[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-661654-105":59,"doc-detail-661654-en":130},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":123,"head_meta":125,"extra_data":127,"updated_unix":129},105,"en","exercise-1611-find-the-largest-possible-value-for-x3","Exercise 1.6.11 - Find the largest possible value for x3","","Exercise 1.6.11 in Linear Algebra (MATH 2130) determines the general flow pattern in a given network with nonnegative flows. Using node flow-balance conditions, it forms a linear system and converts it to RREF via row operations on the augmented matrix. The solution is parameterized by t, then nonnegativity constraints for x1, x2, x3, and x4 restrict the allowable t values. From these bounds, the maximum feasible x3 is identified as 20.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/exercise-1611-find-the-largest-possible-value-for-x3/661654/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/exercise-1611-find-the-largest-possible-value-for-x3/661654.png","ImageObject",300,407,{"name":92,"@type":93},"Bulrr","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-30","2026-10-28",true,{"@type":102,"interactionType":103,"userInteractionCount":8},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"What equations are used to model flow conservation at each node?","Question",{"text":112,"@type":113},"The solution sets up three node balance equations: x1 + x3 = 20, x2 - x3 - x4 = 0, and -x1 - x2 = -80, derived from equal total input and output at each node.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"How is the augmented matrix used in the solution?",{"text":117,"@type":113},"The system is rewritten into an augmented matrix and then reduced to RREF using row operations, yielding a parameterized form of the variables.",{"name":119,"@type":110,"acceptedAnswer":120},"How are nonnegativity constraints applied to find the largest possible x3?",{"text":121,"@type":113},"Nonnegativity requires x1, x2, x3, x4 to be ≥ 0, which translates to bounds on the parameter t. The resulting range for t gives the maximum feasible x3 value as 20.","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},661654,1793173554,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":8,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":19,"language":139,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":67,"update_tm":129,"read_time":52},5909892115043,"https://ap-avatar.wpscdn.com/davatar_6f874abed73319feea01a86fa6f0fab8","Exercise 1.6.11  \nLinear Algebra  \nMATH 2130  \nSEBASTIAN CASALAINA  \nABSTRACT. This is Exercise 1.6.11 from Lay [LLM21, §1.6]:  \nExercise 1.6.11. Find the general flow pattern of the network shown in the figure. Assuming the flows are all nonnegative, what is the largest possible value for x3?  \no 20  􀀏 A  \nOO  \nx1  \n80 C ~~ /~~ 􀀏  \nSolution. We have the following three equations coming from the condition at each node that the total input and output be equal:  \nx1 + x3 􀀀 20 = 0  \nx2 􀀀 x3 􀀀 x4 = 0  \n􀀀x1 􀀀 x2 + 80 = 0  \nWe can rearrange this as:  \nx1 + x3 = 20  \nx2 􀀀 x3 􀀀 x4 = 0  \n􀀀x1 􀀀 x2 = 􀀀80  \nDate: September 8, 2025 .  \nWe then obtain the following augmented matrix:  \n2  \n6 6  \n6  \n4  \n1  \n0  \n􀀀1  \n0  \n1  \n􀀀1  \n1  \n􀀀1  \n0  \n0  \n􀀀1  \n0  \n20  \n0  \n􀀀80  \n3  \n7 7  \n7  \n5  \nWe now use row operations to put the left hand side of the matrix in RREF:  \nFirst we add the first row to the third row:  \n2  \n6 6  \n6  \n4  \n1  \n0  \n0  \n0  \n1  \n􀀀1  \n1  \n􀀀1  \n1  \n0  \n􀀀1  \n0  \n20  \n0  \n􀀀60  \n3  \n7 7  \n7  \n5  \nThen we add the second row to the third row:  \n2  \n6 6  \n6  \n4  \n1  \n0  \n0  \n0  \n1  \n0  \n1  \n􀀀1  \n0  \n0  \n􀀀1  \n􀀀1  \n20  \n0  \n􀀀60  \n3  \n7 7  \n7  \n5  \nThen we multiply the last row by 􀀀1, and we also add that row to the second row:  \n2  \n6 6  \n6  \n4  \n1  \n0  \n0  \n0  \n1  \n0  \n1  \n􀀀1  \n0  \n0  \n0  \n1  \n20  \n60  \n60  \n3  \n7 7  \n7  \n5  \nThe left hand side is in RREF. We then modify the matrix:  \n2  \n6 6  \n6 6  \n6 6  \n6  \n4  \n1  \n0  \n0  \n0  \n0  \n1  \n0  \n0  \n1  \n􀀀1  \n􀀀1  \n0  \n0  \n0  \n0  \n1  \n20  \n60  \n0  \n60  \n3  \n7 7  \n7 7  \n7 7  \n7  \n5  \nThe solutions to the system of equations are therefore given by:  \n2  \n6 6  \n6 6  \n6 6  \n6  \n4  \nx1  \nx2  \nx3  \nx4  \n3 2 7 6 7 6 7 6  \n7 = 6 7 6 7 6 7 6 5 4  \n1  \n􀀀1  \n􀀀1  \n0  \n3 2 7 6 7 6  \n7 t + 6 7 6 7 6 5 4  \n20  \n60  \n0  \n60  \n3  \n77  \n7 , t 2 R. 7  \n7  \n5  \nFor the solutions to the problem, we only want those solutions with x1, x2, x3, x4 􀀕 0. These are the conditions  \nx1 = t + 20 􀀕 0  \nx2 = 􀀀t + 60 􀀕 0  \nx3 = 􀀀t 􀀕 0  \nx4 = 60 􀀕 0.  \nThis translates to  \nt 􀀕 􀀀20  \nt 􀀔 60  \nt 􀀔 0.  \nSo we have  \n􀀀20 􀀔 t 􀀔 0.  \nTherefore, all possible flows are determined by:  \n2 x1 3 2 1 3 220 3  \n6 7 6 7 6 7  \n6 x2 7 6 􀀀1 7 6 60 7  \n6 7 = 6 7 t + 6 7 , 􀀀20 􀀔 t 􀀔 0.  \n6 x3 7 6 􀀀1 7 6 0 7 6 7 6 7 6 7  \n4 x4 5 4 0 5 4 60 5 We can also replace t with 􀀀x3, to obtain:  \n2  \n6 6  \n6 6  \n6 6  \n6  \n4  \nx1  \nx2  \nx3  \nx4  \n3 2 7 6 7 6 7 6  \n7 = 6 7 6 7 6 7 6 5 4  \n􀀀1  \n1  \n1  \n0  \n3 2 7 6 7 6 7 6  \n7 x3 + 6 7 6 7 6 5 4  \nThe largest possible value for x3 is therefore 20 .  \n20  \n60  \n0  \n60  \n3  \n7 7  \n7 , 0 􀀔 x3 􀀔 20.  \n7 7  \n5  \n􀀃  \nREFERENCES  \n[LLM21] David Lay, Stephen Lay, and Judi McDonald, Linear Algebra and its Applications, Sixth edition, Pearson, 2021 .  \nUNIVERSITY OF COLORADO, DEPARTMENT OF MATHEMATICS, CAMPUS BOX 395, BOULDER, CO 80309 Email address: [casa@math.colorado.edu](casa@math.colorado.edu)","cbCaipBs2PHSQR31","https://ap.wps.com/l/cbCaipBs2PHSQR31","pdf",103429,"English","# Exercise 1.6.11\n## Problem statement\n## Node balance equations\n## Augmented matrix and RREF\n## Parameter solution with nonnegativity constraints\n## Maximum value of x3","[{\"question\":\"What equations are used to model flow conservation at each node?\",\"answer\":\"The solution sets up three node balance equations: x1 + x3 = 20, x2 - x3 - x4 = 0, and -x1 - x2 = -80, derived from equal total input and output at each node.\"},{\"question\":\"How is the augmented matrix used in the solution?\",\"answer\":\"The system is rewritten into an augmented matrix and then reduced to RREF using row operations, yielding a parameterized form of the variables.\"},{\"question\":\"How are nonnegativity constraints applied to find the largest possible x3?\",\"answer\":\"Nonnegativity requires x1, x2, x3, x4 to be ≥ 0, which translates to bounds on the parameter t. The resulting range for t gives the maximum feasible x3 value as 20.\"}]","Exercise 1.6.11 - Find the largest possible value for x3 | PDF"]