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# LINR ALG & NUM ANLY AMATH 352

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This 8 page Class Notes was uploaded by Mossie Pfannerstill V on Wednesday September 9, 2015. The Class Notes belongs to AMATH 352 at University of Washington taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/192279/amath-352-university-of-washington in Applied Mathematics at University of Washington.

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Date Created: 09/09/15

AMATH 352 LECTURE 13 JULY 25 2008 1 Code for this lecture The version of gausselimm posted along with this lecture is updated slightly from Lecture 12 First off the statement if maxval gt 0 has been replaced with if maxval gt 1e 10 This xes a problem I overlooked the rst time through Rounding errors will produce small numbers very close zero instead of exact zeros This statement says that for all of our concerns anything with magnitude lt 10 10 is zero You could probably get away with a smaller tolerance than this as rounding errors are generally on the order of 1016 This is just be being really excessively safe The other change is a number of statement that make the program pause after each step of Gaussian elimination and say what it just did This is just so you can follow the process step by step if you want to The other two scripts for today are randomunsolvable m and randomsolvable m These produce A and b so I can use the system Ax b to illustrate certain things 2 Output of Gaussian elimination 21 A system with one solution Last time we gured out how to program Gaussian elimination but we didnt look too hard at its output Let7s x that little problem now Set up a system with the following commands Use the rst statement to get the exact same numbers l7m using here Or leave it out or change the 1000 to a different number to try out some different numbers gtgt rand twister 1000 gtgt A rand44 gtgt b rand41 gtgt aug A b aug 06536 08725 01150 02123 09503 00407 04822 03972 02331 08417 02071 07425 03922 01823 07435 00696 08853 09526 09311 04154 The array aug stores the augmented matrix for this system of equations 06536z1 01150z1 09503z1 04822z1 08725z2 02123z2 00407z2 03972z2 02331z3 08417z3 02071z3 07425z3 Now go back to the command line7 and enter aug2 gausselimaug 03922m4 08853 01823m4 09526 1 07435m4 09311 00696m4 04154 After a detailed run through of the operations7 you7ll end up with aug2 10000 00428 0 10000 0 00000 0 00000 Just like aug7 this new augmented matrix is shorthand for tions 2 02179 01074 10000 0 02179z3 01074z3 7 07824 01412 01530 10000 07824x4 01412x4 09799 02900 09816 21748 a linear system of equa 09799 02900 09816 2 21748 And7 since only Gaussian operations were used in getting from sytem 1 to system 27 the two are completely equivalent That is7 any solution to system 1 is a solution to system 27 and vice versa The point of this7 of course7 is that system 2 is a lot easier to solve The last equation gives me a value for x4 I can take this value and plug it into the third equation to get a value for 3 Then I plug both values into the second equation to solve for 2 And nally I take the three values l7ve found to solve for 1 in the rst equation If you do this you should end up with x1 708858 x2 05274 x3 06488 4 21748 3 22 GaussJordan elimination While solving the linear system 2 is technically pretty easy it can still be a bit of a pain Really we ought to be able to do all the calculations on the computer There7s a particularly nice way to do this systematically called Gauss Jordan elimination This follows traditional Gaussian elimination with another sequence of operations l7ll call this sequence the Jordan step7 7 I dont think this is con ventional at all but I want a name for it and this name makes as much sense as anything After I applied Gaussian elimination to my augmented matrix I got back a matrix in which the rst nonzero element in every row is 1 Also every element below one of these leading 17s is zero Jordan step uses pivoting to cancel every element above one of these leading 17s When the system has one solution this gives us the nal answer And when a system has in nitely many solutions this step still makes it a little easier to see whats going on Lets see how this works for the system we7ve been looking at The rst thing PM do is cancel every element above the 1 in row 4 column 4 of aug2 using these operations gtgt aug2 pivotaug2 4 aug234 3 gtgt aug2 pivotaug2 4 aug224 2 gtgt aug2 pivotaug2 4 aug214 1 aug2 10000 00428 02179 0 07218 0 10000 01074 0 05971 0 00000 10000 0 06488 0 0 0000 0 10000 21748 You can remove the semicolon from any step to see its output Next l7ll eliminate the elements above the leading 1 in row 3 column 3 3 gtgt aug2 pivotaug2 3 aug223 2 gtgt aug2 