{"id":218,"date":"2025-04-09T17:29:46","date_gmt":"2025-04-09T17:29:46","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/chapter\/7-7-solving-systems-with-inverses-college-algebra-2e-openstax\/"},"modified":"2025-09-15T19:23:58","modified_gmt":"2025-09-15T19:23:58","slug":"7-7-solving-systems-with-inverses","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/chapter\/7-7-solving-systems-with-inverses\/","title":{"raw":"7.7 Solving Systems with Inverses","rendered":"7.7 Solving Systems with Inverses"},"content":{"raw":"<div id=\"main-content\" class=\"MainContent__ContentStyles-sc-6yy1if-0 NnXKu\" tabindex=\"-1\" data-dynamic-style=\"true\">\r\n<div id=\"page_634d2387-7429-480f-a04e-867c2f9699fb\" class=\"chapter-content-module\" data-type=\"page\" data-book-content=\"true\">\r\n<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<p id=\"para-00001\">In this section, you will:<\/p>\r\n\r\n<ul>\r\n \t<li>Find the inverse of a matrix.<\/li>\r\n<\/ul>\r\n<ul id=\"list-00001\">\r\n \t<li>Solve a system of linear equations using an inverse matrix.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165137640111\">Soriya plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Soriya invest in each bond? What is the best method to solve this problem?<\/p>\r\n<p id=\"fs-id1165133406558\">There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.<\/p>\r\n\r\n<section id=\"fs-id1165135547295\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Finding the Inverse of a Matrix<\/h2>\r\n<p id=\"fs-id1165137422539\">We know that the multiplicative inverse of a real number <em>a<\/em> is [latex] a^{-1} [\/latex] and [latex] aa^{-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] The multiplicative inverse of a matrix is similar in concept, except that the product of matrix <em>A<\/em> and its inverse [latex] A^{-1} [\/latex] equals the identity matrix. 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 <em>n<\/em> represents the dimension of the matrix. Observe the following equations.<\/p>\r\n<p style=\"text-align: center;\">[latex] I_2=\\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] I_3=\\begin{bmatrix} 1 &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165137824188\">The identity matrix acts as a 1 in matrix algebra. For example, [latex] AI=IA=A. [\/latex]<\/p>\r\n<p id=\"fs-id1165137783971\">A matrix that has a multiplicative inverse has the properties<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rl} AA^{-1} &amp;=&amp; I \\\\ A^{-1}A &amp;=&amp; I \\end{array} [\/latex]<\/p>\r\n<p id=\"fs-id1165133354140\">A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex] AA^{-1}=A^{-1}A=I, [\/latex] is a requirement. Not all square matrices have an inverse, but if <em>A<\/em> is invertible, then [latex] A^{-1} [\/latex] is unique. We will look at two methods for finding the inverse of a [latex] 2\\times 2 [\/latex] matrix and a third method that can be used on both [latex] 2\\times 2 [\/latex] and [latex] 3\\times 3 [\/latex] matrices.<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">The Identity Matrix and Multiplicative Inverse<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\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]\u00a0\\begin{array}{rlll} I_2 &amp;=&amp; \\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix} &amp; I_3 &amp;=&amp; \\begin{bmatrix} 1 &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; 1 \\end{bmatrix} \\\\ &amp; 2\\times 2 &amp;&amp; 3\\times 3 \\end{array} [\/latex]<\/p>\r\nIf\u00a0<em>A<\/em> is an [latex] n\\times n [\/latex] matrix and <em>B<\/em> is an\u00a0[latex] n\\times m [\/latex] matrix such that [latex] AB=BA=I_n, [\/latex] then [latex] B=A^{-1}, [\/latex] the <strong><span id=\"term-00008\" data-type=\"term\">multiplicative inverse of a matrix<\/span> <em>A<\/em>.<\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 1: Showing that the Identity Matrix Acts as a 1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nGiven matrix <em data-effect=\"italics\">A<\/em>, show that [latex] AI=IA=A. [\/latex]\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 3 &amp; 4 \\\\ -2 &amp; 5 \\end{bmatrix} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>Use matrix multiplication to show that the product of\u00a0<em>A<\/em> and the identity is equal to the product of the identity and <em data-effect=\"italics\">A.<\/em>\r\n<p style=\"text-align: center;\">[latex] AI= \\begin{bmatrix} 3 &amp; 4 \\\\ -2 &amp; 5 \\end{bmatrix}\u00a0 \\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 3\\cdot 1+4\\cdot 0 &amp; 3\\cdot 0+4\\cdot 1 \\\\ -2\\cdot 1+5\\cdot 0 &amp; -2\\cdot 0+5\\cdot 1 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 3 &amp; 4 \\\\ -2 &amp; 5 \\end{bmatrix} [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] IA= \\begin{bmatrix} 1 &amp; 0\\\\ 0 &amp; 1 \\end{bmatrix}\u00a0 \\begin{bmatrix} 3 &amp; 4 \\\\ -2 &amp; 5 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 1\\cdot 3+0\\cdot (-2) &amp; 1\\cdot 4+0\\cdot 5 \\\\ 0\\cdot 3+1\\cdot (-2) &amp; 0\\cdot 4+1\\cdot 5 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 3 &amp; 4 \\\\ -2 &amp; 5 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<p id=\"fs-id1165134073896\"><strong>Given two matrices, show that one is the multiplicative inverse of the other.\r\n<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135341279\" type=\"1\">\r\n \t<li>Given matrix <em>A<\/em> of order and matrix <em>B<\/em> of order multiply <em>AB<\/em>.<\/li>\r\n \t<li>If then find the product <em>BA<\/em>. If then and<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 2: Showing That Matrix <em>A<\/em> is the Multiplicative Inverse of Matrix <em>B<\/em><\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nShow that the given matrices are multiplicative inverses of each other.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 1 &amp; 5 \\\\ -2 &amp; -9 \\end{bmatrix} , B= \\begin{bmatrix} -9 &amp; -5 \\\\ 2 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>Multiply <em>AB<\/em> and <em>BA<\/em>. 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}{rll} AB &amp;=&amp; \\begin{bmatrix} 1 &amp; 5 \\\\ -2 &amp; -9 \\end{bmatrix} \\cdot \\begin{bmatrix} -9 &amp; -5 \\\\ 2 &amp; 1 \\end{bmatrix}\u00a0 \\\\ &amp;=&amp; \\begin{bmatrix} 1(-9)+5(2) &amp; 1(-5)+5(1) \\\\ -2(-9)-9(2) &amp; -2(-5)-9(1) \\end{bmatrix}\u00a0 \\\\ &amp;=&amp; \\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} BA &amp;=&amp; \\begin{bmatrix} -9 &amp; -5 \\\\ 2 &amp; 1 \\end{bmatrix} \\cdot \\begin{bmatrix} 1 &amp; 5 \\\\ -2 &amp; -9 \\end{bmatrix}\u00a0 \\\\ &amp;=&amp; \\begin{bmatrix} -9(1)-5(02) &amp; -9(5)-5(-9) \\\\ 2(1)+1(-2) &amp; 2(5)+1(-9) \\end{bmatrix}\u00a0 \\\\ &amp;=&amp; \\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\n<em>A<\/em> and <em>B<\/em> are inverses of each other.\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nShow that the following two matrices are inverses of each other.\r\n\r\n[latex] A= \\begin{bmatrix} 1 &amp; 4 \\\\ -1 &amp; -3 \\end{bmatrix} , B= \\begin{bmatrix} -3 &amp; -4 \\\\ 1 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1165137418635\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\r\n<p id=\"fs-id1165131961693\">We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using <span id=\"term-00009\" class=\"no-emphasis\" data-type=\"term\">matrix multiplication<\/span>.<\/p>\r\n\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 3: Finding the Multiplicative Inverse Using Matrix Multiplication<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nUse matrix multiplication to find the inverse of the given matrix.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 1 &amp; -2 \\\\ 2 &amp; -3 \\end{bmatrix} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>For this method, we multiply <em>A<\/em> by a matrix containing unknown constants and set it equal to the identity.\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} 1 &amp; -2 \\\\ 2 &amp; -3 \\end{bmatrix} \\begin{bmatrix} a &amp; b \\\\ c &amp; d \\end{bmatrix} = \\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\nFind the product of the two matrices on the left side of the equal sign.\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} 1 &amp; -2 \\\\ 2 &amp; -3 \\end{bmatrix} \\begin{bmatrix} a &amp; b \\\\ c &amp; d \\end{bmatrix} = \\begin{bmatrix} 1a-2c &amp; 1b-2d \\\\ 2a-3c &amp; 2b-3d \\end{bmatrix} [\/latex]<\/p>\r\nNext, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} 1a-2c &amp;=&amp; 1 &amp; R_1 \\\\ 2a-3c &amp;=&amp; 0 &amp; R_2 \\end{array} [\/latex]<\/p>\r\nUsing row operations, multiply and add as follows: [latex] (-2)R_1+R_2\\rightarrow R_2. [\/latex] Add the equations, and solve for <em>c<\/em>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rl} 1a-2c &amp;=&amp; 1 \\\\ 0+1c &amp;=&amp; -2 \\\\ c &amp;=&amp; -2 \\end{array} [\/latex]<\/p>\r\nBack-substitute to solve for <em>a<\/em>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rl} a-2(-2) &amp;=&amp; 1 \\\\ a+4 &amp;=&amp; 1 \\\\ a &amp;=&amp; -3 \\end{array} [\/latex]<\/p>\r\nWrite another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rl} 1b-2d &amp;=&amp; 0 &amp; R_1 \\\\ 2b-3d &amp;=&amp; 1 &amp; R_2 \\end{array} [\/latex]<\/p>\r\nUsing row operations, multiply and add as follows: [latex] (-2)R_1+R_2=R_2. [\/latex] Add the two equations and solve for <em>d<\/em>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rl} 1b-2d &amp;=&amp; 0 \\\\ \\frac{0+1d=1}{d=1} \\end{array} [\/latex]<\/p>\r\nOnce more, back-substitute and solve for <em>b<\/em>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rl} b-2(1) &amp;=&amp; 0 \\\\ b-2 &amp;=&amp; 0 \\\\ b &amp;=&amp; 2 \\end{array} [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] A^{-1}=\\begin{bmatrix} -3 &amp; 2 \\\\ -2 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137737676\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Finding the Multiplicative Inverse by Augmenting with the Identity<\/h3>\r\n<p id=\"fs-id1165132328759\">Another way to find the <span id=\"term-00010\" class=\"no-emphasis\" data-type=\"term\">multiplicative inverse<\/span> is by augmenting with the identity. When matrix <em>A<\/em> is transformed into <em>I<\/em>, the augmented matrix <em>I<\/em> transforms into [latex] A^{-1}. [\/latex]<\/p>\r\n<p id=\"fs-id1165137893432\">For example, given<\/p>\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 2 &amp; 1 \\\\ 5 &amp; 3 \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165135209378\">augment <em>A<\/em> with the identity<\/p>\r\n<p style=\"text-align: center;\">[latex] \\left[\\begin{array}{rr|rr} 2 &amp; 1 &amp; 1 &amp; 0 \\\\ 5 &amp; 3 &amp; 0 &amp; 1 \\end{array}\\right] [\/latex]<\/p>\r\n<p id=\"fs-id1165137694941\">Perform row operations with the goal of turning <em>A<\/em> into the identity.<\/p>\r\n\r\n<ol id=\"fs-id1165137844094\" type=\"1\">\r\n \t<li>Switch row 1 and row 2.\r\n[latex] \\left[\\begin{array}{rr|rr} 5 &amp; 3 &amp; 0 &amp; 1 \\\\ 2 &amp; 1 &amp; 1 &amp; 0 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>Multiply row 2 by -2 and add to row 1.