[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-484779-105":59,"doc-detail-484779-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","mathematics-practice-problem-29-matrix-algebra-analysis","Mathematics Practice Problem 29 - Matrix Algebra Analysis","","This document presents a detailed mathematical solution for a linear algebra problem involving n x n matrices A and B. The problem asks to determine which property is always true given the condition BA + B^2 = I - BA^2. Through a rigorous step-by-step derivation, the document demonstrates that matrix B is non-singular by rearranging the equation into B(A^2 + A + B) = I. Furthermore, the analysis uses counterexamples for specific matrix values (such as zero and identity matrices) to disprove that A, A+B, or AB must be non-singular, thereby confirming that only the non-singularity of B is a valid conclusion. This content serves as an excellent resource for students reviewing matrix theory, inverse operations, and proof techniques within advanced algebra or collegiate mathematics courses. It effectively highlights the importance of logical verification and the use of counterexamples in determining the properties of matrix equations.",{"@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/mathematics-practice-problem-29-matrix-algebra-analysis/484779/",{"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/mathematics-practice-problem-29-matrix-algebra-analysis/484779.png","ImageObject",300,407,{"name":92,"@type":93},"CatatanPagi","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-28","2026-09-30",true,{"@type":102,"interactionType":103,"userInteractionCount":14},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"What is the given matrix condition in Problem 29?","Question",{"text":112,"@type":113},"The given condition for n x n matrices A and B is BA + B^2 = I - BA^2, where I is the identity matrix.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"Why is matrix B considered non-singular based on the derivation?",{"text":117,"@type":113},"The equation can be rearranged to B(A^2 + A + B) = I. Since the product of B and another matrix equals the identity matrix, B must have an inverse, which confirms that B is non-singular.",{"name":119,"@type":110,"acceptedAnswer":120},"Are matrices A and AB necessarily non-singular under the given condition?",{"text":121,"@type":113},"No. By taking A = 0 and B = I, the condition is satisfied (I*0 + I^2 = I - I*0^2 becomes I = I), but in this case, A and AB are both equal to 0, which is singular.","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},484779,1792876837,{"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":14,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":81,"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":143,"read_time":39,"preview_page_count":4},962090894170,"https://ap-avatar.wpscdn.com/davatar_6f874abed73319feea01a86fa6f0fab8","Njsla algebra 1 standards . Algebra 1 njsla . What is njsla test . What is algebra for beginners .  \nThe NJSLA encompasses a range of subjects across various grade levels: -English language arts for grades 3 to 9-Mathematics for grades 3 to 8-Science for grades 5, 8, and 11-High school level Algebra I, Geometry, and Algebra II, corresponding with the respective courses While the NJSLA is primarily conducted on computers, paper versions are available for students who need them, and some tests are offered in Spanish . The computer-based format includes diverse question types such as multiple choice, constructed response, and technology-enhanced items . High school students in New Jersey are also required to pass the New Jersey Graduation Proficiency Assessment (NJGPA) in English language arts and mathematics to qualify for graduation . This assessment is carried out on the same platform as the NJSLA. For students with significant cognitive disabilities who are unable to participate in the NJSLA with accommodations, the Dynamic Learning Maps (DLM) alternate assessment is an option . This assessment mirrors the NJSLA in terms of covered grades and subjects and is also computer-based . Additionally, the academic English language proficiency of English language learners is evaluated through the ACCESS for ELLs test, with a paper-based version for kindergarteners and a computer-based test for grades 1 to 12 . An alternate version is available for English learners with significant cognitive disabilities . The testing period for the NJSLA and other related assessments in New Jersey typically begins in April . To prepare students for the NJSLA, practice tests can be a valuable resource . bebesu These tests, available on platforms like Pear, help  \nstudents build the skills and confidence necessary to perform well on the actual NJSLA assessments .","cbCaifsgdtgJCwqO","https://ap.wps.com/l/cbCaifsgdtgJCwqO","pdf",211990,"English","# Problem 29 Analysis\n## Matrix Equation Derivation\n## Verification of Properties via Counterexamples","[{\"question\":\"What is the given matrix condition in Problem 29?\",\"answer\":\"The given condition for n x n matrices A and B is BA + B^2 = I - BA^2, where I is the identity matrix.\"},{\"question\":\"Why is matrix B considered non-singular based on the derivation?\",\"answer\":\"The equation can be rearranged to B(A^2 + A + B) = I. Since the product of B and another matrix equals the identity matrix, B must have an inverse, which confirms that B is non-singular.\"},{\"question\":\"Are matrices A and AB necessarily non-singular under the given condition?\",\"answer\":\"No. By taking A = 0 and B = I, the condition is satisfied (I*0 + I^2 = I - I*0^2 becomes I = I), but in this case, A and AB are both equal to 0, which is singular.\"}]","Mathematics Practice Problem 29 - Matrix Algebra Analysis | PDF",1790803061]