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Linear Algebra I

by: Melvina Keeling

Linear Algebra I MATH 369

Melvina Keeling
GPA 3.78

Daniel Bates

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Daniel Bates
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This 3 page Class Notes was uploaded by Melvina Keeling on Monday September 21, 2015. The Class Notes belongs to MATH 369 at Colorado State University taught by Daniel Bates in Fall. Since its upload, it has received 87 views. For similar materials see /class/210084/math-369-colorado-state-university in Mathematics (M) at Colorado State University.


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Date Created: 09/21/15
A complete example of Gaussian elimination Dan Bates Colorado State University Math 369 Fall 2008 The point of Gaussian elimination is to make ef cient use of row operations to transform a given augmented matrix into a row equivalent matrix in reduced row echelon form RREF Recall that every matrix is row equivalent to a unique matrix in RREF De nition A matrix is in RREF if 1 All zero rows rows with no nonzero entries are at the bottom of the matrix 2 The leftmost nonzero entry of each nonzero row is a 1 called a leading Z 3 The leftmost nonzero entry of each nonzero row is the only nonzero entry in its column and r The matrix is in echelon form See page 31 ofthe book for a technical de nition ofthis Simply put7 it means that the leading 17s of the matrix progress from left to right as you move from the top of the matrix to the bottom Algorithmically7 put your pencil on the leading 1 of row 1 Move your pencil to the leading 1 of row 2 If it moved to the left7 then the matrix is not in echelon form If it moved to the right7 proceed to the next row and so on to the last nonzero row of the matrix Note that a zero row cannot have a leading 1 Recall that we are allowed to use in any order the following three row operations to get to the form we want 1 Swap any two rows lt gt Rj 2 Multiply any row by a nonzero constant a 31 0 04R a Bi and 3 Add to any row a nonzero multiple a 31 0 of another row R 04R a Rj So7 how do we proceed We could just guess what to do from one step to the next7 but that may not be terribly ef cient Gaussian elimination is an algorithm recipe for doing this other pseudocode for which may be found on pages 33 34 of the text Algorithm Gaussian elimination Input A matrix A with m rows and 71 columns Output A matrix that is both rowequivalent to A and in RREF For 239 from 1 to m 239 is just the row number 0 Choose a row at or below row 239 which has the leftmost nonzero entry and if it is not row 239 swap that row with row 239 If only zero rows rows with all entries zero remain7 then stop 7 you are done 0 Make the leading nonzero entry of row 239 a 1 by multiplying row 239 by the reciprocal of the leftmost nonzero entry 0 Use the leading 1 of row 239 to make 0 all other nonzero entries in the same pivot column as the leading 1 o Swap any new zero rows down to the bottom of the matrix maintaining echelon form Now for the example Suppose we want to transform the following augmented matrix to reduced row echelon form 1 1 12 7 7 1 7 2 1 1 2 1 3 41 41 44 76 78 42 0 414 429 430 Following the algorithm above7 we start with 239 1 ie7 we rst consider the rst row Although the result is unique7 the way in which we obtain it is not We could multiply row 1 by 7 to get a leading 17 or we could just swap rows 1 and 2 PH opt for the former option 7 i 71 7 i2 1 1 7 12 14 1 1 2 1 3 771314131 1 1 2 1 3 i1 i1 i4 76 is a 71 i1 i4 76 is 72 0 714 729 730 i2 0 714 729 730 From here7 we just use the new leading 1 row 17 column 1 to make 0 everything else in column 1 1 1 7 12 14 1 1 7 12 14 1 1 2 1 3 RgiRl RZ 0 0 75 711 711 71 71 74 76 78 a 71 71 74 76 78 72 0 714 729 730 72 0 714 729 730 1 1 7 12 14 1 1 7 12 14 R3Rl Rg 0 0 75 711 711 R42R1gtR4 0 0 75 711 711 0 0 3 6 6 a 0 0 3 6 6 72 0 714 729 730 0 2 0 75 72 Things are looking good at this point7 at least in column 1 We have our leading 1 in row 1 column 1 is the pivot column corresponding to that leading 17 and the rest of the column has been zeroed out Now we increment 239 to 2 We want the leftmost nonzero entry of row 2 to be as far left as possible using swapping Of rows 27 37 and 47 row 4 has the leftmost nonzero entry7 so we swap that up to row 2 1 1 7 12 14 1 1 7 12 14 0 0 75 711 711 R2 lt gt R4 0 2 0 75 72 0 0 3 6 6 a 0 0 3 6 6 0 2 0 75 72 0 0 75 711 711 Multiplying by makes the leading entry of row 2 a 1 After that7 we need to make zero only the 1 directly above our new leading 1 Notice that the 0 before the leading 1 in row 2 means that we will not damage the special structure of column 1 while making column 2 into the format that we want The fact that the two entries directly below our new leading 1 are both zero already is just dumb luck 1 1 7 12 14 1 0 7 15 RZaRZ 0 1 0 g 71 12171224121 0 1 0 g 71 a 0 0 3 6 6 a 0 0 3 6 6 0 0 i5 711 711 0 0 i5 711 711 The leading 1 of row 2 satis es all of the properties we need so we increment 239 again to 3 For row 3 we can choose to keep it or swap for row 4 never up 7 only downl There is no point in doing that so we just make the leading nonzero entry a 1 and eliminate all nonzero entries above and below this new leading 1 Once again we have a fortunate uke this time a zero in the 3 2 coordinate 107115 100 11 51234113 0103371 131771234121 010771 6 0012 2 a 0012 2 0075 711711 0075 711711 1001 R45R34R4 010 7 71 a 00122 0007171 Almost done Before we tackle that last row pretend that it is a zero row In that case we would be done and there would be in nitely many solutions since there are non pivot columns Now pretend that row 4 instead of a zero row is a row with zeros to the left of the bar and a 1 to the right In that case we would still need to use that 1 a leading 1 to zero out the remainder of column 5 However the point is that such a row would correspond to the linear equation 0 1 This has no solutions so in this case the solution set of the linear system corresponding to the augmented matrix would have no solutions Enough pretending 7 back to our problem Clearly we increment 239 to 4 multiply row 4 by 71 and use our new leading 1 to zero out all other entries of column 4 1001 1000 71244124 0107 i1 Rli R4aR1 0107 71 a 00122 a 00122 00011 00011 10001 10001 R23R44R2 0100 123721344123 0100 a 00122 a 00100 00011 00011 We are done At this point it would be wise to check that this is indeed reduced row echelon form The unique solution of the linear system corresponding to the original augmented matrix is then i g 01


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