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 1 &amp; -2 &amp; 1 \\\\ 2 &amp; 1 &amp; 1 &amp; 0 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>Multiply row 1 by -2 and add to row 2.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 1 &amp; -2 &amp; 1 \\\\ 0 &amp; -1 &amp; 5 &amp; -2 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>Add row 2 to row 1.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 0 &amp; 3 &amp; -1 \\\\ 0 &amp; -1 &amp; 5 &amp; -2 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>Multiply row 2 by -1.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 0 &amp; 3 &amp; -1 \\\\ 0 &amp; 1 &amp; -5 &amp; 2 \\end{array}\\right] [\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"eip-702\">The matrix we have found is [latex] A^{-1}. [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] A^{-1}= \\begin{bmatrix} 3 &amp; -1 \\\\ -5 &amp; 2 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/section><section id=\"fs-id1165132300248\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Finding the Multiplicative Inverse of [latex] 2\\times 2 [\/latex] Matrices Using a Formula<\/h3>\r\n<p id=\"fs-id1165135688796\">When we need to find the multiplicative inverse of a [latex] 2\\times 2 [\/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.<\/p>\r\n<p id=\"fs-id1165134047623\">If <em>A<\/em> is a [latex] 2\\times 2 [\/latex] matrix, such as<\/p>\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} a &amp; b \\\\ c &amp; d \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165137459546\">the multiplicative inverse of <em>A<\/em> is given by the formula<\/p>\r\n<p style=\"text-align: center;\">[latex] A^{-1}=\\frac{1}{ad-bc} \\begin{bmatrix}d &amp; -b \\\\ -c &amp; a \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165137930360\">where [latex] ad-bc\\not=0. [\/latex] If [latex] ad-bc=0, [\/latex] then <em>A<\/em> has no inverse.<\/p>\r\n\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 4: Using the Formula to Find the Multiplicative Inverse of Matrix <em>A<\/em><\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nUse the formula to find the multiplicative inverse of\r\n<p style=\"text-align: center;\">[latex] A=\\begin{bmatrix} 1 &amp; -2 \\\\ 2 &amp; -3 \\end{bmatrix} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>Using the formula, we have\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} A^{-1} &amp;=&amp; \\frac{1}{(1)(-3)-(-2)(2)} \\begin{bmatrix} -3 &amp; 2 \\\\ -2 &amp; 1 \\end{bmatrix} \\\\ &amp;=&amp; \\frac{1}{-3+4} \\begin{bmatrix} -3 &amp; 2 \\\\ -2 &amp; 1 \\end{bmatrix} \\\\ &amp;=&amp; \\begin{bmatrix} -3 &amp; 2 \\\\ -2 &amp; 1 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\nWe can check that our formula works by using one of the other methods to calculate the inverse. Let\u2019s augment A with the identity.\r\n<p style=\"text-align: center;\">[latex] \\left[\\begin{array}{rr|rr} 1 &amp; -2 &amp; 1 &amp; 0\\\\ 2 &amp; 3 &amp; 0 &amp; 1 \\end{array}\\right] [\/latex]<\/p>\r\nPerform <span id=\"term-00013\" class=\"no-emphasis\" data-type=\"term\">row operations<\/span> with the goal of turning A into the identity.\r\n<ol id=\"fs-id1165134550520\" type=\"1\">\r\n \t<li>Multiple row 1 by -2 and add to row 2.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; -2 &amp; 1 &amp; 0 \\\\ 0 &amp; 1 &amp; -2 &amp; 1 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>Multiply row 2 by -2 and add to row 1.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 0 &amp; -3 &amp; 2 \\\\ 0 &amp; 1 &amp; -2 &amp; 1 \\end{array}\\right] [\/latex]<\/li>\r\n<\/ol>\r\nSo, we have verified our original solution.\r\n<p style=\"text-align: center;\">[latex] A^{-1}= \\begin{bmatrix} -3 &amp; 2 \\\\ -2 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nUse the formula to find the inverse of matrix <em>A<\/em>. Verify your answer by augmenting with the identity matrix.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 1 &amp; -1 \\\\ 2 &amp; 3 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 5: Finding the Inverse of the Matrix, if it Exists<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the inverse, if it exists, of the given matrix.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 3 &amp; 6 \\\\ 1 &amp; 2 \\end{bmatrix} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>We will use the method of augmenting with the identity.\r\n<p style=\"text-align: center;\">[latex] \\left[\\begin{array}{rr|rr} 3 &amp; 6 &amp; 1 &amp; 0\\\\ 1 &amp; 2 &amp; 0 &amp; 1\u00a0 \\end{array}\\right] [\/latex]<\/p>\r\n\r\n<ol>\r\n \t<li>Switch row 1 and row 2.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 3 &amp; 01 &amp; 1 \\\\ 3 &amp; 2 &amp; 1 &amp; 0\u00a0 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>Multiply row 1 by -3 and add to row 2.\r\n[latex] \\left[\\begin{array}{rr|rr} 1 &amp; 2 &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; -3 &amp; 1 \\end{array}\\right] [\/latex]<\/li>\r\n \t<li>There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/li>\r\n<\/ol>\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<section id=\"fs-id1165137680361\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Finding the Multiplicative Inverse of [latex] 3\\times 3 [\/latex] Matrices<\/h3>\r\n<p id=\"fs-id1165134238766\">Unfortunately, we do not have a formula similar to the one for a [latex] 2\\times 2 [\/latex] matrix to find the inverse of a [latex] 3\\times 3 [\/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.<\/p>\r\n<p id=\"fs-id1165134435528\">Given a [latex] 3\\times 3 [\/latex] matrix<\/p>\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 2 &amp; 3 &amp; 1 \\\\ 3 &amp; 3 &amp; 1 \\\\ 2 &amp; 4 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165133394718\">augment <em>A<\/em> with the identity matrix<\/p>\r\n<p style=\"text-align: center;\">[latex] A|I = \\left[\\begin{array}{ccc|ccc} 2 &amp; 3 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\ 3 &amp; 3 &amp; 1 &amp; 0 &amp; 1 &amp; 0 \\\\ 2 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right] [\/latex]<\/p>\r\n<p id=\"fs-id1165134357582\">To begin, we write the augmented matrix with the identity on the right and <em>A<\/em> on the left. Performing elementary <span id=\"term-00016\" class=\"no-emphasis\" data-type=\"term\">row operations<\/span> so that the <span id=\"term-00017\" class=\"no-emphasis\" data-type=\"term\">identity matrix<\/span> appears on the left, we will obtain the <span id=\"term-00018\" class=\"no-emphasis\" data-type=\"term\">inverse matrix<\/span> on the right. We will find the inverse of this matrix in the next example.<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Given a [latex] 3\\times 3 [\/latex] matrix, find the inverse<\/strong>\r\n<ol>\r\n \t<li>Write the original matrix augmented with the identity matrix on the right.<\/li>\r\n \t<li>Use elementary row operations so that the identity appears on the left.<\/li>\r\n \t<li>What is obtained on the right is the inverse of the original matrix.<\/li>\r\n \t<li>Use matrix multiplication to show that [latex] AA^{-1}=I [\/latex] and [latex] A^{-1}A=I. [\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 6: Finding the Inverse of a [latex] 3\\times 3 [\/latex] Matrix<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nGiven the [latex] 3\\times 3 [\/latex] matrix <em>A<\/em>, find the inverse.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 2 &amp; 3 &amp; 1 \\\\ 3 &amp; 3 &amp; 1 \\\\ 2 &amp; 4 &amp; 1 \\end{bmatrix} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>Augment <em>A<\/em> with the identity matrix, and then begin row operations until the identity matrix replaces <em>A.<\/em> The matrix on the right will be the inverse of <em>A<\/em>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} \\left[\\begin{array}{rrr|rrr} 2 &amp; 3 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\ 3 &amp; 3 &amp; 1 &amp; 0 &amp; 1 &amp; 0 \\\\ 2 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 1\u00a0 \\end{array}\\right] &amp;\\overset{\\text{Interchange} \\quad R_2 \\quad {and} \\quad R_1}{\\rightarrow}&amp; \\left[\\begin{array}{rrr|rrr} 3 &amp; 3 &amp; 1 &amp; 0 &amp; 1 &amp; 0 \\\\ 2 &amp; 3 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\ 2 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array}\\right]\u00a0 \\\\ -R_2+R_1=R_1 &amp;\\rightarrow &amp; \\left[\\begin{array}{rrr|rrr} 1 &amp; 0 &amp; 0 &amp; -1 &amp; 1 &amp; 0 \\\\ 2 &amp; 3 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\ 2 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array}\\right]\u00a0 \\\\ -R_2+R_3=R_3 &amp;\\rightarrow &amp; \\left[\\begin{array}{rrr|rrr} 1 &amp; 0 &amp; 0 &amp; -1 &amp; 1 &amp; 0 \\\\ 2 &amp; 3 &amp; 1 &amp; 1 &amp; 0 &amp; 00 \\\\ 0 &amp; 1 &amp;\u00a0 0 &amp; -1 &amp; 0 &amp; 1 \\end{array}\\right]\u00a0 \\\\ R_3 \\leftrightarrow R_2 &amp;\\rightarrow &amp; \\left[\\begin{array}{rrr|rrr} 1 &amp; 0 &amp; 0 &amp; -1 &amp; 1 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 &amp; -1 &amp; 0 &amp; 1 \\\\ 2 &amp; 3 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\end{array}\\right]\u00a0 \\\\ -2R_1+R_3=R_3 &amp;\\rightarrow &amp; \\left[\\begin{array}{rrr|rrr} 1 &amp; 0 &amp; 0 &amp; -1 &amp; 1 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 &amp; -1 &amp; 0 &amp; 1 \\\\ 0 &amp; 3 &amp; 1 &amp; 3 &amp; -2 &amp; 0 \\end{array}\\right]\u00a0 \\\\ -3R_2+R_3=R_3&amp;\\rightarrow &amp; \\left[\\begin{array}{rrr|rrr} 1 &amp; 0 &amp; 0 &amp; -1 &amp; 1 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 &amp; -1 &amp; 0 &amp; 1 \\\\ 0 &amp; 0 &amp; 1 &amp; 6 &amp; -2 &amp; -3 \\end{array}\\right] \\end{array} [\/latex]<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[latex] A^{-1}=B= \\begin{bmatrix} -1 &amp; 1 &amp; 0 \\\\ -1 &amp; 0 &amp; 1 &amp; 6 &amp; -2 &amp; -3 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\nTo prove that [latex] B=A^{-1}, [\/latex] let\u2019s multiply the two matrices together to see if the product equals the identity, if [latex] AA^{-1}=I [\/latex] and [latex] A^{-1}A=I. [\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} AA^{-1} &amp;=&amp; \\begin{bmatrix} 2 &amp; 3 &amp; 1 \\\\ 3 &amp; 3 &amp; 1 \\\\ 2 &amp; 4 &amp; 1 \\end{bmatrix} \\begin{bmatrix} -1 &amp; 1 &amp; 0\\\\ -1 &amp; 0 &amp; 1 \\\\ 6 &amp; -2 &amp; -3 \\end{bmatrix} \\\\ &amp;=&amp; \\begin{bmatrix} 2(-1)+3(-1)+1(6) &amp; 2(1)+3(0)+1(-2) &amp; 2(0)+3(1)+1(-3) \\\\ 3(-1)+3(-1)+1(6) &amp; 3(1)+3(0)+1(-2) &amp; 3(0)+3(1)+1(-3) \\\\ 2(-1)+4(-1)+1(6) &amp; 2(1)+4(0)+1(-2) &amp; 2(0)+4(1)+1(-3) \\end{bmatrix} \\\\ &amp;=&amp; \\begin{bmatrix} 1 &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; 1 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} A^{-1}A &amp;=&amp; \\begin{bmatrix} -1 &amp; 1 &amp; 0 \\\\ -1 &amp; 0 &amp; 1 \\\\ 6 &amp; -2 &amp; -3 \\end{bmatrix} \\begin{bmatrix} 2 &amp; 3 &amp; 1 \\\\ 3 &amp; 3 &amp; 1 \\\\ 2 4 &amp; 1 \\end{bmatrix} \\\\ &amp;=&amp; \\begin{bmatrix} -1(2)+1(3)+0(2) &amp; -1(3)+1(3)+0(4) &amp; -1(1)+1(1)+0(1) \\\\ -1(2)+0(3)+1(2) &amp; -1(3)+0(3)+1(4) &amp; -1(1)+0(1)+1(01 \\\\ 6(2)+-2(3)+-3(2) &amp; 6(3)+-2(3)+-3(4) &amp; 6(1)+-2(1)+-3(1) \\end{bmatrix} \\\\ &amp;=&amp; \\begin{bmatrix} 1 &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 \\\\ 0 &amp; ) &amp; 1 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the inverse of the [latex] 3\\times 3 [\/latex] matrix.