[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-347710-105":59,"doc-detail-347710-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","linearalgebrasolutionscomplete","linearAlgebraSolutionsComplete","","Worked linear algebra solutions for systems of linear equations using standard row operations. The notes explain how to replace rows (e.g., R2 with R2 + kR1), scale rows, and perform eliminations to reach triangular or row-echelon forms. Multiple numbered examples compute solution sets for variables, including unique solutions such as (−8, 3), (12, −7), (4/7, 9/7), and cases where the system is inconsistent and has no solution, shown via an equation of the form 0 = 1.",{"@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/linearalgebrasolutionscomplete/347710/",{"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/linearalgebrasolutionscomplete/347710.png","ImageObject",300,407,{"name":92,"@type":93},"Elsa","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-09-24","2026-09-22",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 do R1, R2, and similar symbols mean in these solutions?","Question",{"text":112,"@type":113},"R1, R2, etc. denote row 1, row 2 (or the corresponding equation/row), which are manipulated during elimination.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"How are row operations applied to simplify a linear system?",{"text":117,"@type":113},"The procedures replace one row with itself plus a multiple of another row, and scale rows to create zeros and move toward triangular form.",{"name":119,"@type":110,"acceptedAnswer":120},"How can the solutions determine when a system has no solution?",{"text":121,"@type":113},"An inconsistency is detected when row operations produce an equation equivalent to 0 = 1, indicating the system cannot be satisfied.","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},347710,1790293614,{"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":139,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":145},137455077381,"https://ap-avatar.wpscdn.com/davatar_994ba38a5ba835b3df7d355c54d3ed8d","1.1 SOLUTIONS    \nNotes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.  \nx1 + 5x2 = 7 􀂪 1 5 7􀂺  \n1 . 􀂫 􀂻  \n−2x1 − 7x2 = −5 􀂬−2 −7 −5􀂼  \nReplace R2 by R2 + (2)R1 and obtain:  \nScale R2 by 1/3:  \nReplace R1 by R1 + (–5)R2:  \nThe solution is (x1, x2) = (–8, 3), or simply (–8, 3) .  \n2. 2x1 + 4x2 = −4 􀂪 2 4 −4􀂺  \n5x1 + 7x2 = 11 􀂬􀂫 5 7 11 􀂼􀂻 Scale R1 by 1/2 and obtain:  \nReplace R2 by R2 + (–5)R1:  \nScale R2 by –1/3:  \nReplace R1 by R1 + (–2)R2:  \nx + 5x = 7  \n1 2  \n3x = 9 2  \nx + 5x = 7  \n1 2 x = 3  \n2 x = −8  \n1  \nx = 3 2  \nx + 2x = −2 1 2  \n5x + 7x = 11 1 2  \nx + 2x = −2 1 2  \n−3x = 21  \n2  \nx + 2x = −2 1 2  \nx2 = −7  \nx = 12  \n1  \nx = −7 2  \n􀂪 1 5 7 􀂺  \n􀂬􀂫 0 3 9􀂼􀂻  \n􀂪 1 5 7 􀂺  \n􀂬􀂫 0 1 3􀂼􀂻  \n􀂪 1 0 −8􀂺􀂬􀂫 0 1 3􀂼􀂻  \n􀂪 1 2 −2􀂺􀂬􀂫 5 7 11 􀂼􀂻􀂪 1 2 −2􀂺􀂬􀂫 0 −3 21􀂼􀂻􀂪 1 2 −2􀂺􀂬􀂫 0 1 −7􀂼􀂻􀂪 1 0 12􀂺􀂬􀂫 0 1 −7􀂼􀂻  \nThe solution is (x1, x2) = (12,–7), or simply (12,–7) .  \n2 CHAPTER 1 • Linear Equations in Linear Algebra  \n3. The point of intersection satisfies the system of two linear equations:  \nx + 5x = 7  \n1 2  \nx − 2x = −2  \n1 2  \nReplace R2 by R2 +  \nScale R2 by –1/7:  \nReplace R1 by R1 +  \n􀂪 1  \n􀂬􀂫 1  \n5  \n−2  \n7 􀂺  \n−2􀂼􀂻  \n(–1)R1 and obtain:  \n(–5)R2:  \nx + 5x = 7 1 2  \n−7x = −9  \n2  \nx + 5x = 7 1 2  \nx2 = 9/7  \nx1 = 4/7 x2 = 9/7  \nThe point of intersection is (x1, x2) = (4/7, 9/7) .  \n4. The point of intersection satisfies the system of two linear equations:  \nx − 5x = 1 1 2  \n3x − 7x = 5 1 2  \n􀂪 1  \n􀂬􀂫 3  \n−5  \n−7  \n1􀂺  \n5􀂼􀂻  \nx − 5x = 1  \nReplace R2 by R2 + (–3)R1 and obtain: 1 2  \n8x = 2 2  \nScale R2 by 1/8:  \nReplace R1 by R1 + (5)R2:  \nx − 5x = 1 1 2  \nx2 = 1/4 x = 9/4  \n1  \nx2 = 1/4  \n􀂪 1  \n􀂬􀂫 0 􀂪 1  \n􀂬􀂫 0 􀂪 1  \n􀂬􀂫 0  \n􀂪 1  \n􀂬􀂫 0 􀂪 1  \n􀂬􀂫 0 􀂪 1  \n􀂬􀂫 0  \n5  \n−7  \n5  \n1  \n0  \n1  \n−5  \n8  \n−5  \n1  \n0  \n1  \n7 􀂺  \n−9􀂼􀂻  \n7 􀂺 9/7 􀂼􀂻 4/7 􀂺 9/7 􀂼􀂻  \n1􀂺  \n2􀂼􀂻  \n1􀂺 1/4 􀂼􀂻 9/4 􀂺 1/4 􀂼􀂻  \nThe point of intersection is (x1, x2) = (9/4, 1/4) .  \n5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3 .  \n6. One more step will put the system in triangular form. Replace R4 by its sum with –3 times R3, which  \n􀂪 1 􀂫  \nproduces 􀂫 00 􀂫􀂬 0  \n−6  \n2  \n0  \n0  \n4  \n−7  \n1  \n0  \n0  \n0  \n2  \n−5  \n−1􀂺  \n4 􀂻  \n􀂻 . After that, the next step is to scale the fourth row by –1/5 .−3 􀂻  \n􀂻  \n15􀂼  \n7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0 x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident.  \nThe solution set is empty.  \n1.1 • Solutions 3  \n8. The standard row operations are:  \n􀂪 1  \n􀂫 0  \n􀂫􀂬 0  \n−4  \n1  \n0  \n9  \n7  \n2  \n0􀂺 􀂪 1  \n0 􀂻 ~ 􀂫 0 0􀂼􀂻 􀂫􀂬 0  \n−4  \n1  \n0  \n9  \n7  \n1  \n0􀂺 􀂪 1  \n0 􀂻 ~ 􀂫 0 0􀂼􀂻 􀂫􀂬 0  \n−4  \n1  \n0  \n0  \n0  \n1  \n0􀂺 􀂪 1  \n0 􀂻 ~ 􀂫 0 0􀂼􀂻 􀂫􀂬 0  \n0  \n1  \n0  \n0  \n0  \n1  \n0􀂺  \n0 􀂻  \n0􀂼􀂻  \nThe solution set contains one solution: (0, 0, 0) .  \n9. The system has already been reduced to triangular form. Begin by scaling the fourth row by 1/2 and then replacing R3 by R3 + (3)R4:  \n􀂪 1  \n􀂫 0  \n􀂫  \n􀂫 0  \n􀂫  \n􀂬 0  \n−1 0 0  \n1 −3 0  \n0 1 −3  \n0 0 2  \n−4􀂺 􀂪 1 −7 􀂻 ~ 􀂫 0 −1 􀂻 􀂫 0  \n􀂻 􀂫  \n4􀂼 􀂬 0  \n−1 0 0  \n1 −3 0  \n0 1 −3  \n0 0 1  \n−4􀂺 􀂪 1 7 􀂻 􀂫 0  \n􀂻 ~ 􀂫  \n−1 􀂻 􀂫 0 􀂻 􀂫  \n2􀂼 􀂬 0  \n−1  \n1  \n0  \n0  \n0  \n−3  \n1  \n0  \n0  \n0  \n0  \n1  \n−4􀂺  \n−7 􀂻  \n5 􀂻  \n􀂻  \n2􀂼  \nNext, replace R2 by R2 + (3)R3 . Finally, replace R1 by R1 + R2:  \n􀂪 1 −1 0 0 −4􀂺 􀂪 1 􀂫 0 1 0 0 8 􀂻 ~ 􀂫 0  \n~  \n􀂫 0 0 1 0 5 􀂻 􀂫 0  \n􀂫 􀂻 􀂫  \n􀂬 0 0 0 1 2􀂼 􀂬 0  \n0  \n1","cbCaijW25Zn3T27D","https://ap.wps.com/l/cbCaijW25Zn3T27D","pdf",5812947,423,"English","# Solutions\n## Notes and row-operation conventions\n## Solving systems and identifying intersections\n## Triangular form and inconsistency cases\n## Standard row operations","[{\"question\":\"What do R1, R2, and similar symbols mean in these solutions?\",\"answer\":\"R1, R2, etc. denote row 1, row 2 (or the corresponding equation/row), which are manipulated during elimination.\"},{\"question\":\"How are row operations applied to simplify a linear system?\",\"answer\":\"The procedures replace one row with itself plus a multiple of another row, and scale rows to create zeros and move toward triangular form.\"},{\"question\":\"How can the solutions determine when a system has no solution?\",\"answer\":\"An inconsistency is detected when row operations produce an equation equivalent to 0 = 1, indicating the system cannot be satisfied.\"}]","linearAlgebraSolutionsComplete | PDF",1790068680,1066]