{"id":138,"date":"2025-02-13T22:44:13","date_gmt":"2025-02-13T22:44:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-inverses\/"},"modified":"2026-03-24T17:39:12","modified_gmt":"2026-03-24T17:39:12","slug":"solving-systems-with-inverses","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-inverses\/","title":{"raw":"Solving Systems with Inverses: Learn It 1","rendered":"Solving Systems with Inverses: Learn It 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find the inverse of a matrix.<\/li>\r\n \t<li>Solve a system of linear equations using an inverse matrix.<\/li>\r\n<\/ul>\r\n<\/section>In this section, we will explore how to solve systems of linear equations using the inverse of a matrix. This method is particularly useful for handling larger systems and provides a systematic and efficient approach to finding solutions. We will discuss the advantages of using matrix inversion and outline the steps involved in this process. Understanding this method will enhance your ability to solve complex systems and apply these techniques in various practical scenarios.\r\n<h2>Finding the Inverse of a Matrix<\/h2>\r\nWe know that the multiplicative inverse of a real number [latex]a[\/latex] is [latex]{a}^{-1}[\/latex] and [latex]a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1[\/latex]. For example, [latex]{2}^{-1}=\\frac{1}{2}[\/latex] and [latex]\\left(\\frac{1}{2}\\right)2=1[\/latex].\r\n\r\nThe <strong>multiplicative inverse of a matrix<\/strong> is similar in concept, except that the product of matrix [latex]A[\/latex] and its inverse [latex]{A}^{-1}[\/latex] equals the <strong>identity matrix<\/strong>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by [latex]{I}_{n}[\/latex] where [latex]n[\/latex] represents the dimension of the matrix. The equations below\u00a0are the identity matrices for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and [latex]3\\text{}\\times \\text{}3[\/latex] matrix, respectively.\r\n<p style=\"text-align: center;\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nThe identity matrix acts as a [latex]1[\/latex] in matrix algebra. For example, [latex]AI=IA=A[\/latex].\r\n\r\nA matrix that has a multiplicative inverse has the properties\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/p>\r\nA matrix that has a multiplicative inverse is called an <strong>invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>the identity matrix and multiplicative inverse<\/h3>\r\nThe <strong>identity matrix<\/strong>, [latex]{I}_{n}[\/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2} &amp; =\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\begin{array}{cccc}&amp; &amp; &amp; \\end{array} &amp; {I}_{3} &amp; =\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\hfill \\\\ &amp; \\text{}2\\times 2 &amp; &amp; \\text{3}\\times 3\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nIf [latex]A[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix and [latex]B[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix such that [latex]AB=BA={I}_{n}[\/latex], then [latex]B={A}^{-1}[\/latex], the <strong>multiplicative inverse of a matrix<\/strong> [latex]A[\/latex].\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321752[\/ohm_question]<\/section><section class=\"textbox example\" aria-label=\"Example\">Given matrix [latex]A[\/latex], show that [latex]AI=IA=A[\/latex].\r\n<center>[latex]A=\\left[\\begin{array}{cc}3&amp; 4\\\\ -2&amp; 5\\end{array}\\right][\/latex]<\/center>[reveal-answer q=\"38185\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"38185\"]Use matrix multiplication to show that the product of [latex]A[\/latex] and the identity is equal to the product of the identity and [latex]A[\/latex]<em>.<\/em>\r\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given two matrices, show that one is the multiplicative inverse of the other\r\n<\/strong>\r\n<ol>\r\n \t<li>Given matrix [latex]A[\/latex] of order [latex]n\\times n[\/latex] and matrix [latex]B[\/latex] of order [latex]n\\times n[\/latex] multiply [latex]AB[\/latex].<\/li>\r\n \t<li>If [latex]AB=I[\/latex], then find the product [latex]BA[\/latex]. If [latex]BA=I[\/latex], then [latex]B={A}^{-1}[\/latex] and [latex]A={B}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Show that the given matrices are multiplicative inverses of each other.\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"812081\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812081\"]\r\n\r\nMultiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}AB &amp; =\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\\\ &amp; =\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)&amp; \\hfill &amp; \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)&amp; \\hfill &amp; \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ &amp; =\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}BA &amp; =\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\\hfill \\\\ &amp; =\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)&amp; \\hfill &amp; \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)&amp; \\hfill &amp; \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ &amp; =\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321753[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321754[\/ohm_question]<\/section><\/div>\r\n<dl id=\"fs-id1165134179626\" class=\"definition\">\r\n \t<dd id=\"fs-id1165135528440\"><\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find the inverse of a matrix.<\/li>\n<li>Solve a system of linear equations using an inverse matrix.<\/li>\n<\/ul>\n<\/section>\n<p>In this section, we will explore how to solve systems of linear equations using the inverse of a matrix. This method is particularly useful for handling larger systems and provides a systematic and efficient approach to finding solutions. We will discuss the advantages of using matrix inversion and outline the steps involved in this process. Understanding this method will enhance your ability to solve complex systems and apply these techniques in various practical scenarios.