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 2 &amp; -17 &amp; 11 \\\\ -1 &amp; 11 &amp; -7 \\\\ 0 &amp; 3 &amp; -2 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137694180\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Solving a System of Linear Equations Using the Inverse of a Matrix<\/h2>\r\n<p id=\"fs-id1165135394318\">Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: <em>X<\/em> is the matrix representing the variables of the system, and <em>B<\/em> is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as<\/p>\r\n<p style=\"text-align: center;\">[latex] AX=B [\/latex]<\/p>\r\n<p id=\"fs-id1165137466371\">To solve a system of linear equations using an inverse matrix, let <em>A<\/em> be the coefficient matrix, let <em>X<\/em> be the variable matrix, and let <em>B<\/em> be the constant matrix. Thus, we want to solve a system [latex] AX=B. [\/latex] For example, look at the following system of equations.<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} a_1x+b_1y &amp;=&amp; c_1 \\\\ a_2x+b_2y &amp;=&amp; c_2 \\end{array} [\/latex]<\/p>\r\n<p id=\"fs-id1165137626840\">From this system, the coefficient matrix is<\/p>\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} a_1 &amp; b_1 \\\\ a_2 &amp; b_2 \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165137628730\">The variable matrix is<\/p>\r\n<p style=\"text-align: center;\">[latex] X= \\begin{bmatrix} x \\\\ y \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165135670279\">And the constant matrix is<\/p>\r\n<p style=\"text-align: center;\">[latex] B= \\begin{bmatrix} c_1 \\\\ c_2 \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165134043634\">Then [latex] AX=B [\/latex] looks like<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} a_1 &amp; b_1 \\\\ a_2 &amp; b_2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} c_1 \\\\ c_2 \\end{bmatrix} [\/latex]<\/p>\r\n<p id=\"fs-id1165137418368\">Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex] \\left(2^{-1}\\right) 2 =\\left(\\frac{1}{2}\\right)2=1\/ [\/latex] To solve a single linear equation [latex] ax=b [\/latex] for <em>x<\/em>, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of <em>a<\/em>. Thus,<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} ax &amp;=&amp; b \\\\ \\left(\\frac{1}{a}\\right)ax &amp;=&amp; \\left(\\frac{1}{a}\\right)b \\\\ \\left(a^{-1}\\right)ax- &amp;=&amp; \\left(a^{-1}\\right)b \\\\ [\\left(a^{-1}\\right)a]x &amp;=&amp; \\left(a^{-1}\\right)b \\\\ 1x &amp;=&amp; \\left(a^{-1}\\right)b \\\\ x &amp;=&amp; \\left(a^{-1}\\right)b \\end{array} [\/latex]<\/p>\r\n<p id=\"fs-id1165137843896\">The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same\u2014to isolate the variable.<\/p>\r\n<p id=\"fs-id1165135487326\">We will investigate this idea in detail, but it is helpful to begin with a [latex] 2\\times 2 [\/latex] system and then move on to a [latex] 3\\times 3 [\/latex] system.<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Solving a System of Equations Using the Inverse of a Matrix<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nGiven a system of equations, write the coefficient matrix <em>A<\/em>, the variable matrix <em>X<\/em>, and the constant matrix B. Then\r\n<p style=\"text-align: center;\">[latex] AX=B [\/latex]<\/p>\r\nMultiply both sides by the inverse of <em>A<\/em> to obtain the solution.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} \\left(A^{-1}\\right)AX &amp;=&amp; \\left(A^{-1}\\right)B \\\\ [\\left(A^{-1}\\right)A]X &amp;=&amp; \\left(A^{-1}\\right)B \\\\ IX &amp;=&amp; \\left(A^{-1}\\right)B \\\\ x &amp;=&amp; \\left(A^{-1}\\right)B \\end{array} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Q&amp;A<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<p id=\"fs-id1165135149799\"><strong>Q: If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/strong><\/p>\r\n<p id=\"fs-id1165135344922\"><em data-effect=\"italics\">A: No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7: Solving a [latex] 2\\times 2 [\/latex] System Using the Inverse of a Matrix<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve the given system of equations using the inverse of a matrix.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} 3x+8y &amp;=&amp; 5 \\\\ 4x+11y &amp;=&amp; 7 \\end{array} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.\r\n<p style=\"text-align: center;\">[latex] A= \\begin{bmatrix} 3 &amp; 8 \\\\ 4 &amp; 11 \\end{bmatrix}, X= \\begin{bmatrix} x \\\\ y \\end{bmatrix}, B= \\begin{bmatrix} 5 \\\\ 7 \\end{bmatrix}\u00a0[\/latex]<\/p>\r\nThen\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} 3 &amp; 8 \\\\ 4 &amp; 11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} 5 \\\\ 7 \\end{bmatrix} [\/latex]<\/p>\r\nFirst, we need to calculate [latex] A^{-1}\/ [\/latex] Using the formula to calculate the inverse of a 2 by 2 matrix, we have:\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} A^{-1} &amp;=&amp; \\frac{1}{ad-bc} \\begin{bmatrix} d &amp; -b \\\\ -c &amp; a \\end{bmatrix} \\\\ &amp;=&amp; \\frac{1}{3(11)-8(4)} \\begin{bmatrix} 11 &amp; -8 \\\\ -4 &amp; 3 \\end{bmatrix} \\\\ &amp;=&amp; \\frac{1}{1} \\begin{bmatrix} 11 &amp; -8 \\\\ -4 &amp; 3 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\nSo,\r\n<p style=\"text-align: center;\">[latex] A^{-1} = \\begin{bmatrix} 11 &amp; -8 \\\\ -4 &amp; 3 \\end{bmatrix} [\/latex]<\/p>\r\nNow we are ready to solve. Multiply both sides of the equation by [latex] A^{-1}\/ [\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} \\left(A^{-1}\\right)AX &amp;=&amp; \\left(A^{-1}\\right)B \\\\ \\begin{bmatrix} 11 &amp; -8 \\\\ -4 &amp; 3 \\end{bmatrix} \\begin{bmatrix} 3 &amp; 8 \\\\ 4 &amp; 11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} &amp;=&amp; \\begin{bmatrix} 11 &amp; -8 \\\\ -4 &amp; 3 \\end{bmatrix} \\begin{bmatrix} 5 \\\\ 7 \\end{bmatrix} \\\\ \\begin{bmatrix} 1 &amp; 0 \\\\ 0 &amp; 1 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} &amp;=&amp; \\begin{bmatrix} 11(5)+(-8)7 \\\\ -4(5)+3(7) \\end{bmatrix} \\\\ \\begin{bmatrix} x \\\\ y \\end{bmatrix} &amp;=&amp; \\begin{bmatrix} -1 \\\\ 1 \\end{bmatrix} \\end{array} [\/latex]<\/p>\r\nThe solution is [latex] (-1, 1). [\/latex]\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Q&amp;A<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Q: Can we solve for <em>X<\/em> by finding the product [latex] BA^{-1} [\/latex]?<\/strong>\r\n\r\n<em data-effect=\"italics\">A: No, recall that matrix multiplication is not commutative, so [latex] A^{-1}B\\not=BA^{-1}. [\/latex] Consider our steps for solving the matrix equation.<\/em>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} \\left(A^{-1}\\right)AX &amp;=&amp; \\left(A^{-1}\\right)B \\\\ [\\left(A^{-1}\\right)]X &amp;=&amp; \\left(A^{-1}\\right)B \\\\ IX &amp;=&amp; \\left(A^{-1}\\right)B \\\\ X &amp;=&amp; \\left(A^{-1}\\right)B \\end{array} [\/latex]<\/p>\r\n<em data-effect=\"italics\">Notice in the first step we multiplied both sides of the equation by [latex] A^{-1}, [\/latex] but the [latex] A^{-1} [\/latex] was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters.<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 8: Solving a [latex] 3\\times 3 [\/latex] System Using the Inverse of a Matrix<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve the following system using the inverse of a matrix.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} 5x+15y+56z &amp;=&amp; 35 \\\\ -4x-11y-41z &amp;=&amp; -26 \\\\ -x-3y-11z &amp;=&amp; -7 \\end{array} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>Write the equation [latex] AX=B. [\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} 5 &amp; 15 &amp; 56 \\\\ -4 &amp; -11 &amp; -41 \\\\ -1 &amp; -3 &amp; -11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{bmatrix} 35 \\\\ -26 \\\\ -7 \\end{bmatrix} [\/latex]<\/p>\r\nFirst, we will find the inverse of <em>A<\/em> by augmenting with the identity.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 5 &amp; 15 &amp; 56 &amp; 1 &amp; 0 &amp; 0 \\\\ -4 &amp; -11 &amp; -41 &amp; 0 &amp; 1 &amp; 0 \\\\ -1 &amp; -3 &amp; -11 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right] [\/latex]<\/p>\r\nMultiply row 1 by [latex] \\frac{1}{5}. [\/latex]\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 3 &amp; \\frac{56}{5} &amp; \\frac{1}{5} &amp; 0 &amp; 0 \\\\ -4 &amp; -11 &amp; -41 &amp; 0 &amp; 1 &amp; 0 \\\\ -1 &amp; -3 &amp; -11 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right] [\/latex]<\/p>\r\nMultiply row 1 by 4 and add to row 2.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 3 &amp; \\frac{56}{5} &amp; \\frac{1}{5} &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; \\frac{19}{5} &amp; \\frac{4}{5} &amp; 1 &amp; 0 \\\\ -1 &amp; -3 &amp; -11 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right] [\/latex]<\/p>\r\nAdd row 1 to row 3.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 3 &amp; \\frac{56}{5} &amp; \\frac{1}{5} &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; \\frac{19}{5} &amp; \\frac{4}{5} &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; \\frac{1}{5} &amp; \\frac{1}{5} &amp; 0 &amp; 1 \\end{array} \\right] [\/latex]<\/p>\r\nMultiply row 2 by \u22123 and add to row 1.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; -\\frac{1}{5} &amp; -\\frac{11}{5} &amp; -3 &amp; 0 \\\\ 0 &amp; 1 &amp; \\frac{19}{5} &amp; \\frac{4}{5} &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; \\frac{1}{5} &amp; \\frac{1}{5} &amp; 0 &amp; 1 \\end{array} \\right] [\/latex]<\/p>\r\nMultiply row 3 by 5.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; -\\frac{1}{5} &amp; -\\frac{11}{5} &amp; -3 &amp; 0 \\\\ 0 &amp; 1 &amp; \\frac{19}{5} &amp; \\frac{4}{5} &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 5 \\end{array} \\right] [\/latex]<\/p>\r\nMultiply row 3 by [latex] \\frac{1}{5} [\/latex] and add to row 1.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 0 &amp; -2 &amp; -3 &amp; 1 \\\\ 0 &amp; 1 &amp; \\frac{19}{5} &amp; \\frac{4}{5} &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 5 \\end{array} \\right] [\/latex]<\/p>\r\nMultiply row 3 by [latex] -\\frac{19}{5} [\/latex] and add to row 2.