 | 8 0 - 1 - 2 - 3 + 4 | 4 0 - 1 - 2 - 3 + 1 | -2 0 - 1 - 2 - 3 + 4 | 4 x y z w u RHS 1 + 2 + 5 + 7 0 | 8 0 + 1 + 2 + 3 - 4 | -4 0 - 1 - 2 - 3 + 1 | -2 0 - 1 - 2 - 3 + 4 | 4 x y z w u RHS 1 0 + 1 + 1 8 | 16 0 + 1 + 2 + 3 - 4 | -4 0 + 0 0 0 - 3 | -6 0 0 0 0 0 | 0 x y z w u RHS 1 0 + 1 + 1 8 | 16 0 + 1 + 2 + 3 - 4 | -4 0 + 0 0 0 1 | 2 0 0 0 0 0 | 0 x y z=a w=b u RHS 1____ 0 + 1 + 1 0 | 0 0 |_1____ 2______3___ 0 | 4 0 + 0 0 0 |_1____|__2 0 0 0 0 0 | 0 The above is the Row Canonical Form. x | 0 | |-1 | |-1 | y | 4 | |-2 | |-3 | z =| 0 |+a| 1 |+b| 0 | w | 0 | | 0 | | 1 | u | 2 | | 0 | | 0 | | 2 + 3 + 8 + 11 + 4 | | 0 -1 -1 | | 20 0 0 | | 1 + 2 + 5 + 7 0 | | 4 -2 -3 | = | 8 0 0 | | 2 + 3 + 8 + 11 + 1 | | 0 1 0 | | 14 0 0 | | 2 + 3 + 8 + 11 + 4 | | 0 0 1 | | 20 0 0 | | 2 0 0 | Solution to Class Problem 2. x y z w u RHS 0 0 1 -1 -1 4 2 4 2 4 2 4 2 4 3 3 3 4 3 6 6 3 6 6 1 2 1 2 1 2 0 0 1 -1 -1 4 2 4 3 3 3 4 3 6 6 3 6 6 1 2 1 2 1 2 0 0 1 -1 -1 4 0 0 1 -1 1 0 0 0 3 -3 3 0 1 2 0 3 2 -2 0 0 1 -1 -1 4 0 0 0 0 2 -4 0 0 0 0 6-12 1 2 0 3 2 -2 0 0 1 -1 -1 4 0 0 0 0 1 -2 0 0 0 0 6-12 x y=a z w=b u 1 2 0 3 0 2 0 0 1 -1 0 2 0 0 0 0 1 -2 0 0 0 0 0 0 | x | | 2 | |-2 | |-3 | | y | | 0 | | 1 | | 0 | | z | = | 2 |+a| 0 |+b| 1 | | w | | 0 | | 0 | | 1 | | u | |-2 | | 0 | | 0 | | 0 0 1 -1 -1 | | 2 -2 -3 | | 4 0 0 | | 2 4 2 4 2 | | 0 1 0 | | 4 0 0 | | 2 4 3 3 3 | | 2 0 1 | = | 4 0 0 | | 3 6 6 3 6 | | 0 0 1 | | 6 0 0 | |-2 0 0 |