{"id":20040,"date":"2024-04-15T05:47:34","date_gmt":"2024-04-15T05:47:34","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20040"},"modified":"2024-04-15T05:47:34","modified_gmt":"2024-04-15T05:47:34","slug":"the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/","title":{"rendered":"The inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&2 \\\\ 5&{ – 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ – 2} \\\\ { – 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&{ – 2} \\\\ { – 5}&{ – 1} \\end{array}} \\right]\\]"},"content":{"rendered":"

[amp_mcq option1=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&2 \\\\ 5&{ – 1} \\end{array}} \\right]\\]” option2=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\]” option3=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ – 2} \\\\ { – 5}&1 \\end{array}} \\right]\\]” option4=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&{ – 2} \\\\ { – 5}&{ – 1} \\end{array}} \\right]\\]” correct=”option3″]<\/p>\n

The correct answer is $\\boxed{\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\ -5&1 \\end{array}} \\right]}$.<\/p>\n

To find the inverse of a 2×2 matrix, we can use the formula $A^{-1} = \\frac{1}{|A|}\\left[ {\\begin{array}{*{20}{c}} \\left| \\begin{array}{cc} a_{2,2} & a_{2,1} \\\\ a_{1,2} & a_{1,1} \\end{array} \\right| & \\left| \\begin{array}{cc} a_{1,1} & a_{1,2} \\\\ a_{2,1} & a_{2,2} \\end{array} \\right| \\\\ \\left| \\begin{array}{cc} a_{2,2} & a_{2,3} \\\\ a_{1,2} & a_{1,3} \\end{array} \\right| & \\left| \\begin{array}{cc} a_{1,3} & a_{1,2} \\\\ a_{2,3} & a_{2,2} \\end{array} \\right| \\\\ \\end{array}} \\right]$, where $a_{i,j}$ is the element of $A$ in the $i$th row and $j$th column.<\/p>\n

In this case, $A = \\left[ {\\begin{array}{{20}{c}} 1&2 \\ 5&7 \\end{array}} \\right]$, so $|A| = \\left| \\begin{array}{cc} 7 & 2 \\\\ 5 & 1 \\end{array} \\right| = 3$. Therefore, the inverse of $A$ is $\\frac{1}{3}\\left[ {\\begin{array}{<\/em>{20}{c}} 7&2 \\ -5&1 \\end{array}} \\right]$.<\/p>\n

Here is a step-by-step solution:<\/p>\n

    \n
  1. Find $|A|$.<\/li>\n<\/ol>\n

    $|A| = \\left| \\begin{array}{cc} 7 & 2 \\\\ 5 & 1 \\end{array} \\right| = 3$<\/p>\n

      \n
    1. Divide $|A|$ by itself to find $A^{-1}$.<\/li>\n<\/ol>\n

      $A^{-1} = \\frac{1}{|A|}\\left[ {\\begin{array}{{20}{c}} 7&2 \\ -5&1 \\end{array}} \\right] = \\frac{1}{3}\\left[ {\\begin{array}{<\/em>{20}{c}} 7&2 \\ -5&1 \\end{array}} \\right]$<\/p>\n","protected":false},"excerpt":{"rendered":"

      [amp_mcq option1=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&2 \\\\ 5&{ – 1} \\end{array}} \\right]\\]” option2=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\]” option3=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ – 2} \\\\ { – 5}&1 \\end{array}} \\right]\\]” option4=”\\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&{ – 2} \\\\ { – 5}&{ – 1} \\end{array}} \\right]\\]” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-20040","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\nThe inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&2 \\\\ 5&{ - 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ - 2} \\\\ { - 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&{ - 2} \\\\ { - 5}&{ - 1} \\end{array}} \\right]\\]<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a\u0097-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&2 \\\\ 5&{ - 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ - 2} \\\\ { - 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&{ - 2} \\\\ { - 5}&{ - 1} \\end{array}} \\right]\\]\" \/>\n<meta property=\"og:description\" content=\"[amp_mcq option1=”[frac{1}{3}left[ {begin{array}{*{20}{c}} { – 7}&2 \\ 5&{ – 1} end{array}} right]]” option2=”[frac{1}{3}left[ {begin{array}{*{20}{c}} 7&2 \\ 5&1 end{array}} right]]” option3=”[frac{1}{3}left[ {begin{array}{*{20}{c}} 7&{ – 2} \\ { – 5}&1 end{array}} right]]” option4=”[frac{1}{3}left[ {begin{array}{*{20}{c}} { – 7}&{ – 2} \\ { – 5}&{ – 1} end{array}} right]]” correct=”option3″]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a\u0097-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2024-04-15T05:47:34+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"The inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&2 \\\\ 5&{ - 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ - 2} \\\\ { - 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&{ - 2} \\\\ { - 5}&{ - 1} \\end{array}} \\right]\\]","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a\u0097-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/","og_locale":"en_US","og_type":"article","og_title":"The inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&2 \\\\ 5&{ - 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ - 2} \\\\ { - 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&{ - 2} \\\\ { - 5}&{ - 1} \\end{array}} \\right]\\]","og_description":"[amp_mcq option1=”[frac{1}{3}left[ {begin{array}{*{20}{c}} { – 7}&2 \\ 5&{ – 1} end{array}} right]]” option2=”[frac{1}{3}left[ {begin{array}{*{20}{c}} 7&2 \\ 5&1 end{array}} right]]” option3=”[frac{1}{3}left[ {begin{array}{*{20}{c}} 7&{ – 2} \\ { – 5}&1 end{array}} right]]” option4=”[frac{1}{3}left[ {begin{array}{*{20}{c}} { – 7}&{ – 2} \\ { – 5}&{ – 1} end{array}} right]]” correct=”option3″]","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a\u0097-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2024-04-15T05:47:34+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/","name":"The inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&2 \\\\ 5&{ - 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ - 2} \\\\ { - 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { - 7}&{ - 2} \\\\ { - 5}&{ - 1} \\end{array}} \\right]\\]","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2024-04-15T05:47:34+00:00","dateModified":"2024-04-15T05:47:34+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-inverse-of-the-2-a%c2%97-2-matrix-left-beginarray20c-12-57-endarray-right-is-a-frac13left-beginarray20c-72-5-1-endar\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"mcq","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/mcq\/"},{"@type":"ListItem","position":3,"name":"Linear Algebra","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/mcq\/linear-algebra\/"},{"@type":"ListItem","position":4,"name":"The inverse of the 2 \u00c3\u0097 2 matrix \\[\\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 5&7 \\end{array}} \\right]\\] is A. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&2 \\\\ 5&{ – 1} \\end{array}} \\right]\\] B. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&2 \\\\ 5&1 \\end{array}} \\right]\\] C. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} 7&{ – 2} \\\\ { – 5}&1 \\end{array}} \\right]\\] D. \\[\\frac{1}{3}\\left[ {\\begin{array}{*{20}{c}} { – 7}&{ – 2} \\\\ { – 5}&{ – 1} \\end{array}} \\right]\\]"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/20040","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=20040"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/20040\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=20040"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=20040"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=20040"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}