\r\n<p style=\"text-align: center;\">[latex] \\left[ \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 0 &amp; -2 &amp; -3 &amp; 1 \\\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 1 &amp; -19 \\\\ 0 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 5 \\end{array} \\right] [\/latex]<\/p>\r\nSo,\r\n<p style=\"text-align: center;\">[latex] A^{-1}= \\begin{bmatrix} -2 &amp; 03 &amp; 1 \\\\ -3 &amp; 1 &amp; -19 \\\\ 1 &amp; 0 &amp; 5 \\end{bmatrix} [\/latex]<\/p>\r\nMultiply both sides of the equation by [latex] A^{-1}. [\/latex] We want [latex] A^{-1}AX=A^{-1}B : [\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} -2 &amp; -3 &amp; 1 \\\\ -3 &amp; 1 &amp; -19 \\\\ 1 &amp; 0 &amp; 5 \\end{bmatrix} \\begin{bmatrix} 5 &amp; 15 &amp; 56 \\\\ -4 &amp; -11 &amp; -41 \\\\ -1 &amp; -3 &amp; -11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{bmatrix} -2 &amp; -3 &amp; 1 \\\\ -3 &amp; 1 &amp; -19 \\\\ 1 &amp; 0 &amp; 5 \\end{bmatrix} \\begin{bmatrix} 35 \\\\ -26 \\\\ -7 \\end{bmatrix} [\/latex]<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[latex] A^{-1}B= \\begin{bmatrix} -70+78-7 \\\\ -105-26+133 \\\\ 35+0-35 \\end{bmatrix} = \\begin{bmatrix} 1 \\\\ 2 \\\\0 \\end{bmatrix} [\/latex]<\/p>\r\nThe solution is [latex] (1, 2, 0). [\/latex]\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve the system using the inverse of the coefficient matrix.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} 2x-17y+11z &amp;=&amp; 0 \\\\ -x+11y-7z &amp;=&amp; 8 \\\\ 3y-2z &amp;=&amp; -2 \\end{array} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<p id=\"fs-id1165135503565\"><strong>Given a system of equations, solve with matrix inverses using a calculator.\r\n<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135503570\" type=\"1\">\r\n \t<li>Save the coefficient matrix and the constant matrix as matrix variables [latex] [A] [\/latex] and [latex] [B]. [\/latex]<\/li>\r\n \t<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\r\n \t<li>If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve the system of equations with matrix inverses using a calculator\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rll} 2x+3y+z &amp;=&amp; 32 \\\\ 3x+3y+z &amp;=&amp; -27 \\\\ 2x+4y+z &amp;=&amp; -2 \\end{array} [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>On the matrix page of the calculator, enter the <span id=\"term-00023\" class=\"no-emphasis\" data-type=\"term\">coefficient matrix<\/span> as the matrix variable [latex] [A], [\/latex] and enter the constant matrix as the matrix variable [latex] [B]. [\/latex]\r\n<p style=\"text-align: center;\">[latex] [A]= \\begin{bmatrix} 2 &amp; 3 &amp; 1 \\\\ 3 &amp; 3 &amp; 1 \\\\ 2 &amp; 4 &amp; 1 \\end{bmatrix}, [B]= \\begin{bmatrix} 32 \\\\ -27 \\\\ -2 \\end{bmatrix} [\/latex]<\/p>\r\nOn the home screen of the calculator, type in the multiplication to solve for <em>X<\/em>, calling up each matrix variable as needed.\r\n<p style=\"text-align: center;\">[latex] [A]^{-1}\\times [B] [\/latex]<\/p>\r\nEvaluate the expression.\r\n<p style=\"text-align: center;\">[latex] \\begin{bmatrix} =59 \\\\ -34 \\\\ 252 \\end{bmatrix} [\/latex]<\/p>\r\n\r\n<\/details><section>\r\n<div><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Media<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<p id=\"fs-id1165137836914\">Access these online resources for additional instruction and practice with solving systems with inverses.<\/p>\r\n\r\n<ul id=\"fs-id1165137836918\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/identmatrix\" target=\"_blank\" rel=\"noopener nofollow\">The Identity Matrix<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/inversematrix\" target=\"_blank\" rel=\"noopener nofollow\">Determining Inverse Matrices<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/matrixsystem\" target=\"_blank\" rel=\"noopener nofollow\">Using a Matrix Equation to Solve a System of Equations<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"os-eos os-section-exercises-container\" data-uuid-key=\".section-exercises\">\r\n<h2 data-type=\"document-title\" data-rex-keep=\"true\"><span class=\"os-text\">7.7 Section Exercises<\/span><\/h2>\r\n<section id=\"fs-id1165135446624\" class=\"section-exercises\" data-depth=\"1\"><section id=\"fs-id1165137548736\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Verbal<\/h3>\r\n<div id=\"fs-id1165137548741\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137548742\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137548741-solution\">1<\/a><span class=\"os-divider\">. <\/span>In a previous section, we showed that matrix multiplication is not commutative, that is, [latex] AB\\not=BA [\/latex] in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is, [latex] A^{-1}A=AA^{-1}? [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/section><\/div>\r\n<div id=\"fs-id1165134493452\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134493453\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">2<\/span><span class=\"os-divider\">. <\/span>Does every [latex] 2\\times 2 [\/latex] matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135547596\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135547597\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135547596-solution\">3<\/a><span class=\"os-divider\">. <\/span>Can you explain whether a [latex] 2\\times 2 [\/latex] matrix with an entire row of zeros can have an inverse?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133401600\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135702541\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">4<\/span><span class=\"os-divider\">. <\/span>Can a matrix with an entire column of zeros have an inverse? Explain why or why not.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135702546\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135702547\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135702546-solution\">5<\/a><span class=\"os-divider\">. <\/span>Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a [latex] 2\\times 2 [\/latex] matrix.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<section id=\"fs-id1165133075017\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Algebraic<\/h3>\r\n<p id=\"fs-id1165132059611\">In the following exercises, show that matrix <em>A<\/em> is the inverse of matrix <em>B<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1165137804955\" class=\"material-set-2\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137804956\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">6<\/span><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix} 1 &amp; 0 \\\\ -1 &amp; 1 \\end{bmatrix}, B= \\begin{bmatrix} 1 &amp; 0 \\\\ 1 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n<div id=\"fs-id1165137629353\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137629354\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137629353-solution\">7<\/a><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix} 1 &amp; 2 \\\\ 3 &amp; 4 \\end{bmatrix}, B= \\begin{bmatrix} -2 &amp; 1 \\\\ \\frac{3}{2} &amp; -\\frac{1}{2} \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137832785\" class=\"material-set-2\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137832786\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">8<\/span><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix} 4 &amp; 5 \\\\ 7 &amp; 0 \\end{bmatrix}, B= \\begin{bmatrix} 0 &amp; \\frac{1}{7} \\\\ \\frac{1}{5} &amp; -\\frac{4}{35} \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137417445\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137417446\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137417445-solution\">9<\/a><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix} -2 &amp; \\frac{1}{2} \\\\ 3 &amp; -1 \\end{bmatrix}, B= \\begin{bmatrix} -2 &amp; 01 \\\\ -6 &amp; -4 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137810179\" class=\"material-set-2\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137810180\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">10<\/span><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix} 1 &amp; 0 &amp; 1 \\\\ 0 &amp; 1 &amp; -1 \\\\ 0 &amp; 1 &amp; 1 \\end{bmatrix}, B= \\frac{1}{2}\\begin{bmatrix} 2 &amp; 1 &amp; -1 \\\\ 0 &amp; 1 &amp; 1 \\\\ 0 &amp; -1 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135481914\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135481915\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135481914-solution\">11<\/a><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix}\u00a0 1 &amp; 2 &amp; 3 \\\\ 4 &amp; 0 &amp; 2 \\\\ 1 &amp; 6 &amp; 9\\end{bmatrix}, B= \\frac{1}{4}\\begin{bmatrix} 6 &amp; 0 &amp; -2 \\\\ 17 &amp; -3 &amp; -5 \\\\ -12 &amp; 2 &amp; 4 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134371186\" class=\"material-set-2\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134371187\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">12<\/span><span class=\"os-divider\">. <\/span> [latex] A= \\begin{bmatrix} 3 &amp; 8 &amp; 2 \\\\ 1 &amp; 1 &amp; 1 \\\\ 5 &amp; 6 &amp; 12 \\end{bmatrix}, B= \\frac{1}{36}\\begin{bmatrix} -6 &amp; 84 &amp; -6 \\\\ 7 &amp; -26 &amp; 1 \\\\ -1 &amp; -22 &amp; 5 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165134282177\">For the following exercises, find the multiplicative inverse of each matrix, if it exists.<\/p>\r\n\r\n<div id=\"fs-id1165134282180\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134282181\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134282180-solution\">13<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 3 &amp; -2 \\\\ 1 &amp; 9 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135445702\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135445703\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">14<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} -2 &amp; 2 \\\\ 3 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137387518\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137387519\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137387518-solution\">15<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} -3 &amp; 7 \\\\ 9 &amp; 2 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134339996\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134339997\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">16<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} -4 &amp; -3 \\\\ -5 &amp; 8 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133103275\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133103276\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165133103275-solution\">17<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 1 \\\\ 2 &amp; 2 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165132957188\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165132957189\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">18<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 0 &amp; 1 \\\\ 1 &amp; 0 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134185464\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134185465\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134185464-solution\">19<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 0.5 &amp; 1.5 \\\\ 1 &amp; -0.5 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134152537\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134152538\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">20<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 0 &amp; 6 \\\\ 02 &amp; 1 &amp; 7 \\\\ 3 &amp; 0 &amp; 2 