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a. Find the general flow pattern in the network shown in the figure.b. Assuming that the flow must be in the directions indicated, find the minimum flows in the branches denoted by $x_{2}, x_{3}, x_{4},$ and $x_{5} .$figure can't copy

a)$x_{1}=-40+x_{3}$$x_{2}=10+x_{3}$$x_{4}=50+x_{6}$$x_{5}=60+x_{6}$$x_{3}, x_{6}$ are free variables

b)$x_{2}=50$$x_{3}=40$$x_{4}=50$$x_{5}=60$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 6

Applications of Linear Systems

Introduction to Matrices

Angel E.

October 1, 2020

Missouri State University

Baylor University

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

11:48

Find the general flow patt…

12:42

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06:01

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02:58

Network Analysis The flow …

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09:45

How much current flows thr…

19:24

Determine the values of x1…

02:18

Circulation and flux For t…

06:43

Determine the rate of flow…

in problem. 13. We want to get the general flu better in this network, knowing that we have here x one in this direction. X two Isn't this direction Exit three in this direction? X four this direction. And finally, we have x five in this direction. Not finally, because we have Year X six. Then we have sex or months, and we can get the equations from the intersection. The flow in tow. Each intersection equals flow out. We have five intercept, 600 mhm be C, D and E. Let's construct all intersections at any we have flow in is 30 plus x two equals flew out equals X one. Let's eat and boy rearranging the equation. We have x one winds x two x one minus x two equals minus 50 from being we have the flow in is x five x three blocks X five and the flow out is x two plus x four. By arranging the equations, we have minus x two or x two lost X four minus x three minus X five equals zero. But see over intersection see we have flow in is X six plus 100 equal flow out his X five plus 40 by rearranging the equations. We have x five plus six six minus x x Sorry minus X six. My equals 60 from Intersection D. We have floating is X four plus 40 equals flow out, which is ex sex plus 90. Then we have x four minus X six equals 50. Finally, from Intersection E. We have we have X one plus 60 as a flow in and the flow out is exit three plus 20. Then we have x one minus x three. We can rearrange x three and x four not to make any problems. You have your minus X three minus x three plus X four and we make Make it here x four and we have here X one minus 63 equals minus 40. These are the five equations we have and we have six unknowns, which means we have a three very at least one flavor. Let's construct the augmented metrics for these For this system, we have one minus 1000 minus 50 second equation. We have 01 minus 11 minus one. We have six a month story. Then we have here another zero then minus 50. And we have here. Zero zero 0000 Then we have one minus one 16. We have 000 10 minus 1 50. And finally, we have one zero minus one, zero 00 and minus 40. We can make it this momentous metrics, Toby. In the fall off the reduced ancient form we have here, the first role isn't the phone. The second group is in the full. The third row is in the fall, but it has a leading value in the fifth Column. We can make it in the fourth. We want to have year one, and then we have your own. Then we can make it here in the fifth. In the fifth through we switch R three and R five to make another minted metrics which you have one minus one 0000 minus 50 01 minus 11 minus 100 We have zero. We have one because we have here the fifth row 10 minus 1000 minus 40. And we have the fourth row as it is. Then we have 00 zero. Is you one minus 1 60 We can see that we can see that the augmented metrics in the ancient form except for the third row, third row. Here we can make this number a zero boy multiplying the for the first row by minus one, then added to the third row. Then we have we have one minus 10000 minus 50 01 minus 11 minus 100 We have zero minus one and it's 10 calls one minus 1000 minus 50. What a bob. I minus one equals 50 plus minus 40 equals 10. Then we have the other roads as it is as they are. Then we want to make this as zero. We want to blow the second group by minus one. Then edit with the third row. Then we have the augmented metrics. One minus 10000 minus 50 01 minus 11 minus 100 Then we have zero zero one minus one gives another zero minus one, minus one minus one post of one. Then we have zero then and we have here the same enemies. We can see that we are getting you close to the petulant four. We can continue by adding the second rule the third row to the fourth row and replaced the four throat. And we can then divide the third rule by minus one to make this number one. Let's make it The equivalent metrics is the first through as it is one and this 10000 minus 50 01 minus one one minus 100 000 We should hear. Start by the fourth room to get here zero one minus one 15. And when we divide the third row by minus one, we have one minus 10 minus 10 and the fifth row is zero 00 another deer one. When this 16 we have here a little mistake. We have 10 plus 50 equals six. Then we can see that by subtracting are five and our six we can get the final four of the reduced action on 41 minus 10000 minus 50 01 minus 11 minus 100 0001 minus 10 minus 10 00001 minus 1 60 the text. The fifth row is all zeros, which means we have another flavor. We have four equations, and the fifth equation gives another free variable, and we have only six months, which means we have to free variable. We can make X five and x six as free variables. Then from the fourth equation can get X four, which equals X five. No, we don't have here. X four equals x five we have from the fourth equation. We have X five equals x six, which means we can not choose X five and X X as free verb. We can choose only one. Then we will not to choose XX as a free variable x five. We will only choose now xx as a river. Then from the third equation we get, we have from the fourth equation we have x five equals X six plus 16. Don't Don't forget 60. And from third equation, we have x four equals X five, which is egg six plus 60. I understand equals x six lost 50. And from the second equation, we have X three equals from the second equation. We have X two. Sorry, we have x two have X two equals Exit three minus X four, which is now here x six plus 50 plus x five, which is given here x six plus 60 and this equals zero now. We don't have X three. Then we will. The place x five year by exit three xx as free variable and exit three as free very. Then it equals Exit three minus two x six No x minus x x plus x six equals zero. It's a three plus zero, minus 50 plus 60 equals plus thin. And from the first equation we have x one. Exxon equals X two minus 50 and we have x two as excessively Boston equals accessory plus 10 minus 50 equals accessory. My in US 40. Now we can get the general solution or the general flow better in the network as X equals accessory multiplied by Victor Plus x six multiplied by a victor plus the free victor excess three Multiplied boy for X one. It's multiplied by on for X two. It's multiplied by on we don't see x one and X for X one. Of course it will be X one for excess. Three will be will be one accessory equals accessory. And for X four, we don't have exit three for ex wife. We don't have X three for XX. We don't have X three. And for XX, we can see XX in x one and x two or X three, then 000 and we have one in X four, one in X five and, of course, one in x six. Plus the free. The values we have minus 40 Forex. Do we have 10 for X one? We have minus 40 because minus 50 plus that then we have 10 for X three. We have zero because it's a privilege, X four will be 50 x five will be 60 and for X X equals zero because it's free. Vary. This is the general flow button this football A football TV. We want to assume that the floor must be in the directions indicated we want to find the minimum flows in the branches, denoted by x two x four x three x four and x five x two next to 63 x four and x five in the network for bar to be the network we can't have, we can't get negative values because all the directions are the same as indicated. And for that we can see that X one x one equals excess three minus 40 which means X three must be equals 40 or more than four than the minimum values. The minimum values for X three will be 40. So make X one equals zero as the minimum, and by booting exit three equals 40 as a minimum, we can see that exit, too, equals exist. Three. Austin 63 Boston This is the minimum value off Exit 3 40 which means extra, too, as a minimum will be 40. Boston equals 15, and for X four we can see that X four equals x six plus 50. The minimum value of X X is zero. There is no flow in X in the direction off x x then x four will be 50 x four will be 50 as minimum and x five To be minimum, we have x five equals X six plus 60. The minimum body off xx zero, then x five as minimum is six and this is the final answer off our problem involved to be. And this is the final answer off part a

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