[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-661661-105":3,"detail-sidebar-cat-0-en-105":80,"doc-detail-661661-en":130},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":73,"head_meta":75,"extra_data":77,"updated_unix":79},105,"en","introduction-to-linear-algebra-sixth-edition-solutions-to-problem-sets-problem-set-11","Introduction to Linear Algebra - Sixth Edition - Solutions to Problem Sets - Problem Set 1.1","","Solutions to Problem Sets for “Introduction to Linear Algebra” (Sixth Edition) by Gilbert Strang, focusing on Problem Set 1.1. The solutions derive relationships among vectors, points, and matrices by using rank conditions for 2×2 matrices, geometric interpretations of parallelograms and triangles, and combination rules in R^2 and R^3. They also cover component-wise constraints, lattice and cone/wedge filling regions, and systems of linear equations, including vector-sum identities and 3D coverage arguments.",{"@graph":14,"@context":72},[15,34,55],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/introduction-to-linear-algebra-sixth-edition-solutions-to-problem-sets-problem-set-11/661661/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/introduction-to-linear-algebra-sixth-edition-solutions-to-problem-sets-problem-set-11/661661.png","ImageObject",300,407,{"name":42,"@type":43},"awa","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-29","2026-10-28",true,{"@type":52,"interactionType":53,"userInteractionCount":26},"InteractionCounter",{"@type":54},"ViewAction",{"@type":56,"mainEntity":57},"FAQPage",[58,64,68],{"name":59,"@type":60,"acceptedAnswer":61},"How do the solutions interpret the condition ad = bc in a 2×2 matrix context?","Question",{"text":62,"@type":63},"With no zero entries, ad = bc is used as the equation ensuring a 2×2 matrix has rank 1.","Answer",{"name":65,"@type":60,"acceptedAnswer":66},"How are parallelogram sides and diagonals represented in the solutions?",{"text":67,"@type":63},"The solutions treat v+w and v−w as diagonals, then use them to reconstruct the side vectors v and w for the parallelogram.",{"name":69,"@type":60,"acceptedAnswer":70},"What does the problem set explain about vector combinations filling regions such as wedges or cones?",{"text":71,"@type":63},"By constraining coefficients (e.g., c ≥ 0 and d ≥ 0), combinations cv + dw fill a wedge/cone between v and w; special cases like w = −v reduce the region to a half-space or a line.","https://schema.org",{"og:url":32,"og:type":74,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":76,"canonical":32},"index,follow",{"doc_id":78,"site_id":7},661661,1793173569,{"code":4,"msg":81,"data":82},"success",[83,87,91,94,99,104,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":84,"show_sort_weight":85,"slug":86},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":88,"show_sort_weight":89,"slug":90},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":92,"slug":93},70,"exam",{"id":95,"doc_module":4,"doc_module_name":25,"category_name":96,"show_sort_weight":97,"slug":98},5,"Comic",60,"comic",{"id":100,"doc_module":4,"doc_module_name":25,"category_name":101,"show_sort_weight":102,"slug":103},6,"Technology",50,"technology",{"id":105,"doc_module":4,"doc_module_name":25,"category_name":106,"show_sort_weight":107,"slug":108},7,"Healthcare",40,"healthcare",{"id":110,"doc_module":4,"doc_module_name":25,"category_name":111,"show_sort_weight":112,"slug":113},8,"Research & Report",30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":95,"slug":129},19,"General","general",{"code":4,"msg":81,"data":131},{"doc_id":78,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":26,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":139,"language":140,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":12,"update_tm":79,"read_time":144},3985747858093,"https://ap-avatar.wpscdn.com/davatar_3d24733baf745e90a7e4bdd5f77d97b2","INTRODUCTION TO LINEAR ALGEBRA  \nSixth Edition  \nSOLUTIONS TO PROBLEM SETS  \nGilbert Strang  \nMassachusetts Institute of Technology [math.mit.edu/weborder.php](math.mit.edu/weborder.php) (orders) [math.mit.edu/linearalgebra](math.mit.edu/linearalgebra) (book website)  \n[math.mit.edu/](math.mit.edu/)∼gs (author website) [www.wellesleycambridge.com](www.wellesleycambridge.com) (all books)  \nWellesley-Cambridge Press  \nBox 812060  \nWellesley, Massachusetts 02482  \nProblem Set 1.1, page 6  \n1 c = ma and d = mb lead to ad = amb = bc. With no zeros, ad = bc is the equation for a 2 × 2 matrix to have rank 1.  \n2 The three edges going around the triangle are u = (5 , 0) , v = (−5 , 12) , w = (0 , −12) . Their sum is u + v + w = (0 , 0) . Their lengths are ||u|| = 5 , ||v|| = 13 , ||w|| = 12 . This is a 5 − 12 − 13 right triangle with 52 + 122 = 25 + 144 = 169 = 132—the best numbers after the 3 − 4 − 5 right triangle if we don’t count 6 − 8 − 10.  \n3 The combinations give (a) a line in R3 (b) a plane in R3 (c) all of R3.  \n4 v + w = (2 , 3) and v − w = (6 , −1) will be the diagonals of the parallelogram with  \nv and w as two sides going out from (0 , 0) .  \nv + w = \" 23\\# w = \" − 22\\# v = \" 41\\#  \nv − w = \" − 61\\#−w  \n5 This problem gives the diagonals v + w = (5 , 1) and v − w = (1 , 5) of the parallelogram and asks for the sides v and w : The opposite of Problem 4 . In this example v = (3 , 3) and w = (2 , −2) . Those come from v = 12 (v + w) + 12 (v − w) and w = 12 (v + w) − 12 (v − w) .  \nw  \n2 Solutions to Problem Sets  \n6 3v + w = (7 , 5) and cv + dw = (2c + d, c + 2d) .  \n7 u+v = (−2 , 3 , 1) and u+v+w = (0 , 0 , 0) and 2u+2v+w = ( add first answers) =(−2 , 3 , 1) . The vectors u, v , w are in the same plane because a combination u+v+w gives (0 , 0 , 0) . Stated another way: u = −v − w is in the plane of v and w.  \n8 The components of every cv+dw add to zero because the components of v = (1 , −2 , 1) and of w = (0 , 1 , −1) add to zero. c = 3 and d = 9 give 3v + 9w = (3 , 3 , −6) . Thereis no solution to cv + dw = (3 , 3 , 6) because 3 + 3 + 6 is not zero.  \n9 The nine combinations c(2 , 1) + d(0 , 1) with c = 0 , 1 , 2 and d = 0 , 1 , 2 will lie on a  \nlattice. If we took all whole numbers c and d, the lattice would lie over the whole plane.  \nc = 2, d = 2  \nc = 0, d = 2 c = 2, d = 0  \nc = 0, d = 1  \nc = 0, d = 0  \n10 The question is whether (a, b, c) is a combination x1 u + x2 v. Can we solve  \n􀀲 1 􀀳 􀀲 0 􀀳 􀀲 a 􀀳  \n􀀶 􀀷 􀀶 􀀷 􀀶 􀀷 x 1  1  + x2  1  =  b  ?  \n􀀴 0 􀀵 􀀴 1 􀀵 􀀴 c 􀀵  \nCertainly x1 has to be a. Certainly x2 has to be c. So the middle components give the requirement a + c = b.  \n11 The fourth corner can be (4 , 4) or (4 , 0) or (−2 , 2) . Draw 3 possible parallelograms!  \n12 Four more corners (1 , 1 , 0) ,(1 , 0 , 1) ,(0 , 1 , 1) ,(1 , 1 , 1) . The center point is 􀀀 12 , 12 , 12 􀀁 . Centers of 6 faces: 􀀀 12 , 12 , 0 􀀁 , 􀀀 12 , 12 , 1 􀀁&􀀀0 , 12 , 12 􀀁 , 􀀀1 , 12 , 12 􀀁&􀀀 12 , 0 , 12 􀀁 , 􀀀 12 , 1 , 12 􀀁 . 12 edges.  \n13 The combinations of i = (1 , 0 , 0) and i + j = (1 , 1 , 0) fill the xy plane in xyz space.  \n14 (a) Sum = zero vector. (b) Sum = −2:00 vector = 8:00 vector.  \n(c) 2:00 is 30◦ from horizontal = (cos π6 , sin π6 ) = ( √3/2 , 1/2) .  \nSolutions to Problem Sets 3  \n15 Moving the origin to 6:00 adds j = (0 , 1) to every vector. So the sum of twelve vectors changes from 0 to 12j = (0 , 12) .  \n16 First part: u, v , w are all in the same direction.  \nSecond part: Some combination of u, v , w gives the zero vector but those 3 vectors are not on a line. Then their combinations fill a plane in 3D.  \n17 The two equations are c + 3d = 14 and 2c + d = 8 . The solution is c = 2 and d = 4 .  \n3   1  \n18 The point v + w is three-fourths of the way to v starting from w. The vector  \n4 4  \n1  1   1  1  \nv + w is halfway to u = v + w. The vector v + w is 2u (the far corner of the  \n4 4 2 2 parallelogram) .  \n19 The combinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelogram with sides v and w. For example, if v = (1 , 0)","cbCaiozTflvb6Aih","https://ap.wps.com/l/cbCaiozTflvb6Aih","pdf",3329939,141,"English","# Problem Set 1.1\n## Vector equations and rank conditions\n## Geometry of triangles, parallelograms, and cones\n## Combinations in R3 and linear systems","[{\"question\":\"How do the solutions interpret the condition ad = bc in a 2×2 matrix context?\",\"answer\":\"With no zero entries, ad = bc is used as the equation ensuring a 2×2 matrix has rank 1.\"},{\"question\":\"How are parallelogram sides and diagonals represented in the solutions?\",\"answer\":\"The solutions treat v+w and v−w as diagonals, then use them to reconstruct the side vectors v and w for the parallelogram.\"},{\"question\":\"What does the problem set explain about vector combinations filling regions such as wedges or cones?\",\"answer\":\"By constraining coefficients (e.g., c ≥ 0 and d ≥ 0), combinations cv + dw fill a wedge/cone between v and w; special cases like w = −v reduce the region to a half-space or a line.\"}]","Introduction to Linear Algebra - Sixth Edition - Solutions to Problem Sets - Problem Set 1.1 | PDF",355]