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133141439\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135679418\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165133141439-solution\">21<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 0 &amp; 1 &amp; -3 \\\\ 4 &amp; 1 &amp; 0 \\\\ 1 &amp; 0 &amp; 5 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134040470\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134040471\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">22<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 2 &amp; -1 \\\\ -3 &amp; 4 &amp; 1 \\\\ -2 &amp; -4 &amp; -5 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134159676\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134159677\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134159676-solution\">23<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 9 &amp; -3 \\\\ 2 &amp; 5 &amp; 6 \\\\ 4 &amp; -2 &amp; 7 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133210779\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133210780\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">24<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; -2 &amp; 3 \\\\ -4 &amp; 8 &amp; -12 \\\\ 1 &amp; 4 &amp; 2 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135397109\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135397110\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135397109-solution\">25<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} \\frac{1}{2} &amp; \\frac{1}{2} &amp; \\frac{1}{2} \\\\ \\frac{1}{3} &amp; \\frac{1}{4} &amp; \\frac{1}{5} \\\\ \\frac{1}{6} &amp; \\frac{1}{7} &amp; \\frac{1}{8} \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165131818536\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165131818537\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">26<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 2 &amp; 3 \\\\ 4 &amp; 5 &amp; 6 \\\\ 7 &amp; 8 &amp; 9 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165137675157\">For the following exercises, solve the system using the inverse of a [latex] 2\\times 2 [\/latex] matrix.<\/p>\r\n\r\n<div id=\"fs-id1165134319666\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134070732\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134319666-solution\">27<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 5x-6y &amp;=&amp; -61 \\\\ 4x+3y &amp;=&amp; -2 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133012293\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133012294\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">28<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 8x+4y &amp;=&amp; -100 \\\\ 3x-4y &amp;=&amp; 1 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165131884588\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165131884589\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165131884588-solution\">29<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 3x-2y &amp;=&amp; 6 \\\\ -x+5y &amp;=&amp; -2 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133045242\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133045243\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">30<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 5x-4y &amp;=&amp; -5 \\\\ 4x+y &amp;=&amp; 2.3 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135205038\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135205039\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135205038-solution\">31<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll}-3x-4y &amp;=&amp; 9 \\\\ 12x+4y &amp;=&amp; -6 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135518212\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135518213\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">32<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} -2x+3y &amp;=&amp; \\frac{3}{10} \\\\ -x+5y &amp;=&amp; \\frac{1}{2} \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165131958337\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165131958338\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165131958337-solution\">33<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} \\frac{8}{5}x-\\frac{4}{5}y &amp;=&amp; \\frac{2}{5} \\\\ -\\frac{8}{5}x+\\frac{1}{5}y &amp;=&amp; \\frac{7}{10} \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133157406\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135523285\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">34<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} \\frac{1}{2}x+\\frac{1}{5}y &amp;=&amp; -\\frac{1}{4} \\\\ \\frac{1}{2}x-\\frac{3}{5}y &amp;=&amp; -\\frac{9}{4} \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165134386500\">For the following exercises, solve a system using the inverse of a [latex] 3\\times 3 [\/latex] matrix.<\/p>\r\n\r\n<div id=\"fs-id1165132079352\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165132079353\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165132079352-solution\">35<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 3x-2y+5z &amp;=&amp; 21 \\\\ 5x+4y &amp;=&amp; 37 \\\\ x-2y-5z &amp;=&amp; 5 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134177544\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134177545\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">36<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 4x+4y+4z &amp;=&amp; 40 \\\\ 2x-3y+4z &amp;=&amp; -12 \\\\ -x+3y+4z &amp;=&amp; 9 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134226706\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134226707\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134226706-solution\">37<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 6x-5y-z &amp;=&amp; 31 \\\\ -x+2y+z &amp;=&amp; -6 \\\\ 3x+3y+2z &amp;=&amp; 13 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134224075\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134224076\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">38<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 6x-5y+2z &amp;=&amp; -4 \\\\ 2x+5y-z &amp;=&amp; 12 \\\\ 2x+5y+z &amp;=&amp; 12 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134185366\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134185367\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134185366-solution\">39<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 4x-2y+3z &amp;=&amp; -12 \\\\ 2x+2y-9z &amp;=&amp; 33 \\\\ 6y-4z &amp;=&amp; 1 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135207327\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135207328\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">40<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} \\frac{1}{10}x-\\frac{1}{5}y+4z &amp;=&amp; \\frac{-41}{2} \\\\ \\frac{1}{5}x-20y+\\frac{2}{5}z &amp;=&amp; -101 \\\\ \\frac{3}{10}x+4y-\\frac{3}{10}z &amp;=&amp; 23 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135421457\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135421458\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135421457-solution\">41<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} \\frac{1}{2}x-\\frac{1}{5}y+\\frac{1}{5}z &amp;=&amp; \\frac{31}{100} \\\\ -\\frac{3}{4}x-\\frac{1}{4}y+\\frac{1}{2}z &amp;=&amp; \\frac{7}{40} \\\\ -\\frac{4}{5}x-\\frac{1}{2}y+\\frac{3}{2}z &amp;=&amp; \\frac{1}{4} \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165132944845\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165132944846\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">42<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} -.1x+0.2y+0.3z &amp;=&amp; -1.4 \\\\ 0.1x-0.2y+0.3z &amp;=&amp; -.6 \\\\ 0.4y+0.9z &amp;=&amp; -2 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<section id=\"fs-id1165137900020\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Technology<\/h3>\r\n<p id=\"fs-id1165134356163\">For the following exercises, use a calculator to solve the system of equations with matrix inverses.<\/p>\r\n\r\n<div id=\"fs-id1165134356166\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134356167\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134356166-solution\">43<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 2x-y &amp;=&amp; -3 \\\\ -x+2y &amp;=&amp; 2.3 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n<div id=\"fs-id1165135186095\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135186096\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">44<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} -\\frac{1}{2}x-\\frac{3}{2}y &amp;=&amp; -\\frac{43}{20} \\\\ \\frac{5}{2}x+\\frac{11}{5}y &amp;=&amp; \\frac{31}{4} \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135652325\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135652326\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135652325-solution\">45<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 12.3x-2y-2.5z &amp;=&amp; 2 \\\\ 36.9x+7y-7.5z &amp;=&amp; -7 \\\\ 8y-5z &amp;=&amp; -10 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134467729\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134467730\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">46<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{array}{rll} 0.5x-3y+6z &amp;=&amp; -0.8 \\\\ 0.7x-2y &amp;=&amp; -0.06 \\\\ 0.5x+4y+5z &amp;=&amp; 0 \\end{array} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<section id=\"fs-id1165134537745\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Extensions<\/h3>\r\n<p id=\"fs-id1165135339476\">For the following exercises, find the inverse of the given matrix.<\/p>\r\n\r\n<div id=\"fs-id1165135339479\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135339480\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135339479-solution\">47<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 0 &amp; 1 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 &amp; 1 \\\\ 0 &amp; 1 &amp; 1 &amp; 0 \\\\ 0 &amp; 0 &amp; 1 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>\r\n<div id=\"fs-id1165135258897\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135258898\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">48<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} -1 &amp; 0 &amp; 2 &amp; 5 \\\\ 0 &amp; 0 &amp; 0 &amp; 2 \\\\ 0 &amp; 2 &amp; -1 &amp; 0 \\\\ 1 &amp; -3 &amp; 0 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135191959\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135191960\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135191959-solution\">49<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; -2 &amp; 3 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 &amp; 2 \\\\ 1 &amp; 4 &amp; -2 &amp; 3 \\\\ -5 &amp; 0 &amp; 1 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134234210\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134234211\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">50<\/span><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 2 &amp; 0 &amp; 2 &amp; 3 \\\\ 0 &amp; 2 &amp; 1 &amp; 0 &amp; 0 \\\\ 0 &amp; 0 &amp; 3 &amp; 0 &amp; 1 \\\\ 0 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\\\ 0 &amp; 0 &amp; 1 &amp; 2 &amp; 0 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134385542\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134385543\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134385542-solution\">51<\/a><span class=\"os-divider\">. <\/span> [latex] \\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\\\ 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; ) \\\\ 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\\\ 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\ 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\end{bmatrix} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<section id=\"fs-id1165133094400\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Real-World Applications<\/h3>\r\n<p id=\"fs-id1165133094406\">For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix.<\/p>\r\n\r\n<div id=\"fs-id1165133260340\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133260342\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">52<\/span><span class=\"os-divider\">. <\/span>2,400 tickets were sold for a basketball game. If the prices for floor 1 and floor 2 were different, and the total amount of money brought in is $64,000, how much was the price of each ticket?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133260348\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133260349\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165133260348-solution\">53<\/a><span class=\"os-divider\">. <\/span>In the previous exercise, if you were told there were 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133111147\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133111148\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">54<\/span><span class=\"os-divider\">. <\/span>A food drive collected two different types of canned goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134081432\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134081433\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134081432-solution\">55<\/a><span class=\"os-divider\">. <\/span>Students were asked to bring their favorite fruit to class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135705936\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135705937\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">56<\/span><span class=\"os-divider\">. <\/span>The nursing club held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at $0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135705944\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135705945\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135705944-solution\">57<\/a><span class=\"os-divider\">. <\/span>A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at $13.99, the beanie at $7.99, and the cowboy hat at $14.49. If 100 hats were sold this past quarter, $1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133354003\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133354004\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">58<\/span><span class=\"os-divider\">. <\/span>Anna, Percy, and Morgan weigh a combined 370 lb. If Morgan weighs 20 lb more than Percy, and Anna weighs 1.5 times as much as Percy, how much does each person weigh?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137501545\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137501546\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137501545-solution\">59<\/a><span class=\"os-divider\">. <\/span>Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Micah ate twice as many ice cream bars as Joe, and Albert ate three less than Micah, how many ice cream bars did each roommate eat?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135547276\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135547278\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">60<\/span><span class=\"os-divider\">. <\/span>A farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost $2 per square foot, the wood $10 per square foot, and the plywood $5 per square foot. The farmer spent a total of $51, and the total amount of materials used was [latex] 14 ft^2. [\/latex] He used [latex] 3 ft^2 [\/latex] more chicken wire than plywood. How much of each material in did the farmer use?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134531903\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134531904\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134531903-solution\">61<\/a><span class=\"os-divider\">. <\/span>Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>","rendered":"<div id=\"main-content\" class=\"MainContent__ContentStyles-sc-6yy1if-0 NnXKu\" tabindex=\"-1\" data-dynamic-style=\"true\">\n<div id=\"page_634d2387-7429-480f-a04e-867c2f9699fb\" class=\"chapter-content-module\" data-type=\"page\" data-book-content=\"true\">\n<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p id=\"para-00001\">In this section, you will:<\/p>\n<ul>\n<li>Find the inverse of a matrix.<\/li>\n<\/ul>\n<ul id=\"list-00001\">\n<li>Solve a system of linear equations using an inverse matrix.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137640111\">Soriya plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Soriya invest in each bond? What is the best method to solve this problem?<\/p>\n<p id=\"fs-id1165133406558\">There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.<\/p>\n<section id=\"fs-id1165135547295\" data-depth=\"1\">\n<h2 data-type=\"title\">Finding the Inverse of a Matrix<\/h2>\n<p id=\"fs-id1165137422539\">We know that the multiplicative inverse of a real number <em>a<\/em> is [latex]a^{-1}[\/latex] and [latex]aa^{-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] The multiplicative inverse of a matrix is similar in concept, except that the product of matrix <em>A<\/em> and its inverse [latex]A^{-1}[\/latex] equals the identity matrix. 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 <em>n<\/em> represents the dimension of the matrix. Observe the following equations.<\/p>\n<p style=\"text-align: center;\">[latex]I_2=\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]I_3=\\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165137824188\">The identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A.[\/latex]<\/p>\n<p id=\"fs-id1165137783971\">A matrix that has a multiplicative inverse has the properties<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} AA^{-1} &=& I \\\\ A^{-1}A &=& I \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165133354140\">A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]AA^{-1}=A^{-1}A=I,[\/latex] is a requirement. Not all square matrices have an inverse, but if <em>A<\/em> is invertible, then [latex]A^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\times 2[\/latex] matrix and a third method that can be used on both [latex]2\\times 2[\/latex] and [latex]3\\times 3[\/latex] matrices.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">The Identity Matrix and Multiplicative Inverse<\/p>\n<\/header>\n<div class=\"textbox__content\">\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]\u00a0\\begin{array}{rlll} I_2 &=& \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} & I_3 &=& \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\\\ & 2\\times 2 && 3\\times 3 \\end{array}[\/latex]<\/p>\n<p>If\u00a0<em>A<\/em> is an [latex]n\\times n[\/latex] matrix and <em>B<\/em> is an\u00a0[latex]n\\times m[\/latex] matrix such that [latex]AB=BA=I_n,[\/latex] then [latex]B=A^{-1},[\/latex] the <strong><span id=\"term-00008\" data-type=\"term\">multiplicative inverse of a matrix<\/span> <em>A<\/em>.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 1: Showing that the Identity Matrix Acts as a 1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Given matrix <em data-effect=\"italics\">A<\/em>, show that [latex]AI=IA=A.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 3 & 4 \\\\ -2 & 5 \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>Use matrix multiplication to show that the product of\u00a0<em>A<\/em> and the identity is equal to the product of the identity and <em data-effect=\"italics\">A.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]AI= \\begin{bmatrix} 3 & 4 \\\\ -2 & 5 \\end{bmatrix}\u00a0 \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 3\\cdot 1+4\\cdot 0 & 3\\cdot 0+4\\cdot 1 \\\\ -2\\cdot 1+5\\cdot 0 & -2\\cdot 0+5\\cdot 1 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 3 & 4 \\\\ -2 & 5 \\end{bmatrix}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]IA= \\begin{bmatrix} 1 & 0\\\\ 0 & 1 \\end{bmatrix}\u00a0 \\begin{bmatrix} 3 & 4 \\\\ -2 & 5 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 1\\cdot 3+0\\cdot (-2) & 1\\cdot 4+0\\cdot 5 \\\\ 0\\cdot 3+1\\cdot (-2) & 0\\cdot 4+1\\cdot 5 \\end{bmatrix}\u00a0 = \\begin{bmatrix} 3 & 4 \\\\ -2 & 5 \\end{bmatrix}[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p id=\"fs-id1165134073896\"><strong>Given two matrices, show that one is the multiplicative inverse of the other.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1165135341279\" type=\"1\">\n<li>Given matrix <em>A<\/em> of order and matrix <em>B<\/em> of order multiply <em>AB<\/em>.<\/li>\n<li>If then find the product <em>BA<\/em>. If then and<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 2: Showing That Matrix <em>A<\/em> is the Multiplicative Inverse of Matrix <em>B<\/em><\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Show that the given matrices are multiplicative inverses of each other.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 1 & 5 \\\\ -2 & -9 \\end{bmatrix} , B= \\begin{bmatrix} -9 & -5 \\\\ 2 & 1 \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>Multiply <em>AB<\/em> and <em>BA<\/em>. 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}{rll} AB &=& \\begin{bmatrix} 1 & 5 \\\\ -2 & -9 \\end{bmatrix} \\cdot \\begin{bmatrix} -9 & -5 \\\\ 2 & 1 \\end{bmatrix}\u00a0 \\\\ &=& \\begin{bmatrix} 1(-9)+5(2) & 1(-5)+5(1) \\\\ -2(-9)-9(2) & -2(-5)-9(1) \\end{bmatrix}\u00a0 \\\\ &=& \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} BA &=& \\begin{bmatrix} -9 & -5 \\\\ 2 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} 1 & 5 \\\\ -2 & -9 \\end{bmatrix}\u00a0 \\\\ &=& \\begin{bmatrix} -9(1)-5(02) & -9(5)-5(-9) \\\\ 2(1)+1(-2) & 2(5)+1(-9) \\end{bmatrix}\u00a0 \\\\ &=& \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<p><em>A<\/em> and <em>B<\/em> are inverses of each other.<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Show that the following two matrices are inverses of each other.<\/p>\n<p>[latex]A= \\begin{bmatrix} 1 & 4 \\\\ -1 & -3 \\end{bmatrix} , B= \\begin{bmatrix} -3 & -4 \\\\ 1 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137418635\" data-depth=\"2\">\n<h3 data-type=\"title\">Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\n<p id=\"fs-id1165131961693\">We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using <span id=\"term-00009\" class=\"no-emphasis\" data-type=\"term\">matrix multiplication<\/span>.