pivotaug2 3 aug213 1 aug2 10000 00428 0 0 08632 0 10000 0 0 05274 0 00000 10000 0 06488 0 00000 0 10000 21748 Finally l7ll eliminate that one last element above row 2 column 2 gtgt aug2 pivotaug2 2 aug212 1 aug2 10000 00000 0 0 08858 0 10000 0 0 05274 0 00000 10000 0 06488 0 00000 0 10000 21748 This nal augmented matrix still represents a linear system of equations only now its a system that just plain hands the answer to me x1 708858 2 05274 zg 06488 4 4 21748 3 Echelon form Reduced echelon form The special types of matrices that come out of Gaussian elimination and Gauss Jordan elimination have their own names While we7ve seen examples of these already they7re a bit more interesting in the cases where there isnt exactly one solution to the system So it might be worth glancing back at these de nitions after seeing the examples in the rest of this lecture De nition A matrix is in echelon form if it has the following two properties 1 The rst nonzero element of any nonzero row is 1 2 Any element below one of these leading 17s is zero Say that I apply Gaussian elimination to the augmented matrix A l b l and I obtain the reduced augmented matrix Ared I bred Then Ared is in echelon form A matrix in echelon form typically looks something like this with 07s representing nonzero elements and empty space representing zeros l o o o o o o o o l o o o o o o l o o o o o l o o De nition A matrix is in reduced echelon form if it is in echelon form with one additional condition 3 Every element above a leading 1 is also zero Say that I apply the Jordan step of Gauss Jordan elimination to Ared l bred and I get back Aredg l bredg Then Aredg is in reduced echelon form If my generic7 matrix in echelon from from above is converted to reduced echelon form it looks like this Hoe 000 0 HOOD l l 4 Unsolvable systems Gaussian elimination makes it really easy to spot an unsolvable system of equations with no need for the Jordan step Use these commands to get an example Again you can omit the rst line to get different numbers gtgt rand twister 1000 gtgt Ab randomunsolvable gtgt aug A b aug 9 18 12 36 9 9 7 23 58 10 84 8 14 3 27 6 4 24 9 3 5 1 23 13 37 9 8 1 22 10 10 60 2 10 8 Then7 passing aug through gausselimm and fast forwarding through the detailed output7 gtgt aug2 gausselimaug 10000 02222 01481 08889 03333 01111 01852 0 10000 01246 12017 02962 03130 01373 0 0 10000 51157 12052 15000 01651 0 0 0 10000 03333 0 0 1686 0 0 0 0 00000 0 80150 In principle this represents a big long system of equations But I only care about the equation corresponding to the last line 0 80150 5 This means the original system of equations is unsolvable How does this look more generally Say l7ve reduced A l b l to Ared l bred Any nonzero row of Ared won7t lead to an inconsistency 7 these just tell me how to solve for my variables On the other hand7 there may be rows of nothing but zeros underneath the nonzero part of Ared lf bred has a nonzero element in any of these rows7 I end up with 0 sornething nonzero7 a contradiction The generic picture looks like this 1 o o o l o 1 0 O H O O 00 5 Systems with in nitely many solutions Now lets check out a system with in nitely many solutions First set up the exam ple gtgt rand twister 1003 gtgt Ab randomsolvable gtgt aug A b aug 1 25 9 6 19 2 90 15 15 5 4 20 1 81 12 63 13 29 28 6 19 8 22 2 6 35 5 91 10 7 9 24 53 5 5 And then reduce the system aug2 10000 10000 03333 02667 13333 00667 54000 0 10000 03333 06314 08627 01333 08980 0 0 00000 10000 21658 00553 17184 0 0 00000 0 10000 03333 43333 0 0 00000 0 0 00000 0 This augmented matrix corresponds to the system of equations 1 100002 033333 026674 7 133335 006676 754000 2 7 033333 7 063144 086275 7 013336 08980 4 7 216585 7 005536 717184 5 033336 43333 0 0 This time around7 there are two different types of variables First of all7 there are the ones that have a leading 17 as a coef cient in one of the equations These are the exact same variables that were used to pivot when Gaussian elimination was taking place So these are called pivot variables Second7 there are well all the other variables These are called free vari ables because their values are free I can pick any values I like for these7 and PH still be able to nd a solution In this case7 17 27 47 and 5 are the pivot variables Meanwhile7 3 and 6 are the free variables If you were to make a solution to this system7 you could start with the bottom equation and work your way to the top7 following this sequence of events Free variables Pivot variables Pick a value for 6 Solve for 5 Solve for 4 Pick a value for 3 Solve for 2 Solve for 1 Pivot variables and free variables are closely tied with the range and null space of the original matrix A well look at that in the next lecture

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