{"id":2293,"date":"2016-11-03T20:04:12","date_gmt":"2016-11-03T20:04:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2293"},"modified":"2017-04-04T19:55:19","modified_gmt":"2017-04-04T19:55:19","slug":"summary-solve-systems-with-inverses","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/summary-solve-systems-with-inverses\/","title":{"raw":"Summary: Solve Systems With Inverses","rendered":"Summary: Solve Systems With Inverses"},"content":{"raw":"<h2>Key Equations<\/h2>\r\n<table id=\"eip-id1165137848559\" summary=\"..\"><colgroup> <col data-align=\"left\" \/> <col data-align=\"left\" \/><\/colgroup>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td>Identity matrix for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\r\n<td>[latex]{I}_{2}=\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>Identity matrix for a [latex]\\text{3}\\text{}\\times \\text{}3[\/latex] matrix<\/td>\r\n<td>[latex]{I}_{3}=\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>Multiplicative inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\r\n<td>[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d&amp; -b\\\\ -c&amp; a\\end{array}\\right],\\text{ where }ad-bc\\ne 0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>An identity matrix has the property [latex]AI=IA=A[\/latex].<\/li>\r\n \t<li>An invertible matrix has the property [latex]A{A}^{-1}={A}^{-1}A=I[\/latex].<\/li>\r\n \t<li>Use matrix multiplication and the identity to find the inverse of a [latex]2\\times 2[\/latex] matrix.<\/li>\r\n \t<li>The multiplicative inverse can be found using a formula.<\/li>\r\n \t<li>Another method of finding the inverse is by augmenting with the identity.<\/li>\r\n \t<li>We can augment a [latex]3\\times 3[\/latex] matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse.<\/li>\r\n \t<li>Write the system of equations as [latex]AX=B[\/latex], and multiply both sides by the inverse of [latex]A:{A}^{-1}AX={A}^{-1}B[\/latex].<\/li>\r\n \t<li>We can also use a calculator to solve a system of equations with matrix inverses.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<strong>identity matrix<\/strong> a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra\r\n\r\n<strong>multiplicative inverse of a matrix<\/strong> a matrix that, when multiplied by the original, equals the identity matrix","rendered":"<h2>Key Equations<\/h2>\n<table id=\"eip-id1165137848559\" summary=\"..\">\n<colgroup>\n<col data-align=\"left\" \/>\n<col data-align=\"left\" \/><\/colgroup>\n<tbody>\n<tr valign=\"middle\">\n<td>Identity matrix for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\n<td>[latex]{I}_{2}=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Identity matrix for a [latex]\\text{3}\\text{}\\times \\text{}3[\/latex] matrix<\/td>\n<td>[latex]{I}_{3}=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Multiplicative inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\n<td>[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d& -b\\\\ -c& a\\end{array}\\right],\\text{ where }ad-bc\\ne 0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>An identity matrix has the property [latex]AI=IA=A[\/latex].<\/li>\n<li>An invertible matrix has the property [latex]A{A}^{-1}={A}^{-1}A=I[\/latex].<\/li>\n<li>Use matrix multiplication and the identity to find the inverse of a [latex]2\\times 2[\/latex] matrix.<\/li>\n<li>The multiplicative inverse can be found using a formula.<\/li>\n<li>Another method of finding the inverse is by augmenting with the identity.<\/li>\n<li>We can augment a [latex]3\\times 3[\/latex] matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse.<\/li>\n<li>Write the system of equations as [latex]AX=B[\/latex], and multiply both sides by the inverse of [latex]A:{A}^{-1}AX={A}^{-1}B[\/latex].<\/li>\n<li>We can also use a calculator to solve a system of equations with matrix inverses.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<p><strong>identity matrix<\/strong> a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra<\/p>\n<p><strong>multiplicative inverse of a matrix<\/strong> a matrix that, when multiplied by the original, equals the identity matrix<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2293\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"}]","CANDELA_OUTCOMES_GUID":"f39333f2-de91-462f-8a07-dcbf13b6b6f0","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2293","chapter","type-chapter","status-publish","hentry"],"part":2258,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2293","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2293\/revisions"}],"predecessor-version":[{"id":3252,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2293\/revisions\/3252"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/2258"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2293\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=2293"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2293"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=2293"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=2293"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}