<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 3: Finding the Multiplicative Inverse Using Matrix Multiplication<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Use matrix multiplication to find the inverse of the given matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 1 & -2 \\\\ 2 & -3 \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>For this method, we multiply <em>A<\/em> by a matrix containing unknown constants and set it equal to the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} 1 & -2 \\\\ 2 & -3 \\end{bmatrix} \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}[\/latex]<\/p>\n<p>Find the product of the two matrices on the left side of the equal sign.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} 1 & -2 \\\\ 2 & -3 \\end{bmatrix} \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} = \\begin{bmatrix} 1a-2c & 1b-2d \\\\ 2a-3c & 2b-3d \\end{bmatrix}[\/latex]<\/p>\n<p>Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} 1a-2c &=& 1 & R_1 \\\\ 2a-3c &=& 0 & R_2 \\end{array}[\/latex]<\/p>\n<p>Using row operations, multiply and add as follows: [latex](-2)R_1+R_2\\rightarrow R_2.[\/latex] Add the equations, and solve for <em>c<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} 1a-2c &=& 1 \\\\ 0+1c &=& -2 \\\\ c &=& -2 \\end{array}[\/latex]<\/p>\n<p>Back-substitute to solve for <em>a<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} a-2(-2) &=& 1 \\\\ a+4 &=& 1 \\\\ a &=& -3 \\end{array}[\/latex]<\/p>\n<p>Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} 1b-2d &=& 0 & R_1 \\\\ 2b-3d &=& 1 & R_2 \\end{array}[\/latex]<\/p>\n<p>Using row operations, multiply and add as follows: [latex](-2)R_1+R_2=R_2.[\/latex] Add the two equations and solve for <em>d<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} 1b-2d &=& 0 \\\\ \\frac{0+1d=1}{d=1} \\end{array}[\/latex]<\/p>\n<p>Once more, back-substitute and solve for <em>b<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} b-2(1) &=& 0 \\\\ b-2 &=& 0 \\\\ b &=& 2 \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}=\\begin{bmatrix} -3 & 2 \\\\ -2 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137737676\" data-depth=\"2\">\n<h3 data-type=\"title\">Finding the Multiplicative Inverse by Augmenting with the Identity<\/h3>\n<p id=\"fs-id1165132328759\">Another way to find the <span id=\"term-00010\" class=\"no-emphasis\" data-type=\"term\">multiplicative inverse<\/span> is by augmenting with the identity. When matrix <em>A<\/em> is transformed into <em>I<\/em>, the augmented matrix <em>I<\/em> transforms into [latex]A^{-1}.[\/latex]<\/p>\n<p id=\"fs-id1165137893432\">For example, given<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 2 & 1 \\\\ 5 & 3 \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165135209378\">augment <em>A<\/em> with the identity<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr|rr} 2 & 1 & 1 & 0 \\\\ 5 & 3 & 0 & 1 \\end{array}\\right][\/latex]<\/p>\n<p id=\"fs-id1165137694941\">Perform row operations with the goal of turning <em>A<\/em> into the identity.<\/p>\n<ol id=\"fs-id1165137844094\" type=\"1\">\n<li>Switch row 1 and row 2.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 5 & 3 & 0 & 1 \\\\ 2 & 1 & 1 & 0 \\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 2 by -2 and add to row 1.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 1 & -2 & 1 \\\\ 2 & 1 & 1 & 0 \\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 1 by -2 and add to row 2.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 1 & -2 & 1 \\\\ 0 & -1 & 5 & -2 \\end{array}\\right][\/latex]<\/li>\n<li>Add row 2 to row 1.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 0 & 3 & -1 \\\\ 0 & -1 & 5 & -2 \\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 2 by -1.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 0 & 3 & -1 \\\\ 0 & 1 & -5 & 2 \\end{array}\\right][\/latex]<\/li>\n<\/ol>\n<p id=\"eip-702\">The matrix we have found is [latex]A^{-1}.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}= \\begin{bmatrix} 3 & -1 \\\\ -5 & 2 \\end{bmatrix}[\/latex]<\/p>\n<\/section>\n<section id=\"fs-id1165132300248\" data-depth=\"2\">\n<h3 data-type=\"title\">Finding the Multiplicative Inverse of [latex]2\\times 2[\/latex] Matrices Using a Formula<\/h3>\n<p id=\"fs-id1165135688796\">When we need to find the multiplicative inverse of a [latex]2\\times 2[\/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.<\/p>\n<p id=\"fs-id1165134047623\">If <em>A<\/em> is a [latex]2\\times 2[\/latex] matrix, such as<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165137459546\">the multiplicative inverse of <em>A<\/em> is given by the formula<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}=\\frac{1}{ad-bc} \\begin{bmatrix}d & -b \\\\ -c & a \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165137930360\">where [latex]ad-bc\\not=0.[\/latex] If [latex]ad-bc=0,[\/latex] then <em>A<\/em> has no inverse.<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 4: Using the Formula to Find the Multiplicative Inverse of Matrix <em>A<\/em><\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Use the formula to find the multiplicative inverse of<\/p>\n<p style=\"text-align: center;\">[latex]A=\\begin{bmatrix} 1 & -2 \\\\ 2 & -3 \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} A^{-1} &=& \\frac{1}{(1)(-3)-(-2)(2)} \\begin{bmatrix} -3 & 2 \\\\ -2 & 1 \\end{bmatrix} \\\\ &=& \\frac{1}{-3+4} \\begin{bmatrix} -3 & 2 \\\\ -2 & 1 \\end{bmatrix} \\\\ &=& \\begin{bmatrix} -3 & 2 \\\\ -2 & 1 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>We can check that our formula works by using one of the other methods to calculate the inverse. Let\u2019s augment A with the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr|rr} 1 & -2 & 1 & 0\\\\ 2 & 3 & 0 & 1 \\end{array}\\right][\/latex]<\/p>\n<p>Perform <span id=\"term-00013\" class=\"no-emphasis\" data-type=\"term\">row operations<\/span> with the goal of turning A into the identity.<\/p>\n<ol id=\"fs-id1165134550520\" type=\"1\">\n<li>Multiple row 1 by -2 and add to row 2.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & -2 & 1 & 0 \\\\ 0 & 1 & -2 & 1 \\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 2 by -2 and add to row 1.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 0 & -3 & 2 \\\\ 0 & 1 & -2 & 1 \\end{array}\\right][\/latex]<\/li>\n<\/ol>\n<p>So, we have verified our original solution.<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}= \\begin{bmatrix} -3 & 2 \\\\ -2 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Use the formula to find the inverse of matrix <em>A<\/em>. Verify your answer by augmenting with the identity matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 1 & -1 \\\\ 2 & 3 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 5: Finding the Inverse of the Matrix, if it Exists<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the inverse, if it exists, of the given matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 3 & 6 \\\\ 1 & 2 \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>We will use the method of augmenting with the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr|rr} 3 & 6 & 1 & 0\\\\ 1 & 2 & 0 & 1\u00a0 \\end{array}\\right][\/latex]<\/p>\n<ol>\n<li>Switch row 1 and row 2.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 3 & 01 & 1 \\\\ 3 & 2 & 1 & 0\u00a0 \\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 1 by -3 and add to row 2.<br \/>\n[latex]\\left[\\begin{array}{rr|rr} 1 & 2 & 1 & 0 \\\\ 0 & 0 & -3 & 1 \\end{array}\\right][\/latex]<\/li>\n<li>There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/li>\n<\/ol>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137680361\" data-depth=\"2\">\n<h3 data-type=\"title\">Finding the Multiplicative Inverse of [latex]3\\times 3[\/latex] Matrices<\/h3>\n<p id=\"fs-id1165134238766\">Unfortunately, we do not have a formula similar to the one for a [latex]2\\times 2[\/latex] matrix to find the inverse of a [latex]3\\times 3[\/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.<\/p>\n<p id=\"fs-id1165134435528\">Given a [latex]3\\times 3[\/latex] matrix<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 2 & 3 & 1 \\\\ 3 & 3 & 1 \\\\ 2 & 4 & 1 \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165133394718\">augment <em>A<\/em> with the identity matrix<\/p>\n<p style=\"text-align: center;\">[latex]A|I = \\left[\\begin{array}{ccc|ccc} 2 & 3 & 1 & 1 & 0 & 0 \\\\ 3 & 3 & 1 & 0 & 1 & 0 \\\\ 2 & 4 & 1 & 0 & 0 & 1 \\end{array} \\right][\/latex]<\/p>\n<p id=\"fs-id1165134357582\">To begin, we write the augmented matrix with the identity on the right and <em>A<\/em> on the left. Performing elementary <span id=\"term-00016\" class=\"no-emphasis\" data-type=\"term\">row operations<\/span> so that the <span id=\"term-00017\" class=\"no-emphasis\" data-type=\"term\">identity matrix<\/span> appears on the left, we will obtain the <span id=\"term-00018\" class=\"no-emphasis\" data-type=\"term\">inverse matrix<\/span> on the right. We will find the inverse of this matrix in the next example.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Given a [latex]3\\times 3[\/latex] matrix, find the inverse<\/strong><\/p>\n<ol>\n<li>Write the original matrix augmented with the identity matrix on the right.<\/li>\n<li>Use elementary row operations so that the identity appears on the left.<\/li>\n<li>What is obtained on the right is the inverse of the original matrix.<\/li>\n<li>Use matrix multiplication to show that [latex]AA^{-1}=I[\/latex] and [latex]A^{-1}A=I.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 6: Finding the Inverse of a [latex]3\\times 3[\/latex] Matrix<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Given the [latex]3\\times 3[\/latex] matrix <em>A<\/em>, find the inverse.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 2 & 3 & 1 \\\\ 3 & 3 & 1 \\\\ 2 & 4 & 1 \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>Augment <em>A<\/em> with the identity matrix, and then begin row operations until the identity matrix replaces <em>A.<\/em> The matrix on the right will be the inverse of <em>A<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} \\left[\\begin{array}{rrr|rrr} 2 & 3 & 1 & 1 & 0 & 0 \\\\ 3 & 3 & 1 & 0 & 1 & 0 \\\\ 2 & 4 & 1 & 0 & 0 & 1\u00a0 \\end{array}\\right] &\\overset{\\text{Interchange} \\quad R_2 \\quad {and} \\quad R_1}{\\rightarrow}& \\left[\\begin{array}{rrr|rrr} 3 & 3 & 1 & 0 & 1 & 0 \\\\ 2 & 3 & 1 & 1 & 0 & 0 \\\\ 2 & 4 & 1 & 0 & 0 & 1 \\end{array}\\right]\u00a0 \\\\ -R_2+R_1=R_1 &\\rightarrow & \\left[\\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & 0 \\\\ 2 & 3 & 1 & 1 & 0 & 0 \\\\ 2 & 4 & 1 & 0 & 0 & 1 \\end{array}\\right]\u00a0 \\\\ -R_2+R_3=R_3 &\\rightarrow & \\left[\\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & 0 \\\\ 2 & 3 & 1 & 1 & 0 & 00 \\\\ 0 & 1 &\u00a0 0 & -1 & 0 & 1 \\end{array}\\right]\u00a0 \\\\ R_3 \\leftrightarrow R_2 &\\rightarrow & \\left[\\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & 0 \\\\ 0 & 1 & 0 & -1 & 0 & 1 \\\\ 2 & 3 & 1 & 1 & 0 & 0 \\end{array}\\right]\u00a0 \\\\ -2R_1+R_3=R_3 &\\rightarrow & \\left[\\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & 0 \\\\ 0 & 1 & 0 & -1 & 0 & 1 \\\\ 0 & 3 & 1 & 3 & -2 & 0 \\end{array}\\right]\u00a0 \\\\ -3R_2+R_3=R_3&\\rightarrow & \\left[\\begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 1 & 0 \\\\ 0 & 1 & 0 & -1 & 0 & 1 \\\\ 0 & 0 & 1 & 6 & -2 & -3 \\end{array}\\right] \\end{array}[\/latex]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}=B= \\begin{bmatrix} -1 & 1 & 0 \\\\ -1 & 0 & 1 & 6 & -2 & -3 \\end{bmatrix}[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>To prove that [latex]B=A^{-1},[\/latex] let\u2019s multiply the two matrices together to see if the product equals the identity, if [latex]AA^{-1}=I[\/latex] and [latex]A^{-1}A=I.