<\/p>\n<h2>Finding the Inverse of a Matrix<\/h2>\n<p>We know that the multiplicative inverse of a real number [latex]a[\/latex] is [latex]{a}^{-1}[\/latex] and [latex]a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1[\/latex]. For example, [latex]{2}^{-1}=\\frac{1}{2}[\/latex] and [latex]\\left(\\frac{1}{2}\\right)2=1[\/latex].<\/p>\n<p>The <strong>multiplicative inverse of a matrix<\/strong> is similar in concept, except that the product of matrix [latex]A[\/latex] and its inverse [latex]{A}^{-1}[\/latex] equals the <strong>identity matrix<\/strong>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by [latex]{I}_{n}[\/latex] where [latex]n[\/latex] represents the dimension of the matrix. The equations below\u00a0are the identity matrices for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and [latex]3\\text{}\\times \\text{}3[\/latex] matrix, respectively.<\/p>\n<p style=\"text-align: center;\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 0& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>The identity matrix acts as a [latex]1[\/latex] in matrix algebra. For example, [latex]AI=IA=A[\/latex].<\/p>\n<p>A matrix that has a multiplicative inverse has the properties<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/p>\n<p>A matrix that has a multiplicative inverse is called an <strong>invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>the identity matrix and multiplicative inverse<\/h3>\n<p>The <strong>identity matrix<\/strong>, [latex]{I}_{n}[\/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2} & =\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right]\\begin{array}{cccc}& & & \\end{array} & {I}_{3} & =\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right]\\hfill \\\\ & \\text{}2\\times 2 & & \\text{3}\\times 3\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>If [latex]A[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix and [latex]B[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix such that [latex]AB=BA={I}_{n}[\/latex], then [latex]B={A}^{-1}[\/latex], the <strong>multiplicative inverse of a matrix<\/strong> [latex]A[\/latex].<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321752\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321752&theme=lumen&iframe_resize_id=ohm321752&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Given matrix [latex]A[\/latex], show that [latex]AI=IA=A[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3& 4\\\\ -2& 5\\end{array}\\right][\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q38185\">Show Solution<\/button><\/p>\n<div id=\"q38185\" class=\"hidden-answer\" style=\"display: none\">Use matrix multiplication to show that the product of [latex]A[\/latex] and the identity is equal to the product of the identity and [latex]A[\/latex]<em>.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0& \\hfill & \\hfill & \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0& \\hfill & \\hfill & \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given two matrices, show that one is the multiplicative inverse of the other<br \/>\n<\/strong><\/p>\n<ol>\n<li>Given matrix [latex]A[\/latex] of order [latex]n\\times n[\/latex] and matrix [latex]B[\/latex] of order [latex]n\\times n[\/latex] multiply [latex]AB[\/latex].<\/li>\n<li>If [latex]AB=I[\/latex], then find the product [latex]BA[\/latex]. If [latex]BA=I[\/latex], then [latex]B={A}^{-1}[\/latex] and [latex]A={B}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Show that the given matrices are multiplicative inverses of each other.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q812081\">Show Solution<\/button><\/p>\n<div id=\"q812081\" class=\"hidden-answer\" style=\"display: none\">\n<p>Multiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}AB & =\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ & =\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)& \\hfill & \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)& \\hfill & \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ & =\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}BA & =\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\hfill \\\\ & =\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)& \\hfill & \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)& \\hfill & \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ & =\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p>[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321753\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321753&theme=lumen&iframe_resize_id=ohm321753&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321754\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321754&theme=lumen&iframe_resize_id=ohm321754&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<dl id=\"fs-id1165134179626\" class=\"definition\">\n<dd id=\"fs-id1165135528440\"><\/dd>\n<\/dl>\n","protected":false},"author":6,"menu_order":15,"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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":514,"module-header":"learn_it","content_attributions":[{"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":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/138"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/138\/revisions"}],"predecessor-version":[{"id":5994,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/138\/revisions\/5994"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/514"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/138\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=138"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=138"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=138"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=138"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}