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 Wednesday, January 12 Chapter 1.3 Main Idea: How to handle infinitely many solutions. Key Words: X = Xo + a1 X1 + a2 X2 + ... + an Xn A[Xo X1 X2 ... Xn] = [B00...0] Goal: Learn to read the answer off of the row canonical form. Assignment Page 37 Problem 11, 27, 28 ------------------------------------------------------------------ Previous Assignment Page 11 Problem 11 x1 x2 x3 rhs 0 1 4 -5 1 3 5 -2 3 7 7 6 1 3 5 -2 0 1 4 -5 3 7 7 6 1 3 5 -2 0 1 4 -5 0 -2 -8 12 1 0 -7 13 0 1 4 -5 0 0 0 2 <===== No solution 1 0 -8 0 0 1 4 0 0 0 0 1 <===== No solution RowReduce[{{0,1,4,-5},{1,3,5,-2},{3,7,7,6}}]; Page 11 Problem 12 x1 x2 x3 rhs 1 -3 4 -4 3 -7 7 -8 -4 6 -1 7 1 -3 4 -4 0 2 -5 4 0 -6 15 -9 1 -3 4 -4 0 2 -5 4 0 0 0 3 <=== No solution RowReduce[{{1,-3,4,-4},{3,-7,7,-8},{-4,6,-1,7}}] Page 11 Problem 13 x1 x2 x3 rhs 1 0 -3 8 2 2 9 7 0 1 5 -2 1 0 -3 8 0 2 15 -9 0 1 5 -2 1 0 -3 8 0 1 5 -2 0 2 15 -9 1 0 -3 8 0 1 5 -2 0 0 5 -5 1 0 -3 8 0 1 5 -2 0 0 1 -1 1 0 0 5 0 1 0 3 0 0 1 -1 | x1 | | 5 | | x2 | = | 3 | | x3 | | -1 | Check 5 +0 +3 = 8 It checks. 10 +6 -9 = 7 0 +3 -5 = -2 New Material ============ Theory: The relationship between linear equation in their algebraic form, and in their matrix form is obvious. One can uniquely represent the equations in either form and pass easily back and forth from one form to the other. System of Linear Equations algebraic form matrix form x1 x2 xn RHS a11 x1 + a12 x2 + ... a1n xn = b1 | a11 a12 ... a1n b1 | a21 x1 + a22 x2 + ... a2n xn = b2 | a21 a22 ... a2n b2 | a31 x1 + a32 x2 + ... a3n xn = b3 | a31 a32 ... a3n b3 | . | . | . | . | . | . | an1 x1 + an2 x2 + ... ann xn = bn | an1 an2 ... ann bn | The row operations are : (1) interchanging two rows (2) multiplying a row by a nonzero number (3) adding a multiple of one row to another The MOST IMPORTANT things about these three operations is that they do not change the solutions of the system of equations. (1) It does not matter the order in which the equations are written down, a solution to all of them will still be a solution to all of them. This means that (1) preserves solutions. (2) If we multiply one of the equations by any non zero constant, it is still the same equation since it has the same solutions. Written out more complicatedly, If ai1 x1 + ai2 x2 + ... + ain xn = bi then for any non zero constant c, c ai1 x1 + c ai2 x2 + ... + c ain xn = c bi. This shows that (2) preserves solutions. (3) Finally, if we replace one of the equations by a new equation formed by adding a multiple of another equation to it, the solutions to these two equations are unaffected. Thus, if (i) ai1 x1 + ai2 x2 + ... + ain xn = bi (j) aj1 x1 + aj2 x2 + ... + ajn xn = bj then: ai1 x1 + ai2 x2 + ... + ain xn = bi c aj1 x1 + c aj2 x2 + ... + c ajn xn = c bj and adding gives (j') (ai1 + c aj1) x1 + (ai2 + c aj2) x2 + ... + (ain + c ajn) xn = (bi + c bj) Any solution to (i) and (j) is also a solution to (i) and (j'). Further more, the same argument shows that any solution to (i) and (j') is also a solution to (i) and (j') because (j) is generated from (i) and (j') using -c as the multiplying factor. Reading the answer from the Row Canonical Form. For some reason ============================================== this tends to confuse students. It does not seem to make sense for two reasons. (a) The students find it arbitrary, pointless and confusing. or (b) The students find it so obvious that they think they are missing something because the teacher acts like it is difficult. Solve the following system of linear equations and write the answer as X = X + a X + a X + ... a X and check your answer. 0 1 1 2 2 r r x y z w u v RHS | 1 2 0 1 2 3 2 | | 1 3 0 0 1 2 1 | | 2 5 0 1 3 5 3 | | 4 10 0 2 6 11 7 | <--------------------> Matrix of Coefficients <--------------------------> Augmented Matrix <----> Right Hand Side First reduce the augmented matrix to Row Canonical Form. Remember, the row operations do not change the answers. So if we get our answers from the Row Canonical Form, we will be getting the answers to the original system as well. | 1 2 0 1 2 3 2 | | 1 2 0 1 2 3 2 | | 1 3 0 0 1 2 1 | | 0 1 0 -1 -1 -1 -1 | | 2 5 0 1 3 5 3 | | 0 1 0 -1 -1 -1 -1 | | 4 10 0 2 6 11 7 | | 0 2 0 -2 -2 -1 -1 | | 1 0 0 3 4 5 4 | | 1 0 0 3 4 5 4 | | 0 1 0 -1 -1 -1 -1 | | 0 1 0 -1 -1 -1 -1 | | 0 0 0 0 0 0 0 | | 0 0 0 0 0 1 1 | | 0 0 0 0 0 1 1 | | 0 0 0 0 0 0 0 | a b c x y z w u v RHS | 1 0 0 3 4 0 -1 | | |_1_ 0 0 3 4 0 -1 | | 0 1 0 -1 -1 0 0 | | 0 |1__0_-1_-1__0 0 | | 0 0 0 0 0 1 1 | | 0 0 0 0 0 |1___1__| | 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 | Assign the unknowns which are not above the stairstep ones some parameters z=a w=b u = c. The unknowns which are above the stair step ones get their values from the equations. x = -1 -3b -4c y = b + c v = 1 Remember the rest of the unknowns are given at the top of the columns so: x = -1 -3b -4c y = b + c z = a w = b u = c v = 1 We write these compactly as: |x| | -1| |0| |-3| |-4 | |y| | 0| |0| | 1| | 1 | |z| | 0| |1| | 0| | 0 | |w| = | 0|+a |0|+b| 1|+c| 0 | |u| | 0| |0| | 0| | 1 | |v| | 1| |0| | 0| | 0 | Xo X1 X2 X3 When you check, Xo gives the RHS. X1, X2, X3 all give zero. -1 0 0 0 0 1 0 0 1 0 0 0 -3 1 0 1 0 0 -4 1 0 0 1 0 |1 2 0 1 2 3| |2| |1 2 0 1 2 3 | |0| |1 2 0 1 2 3| |0| |1 2 0 1 2 3| |0| |1 3 0 0 1 2|=|1| |1 3 0 0 1 2 |=|0| |1 3 0 0 1 2|=|0| |1 3 0 0 1 2|=|0| |2 5 0 1 3 5| |3| |2 5 0 1 3 5 | |0| |2 5 0 1 3 5| |0| |2 5 0 1 3 5| |0| |4 10 0 2 6 11| |7| |4 10 0 2 6 11 | |0| |4 10 0 2 6 11| |0| |4 10 0 2 6 11| |0| Class Problem: Solve AX = B and write the answer as X = X + a X + a X + ... a X 0 1 1 2 2 r r and check your answer using A[X X X ... X ] = [B 0 0 0 ... 0]. 0 1 2 r | 2 4 4 6 0 | | x | | 16 | | 0 0 1 3 1 | | y | | 5 | | 1 0 2 3 1 | | z | = | 7 | | 0 2 1 3 0 | | w | | 6 | | 1 0 1 0 0 | | u | | 2 | ======================================================================== Solution | x | | -3 | | 3 | | 1 | | y | | 1/2 | | 0 | | 1/2 | | z | = | 5 | +a |-3 | +b | -1 | | u | | 0 | | 1 | | 0 | | v | | 0 | | 0 | | 1 |