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} AA^{-1} &=& \\begin{bmatrix} 2 & 3 & 1 \\\\ 3 & 3 & 1 \\\\ 2 & 4 & 1 \\end{bmatrix} \\begin{bmatrix} -1 & 1 & 0\\\\ -1 & 0 & 1 \\\\ 6 & -2 & -3 \\end{bmatrix} \\\\ &=& \\begin{bmatrix} 2(-1)+3(-1)+1(6) & 2(1)+3(0)+1(-2) & 2(0)+3(1)+1(-3) \\\\ 3(-1)+3(-1)+1(6) & 3(1)+3(0)+1(-2) & 3(0)+3(1)+1(-3) \\\\ 2(-1)+4(-1)+1(6) & 2(1)+4(0)+1(-2) & 2(0)+4(1)+1(-3) \\end{bmatrix} \\\\ &=& \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} A^{-1}A &=& \\begin{bmatrix} -1 & 1 & 0 \\\\ -1 & 0 & 1 \\\\ 6 & -2 & -3 \\end{bmatrix} \\begin{bmatrix} 2 & 3 & 1 \\\\ 3 & 3 & 1 \\\\ 2 4 & 1 \\end{bmatrix} \\\\ &=& \\begin{bmatrix} -1(2)+1(3)+0(2) & -1(3)+1(3)+0(4) & -1(1)+1(1)+0(1) \\\\ -1(2)+0(3)+1(2) & -1(3)+0(3)+1(4) & -1(1)+0(1)+1(01 \\\\ 6(2)+-2(3)+-3(2) & 6(3)+-2(3)+-3(4) & 6(1)+-2(1)+-3(1) \\end{bmatrix} \\\\ &=& \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & ) & 1 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the inverse of the [latex]3\\times 3[\/latex] matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 2 & -17 & 11 \\\\ -1 & 11 & -7 \\\\ 0 & 3 & -2 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137694180\" data-depth=\"1\">\n<h2 data-type=\"title\">Solving a System of Linear Equations Using the Inverse of a Matrix<\/h2>\n<p id=\"fs-id1165135394318\">Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: <em>X<\/em> is the matrix representing the variables of the system, and <em>B<\/em> is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as<\/p>\n<p style=\"text-align: center;\">[latex]AX=B[\/latex]<\/p>\n<p id=\"fs-id1165137466371\">To solve a system of linear equations using an inverse matrix, let <em>A<\/em> be the coefficient matrix, let <em>X<\/em> be the variable matrix, and let <em>B<\/em> be the constant matrix. Thus, we want to solve a system [latex]AX=B.[\/latex] For example, look at the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} a_1x+b_1y &=& c_1 \\\\ a_2x+b_2y &=& c_2 \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165137626840\">From this system, the coefficient matrix is<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165137628730\">The variable matrix is<\/p>\n<p style=\"text-align: center;\">[latex]X= \\begin{bmatrix} x \\\\ y \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165135670279\">And the constant matrix is<\/p>\n<p style=\"text-align: center;\">[latex]B= \\begin{bmatrix} c_1 \\\\ c_2 \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165134043634\">Then [latex]AX=B[\/latex] looks like<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} c_1 \\\\ c_2 \\end{bmatrix}[\/latex]<\/p>\n<p id=\"fs-id1165137418368\">Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex]\\left(2^{-1}\\right) 2 =\\left(\\frac{1}{2}\\right)2=1\/[\/latex] To solve a single linear equation [latex]ax=b[\/latex] for <em>x<\/em>, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of <em>a<\/em>. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} ax &=& b \\\\ \\left(\\frac{1}{a}\\right)ax &=& \\left(\\frac{1}{a}\\right)b \\\\ \\left(a^{-1}\\right)ax- &=& \\left(a^{-1}\\right)b \\\\ [\\left(a^{-1}\\right)a]x &=& \\left(a^{-1}\\right)b \\\\ 1x &=& \\left(a^{-1}\\right)b \\\\ x &=& \\left(a^{-1}\\right)b \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165137843896\">The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same\u2014to isolate the variable.<\/p>\n<p id=\"fs-id1165135487326\">We will investigate this idea in detail, but it is helpful to begin with a [latex]2\\times 2[\/latex] system and then move on to a [latex]3\\times 3[\/latex] system.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Solving a System of Equations Using the Inverse of a Matrix<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Given a system of equations, write the coefficient matrix <em>A<\/em>, the variable matrix <em>X<\/em>, and the constant matrix B. Then<\/p>\n<p style=\"text-align: center;\">[latex]AX=B[\/latex]<\/p>\n<p>Multiply both sides by the inverse of <em>A<\/em> to obtain the solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} \\left(A^{-1}\\right)AX &=& \\left(A^{-1}\\right)B \\\\ [\\left(A^{-1}\\right)A]X &=& \\left(A^{-1}\\right)B \\\\ IX &=& \\left(A^{-1}\\right)B \\\\ x &=& \\left(A^{-1}\\right)B \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Q&amp;A<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p id=\"fs-id1165135149799\"><strong>Q: If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/strong><\/p>\n<p id=\"fs-id1165135344922\"><em data-effect=\"italics\">A: No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.<\/em><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7: Solving a [latex]2\\times 2[\/latex] System Using the Inverse of a Matrix<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve the given system of equations using the inverse of a matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} 3x+8y &=& 5 \\\\ 4x+11y &=& 7 \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A= \\begin{bmatrix} 3 & 8 \\\\ 4 & 11 \\end{bmatrix}, X= \\begin{bmatrix} x \\\\ y \\end{bmatrix}, B= \\begin{bmatrix} 5 \\\\ 7 \\end{bmatrix}\u00a0[\/latex]<\/p>\n<p>Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} 3 & 8 \\\\ 4 & 11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} 5 \\\\ 7 \\end{bmatrix}[\/latex]<\/p>\n<p>First, we need to calculate [latex]A^{-1}\/[\/latex] Using the formula to calculate the inverse of a 2 by 2 matrix, we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} A^{-1} &=& \\frac{1}{ad-bc} \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix} \\\\ &=& \\frac{1}{3(11)-8(4)} \\begin{bmatrix} 11 & -8 \\\\ -4 & 3 \\end{bmatrix} \\\\ &=& \\frac{1}{1} \\begin{bmatrix} 11 & -8 \\\\ -4 & 3 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1} = \\begin{bmatrix} 11 & -8 \\\\ -4 & 3 \\end{bmatrix}[\/latex]<\/p>\n<p>Now we are ready to solve. Multiply both sides of the equation by [latex]A^{-1}\/[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} \\left(A^{-1}\\right)AX &=& \\left(A^{-1}\\right)B \\\\ \\begin{bmatrix} 11 & -8 \\\\ -4 & 3 \\end{bmatrix} \\begin{bmatrix} 3 & 8 \\\\ 4 & 11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} &=& \\begin{bmatrix} 11 & -8 \\\\ -4 & 3 \\end{bmatrix} \\begin{bmatrix} 5 \\\\ 7 \\end{bmatrix} \\\\ \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} &=& \\begin{bmatrix} 11(5)+(-8)7 \\\\ -4(5)+3(7) \\end{bmatrix} \\\\ \\begin{bmatrix} x \\\\ y \\end{bmatrix} &=& \\begin{bmatrix} -1 \\\\ 1 \\end{bmatrix} \\end{array}[\/latex]<\/p>\n<p>The solution is [latex](-1, 1).[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Q&amp;A<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Q: Can we solve for <em>X<\/em> by finding the product [latex]BA^{-1}[\/latex]?<\/strong><\/p>\n<p><em data-effect=\"italics\">A: No, recall that matrix multiplication is not commutative, so [latex]A^{-1}B\\not=BA^{-1}.[\/latex] Consider our steps for solving the matrix equation.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} \\left(A^{-1}\\right)AX &=& \\left(A^{-1}\\right)B \\\\ [\\left(A^{-1}\\right)]X &=& \\left(A^{-1}\\right)B \\\\ IX &=& \\left(A^{-1}\\right)B \\\\ X &=& \\left(A^{-1}\\right)B \\end{array}[\/latex]<\/p>\n<p><em data-effect=\"italics\">Notice in the first step we multiplied both sides of the equation by [latex]A^{-1},[\/latex] but the [latex]A^{-1}[\/latex] was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters.<\/em><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 8: Solving a [latex]3\\times 3[\/latex] System Using the Inverse of a Matrix<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve the following system using the inverse of a matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} 5x+15y+56z &=& 35 \\\\ -4x-11y-41z &=& -26 \\\\ -x-3y-11z &=& -7 \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>Write the equation [latex]AX=B.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} 5 & 15 & 56 \\\\ -4 & -11 & -41 \\\\ -1 & -3 & -11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{bmatrix} 35 \\\\ -26 \\\\ -7 \\end{bmatrix}[\/latex]<\/p>\n<p>First, we will find the inverse of <em>A<\/em> by augmenting with the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 5 & 15 & 56 & 1 & 0 & 0 \\\\ -4 & -11 & -41 & 0 & 1 & 0 \\\\ -1 & -3 & -11 & 0 & 0 & 1 \\end{array} \\right][\/latex]<\/p>\n<p>Multiply row 1 by [latex]\\frac{1}{5}.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 3 & \\frac{56}{5} & \\frac{1}{5} & 0 & 0 \\\\ -4 & -11 & -41 & 0 & 1 & 0 \\\\ -1 & -3 & -11 & 0 & 0 & 1 \\end{array} \\right][\/latex]<\/p>\n<p>Multiply row 1 by 4 and add to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 3 & \\frac{56}{5} & \\frac{1}{5} & 0 & 0 \\\\ 0 & 1 & \\frac{19}{5} & \\frac{4}{5} & 1 & 0 \\\\ -1 & -3 & -11 & 0 & 0 & 1 \\end{array} \\right][\/latex]<\/p>\n<p>Add row 1 to row 3.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 3 & \\frac{56}{5} & \\frac{1}{5} & 0 & 0 \\\\ 0 & 1 & \\frac{19}{5} & \\frac{4}{5} & 1 & 0 \\\\ 0 & 0 & \\frac{1}{5} & \\frac{1}{5} & 0 & 1 \\end{array} \\right][\/latex]<\/p>\n<p>Multiply row 2 by \u22123 and add to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 0 & -\\frac{1}{5} & -\\frac{11}{5} & -3 & 0 \\\\ 0 & 1 & \\frac{19}{5} & \\frac{4}{5} & 1 & 0 \\\\ 0 & 0 & \\frac{1}{5} & \\frac{1}{5} & 0 & 1 \\end{array} \\right][\/latex]<\/p>\n<p>Multiply row 3 by 5.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 0 & -\\frac{1}{5} & -\\frac{11}{5} & -3 & 0 \\\\ 0 & 1 & \\frac{19}{5} & \\frac{4}{5} & 1 & 0 \\\\ 0 & 0 & 1 & 1 & 0 & 5 \\end{array} \\right][\/latex]<\/p>\n<p>Multiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 0 & 0 & -2 & -3 & 1 \\\\ 0 & 1 & \\frac{19}{5} & \\frac{4}{5} & 1 & 0 \\\\ 0 & 0 & 1 & 1 & 0 & 5 \\end{array} \\right][\/latex]<\/p>\n<p>Multiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[ \\begin{array}{ccc|ccc} 1 & 0 & 0 & -2 & -3 & 1 \\\\ 0 & 1 & 0 & -3 & 1 & -19 \\\\ 0 & 0 & 1 & 1 & 0 & 5 \\end{array} \\right][\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}= \\begin{bmatrix} -2 & 03 & 1 \\\\ -3 & 1 & -19 \\\\ 1 & 0 & 5 \\end{bmatrix}[\/latex]<\/p>\n<p>Multiply both sides of the equation by [latex]A^{-1}.[\/latex] We want [latex]A^{-1}AX=A^{-1}B :[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} -2 & -3 & 1 \\\\ -3 & 1 & -19 \\\\ 1 & 0 & 5 \\end{bmatrix} \\begin{bmatrix} 5 & 15 & 56 \\\\ -4 & -11 & -41 \\\\ -1 & -3 & -11 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{bmatrix} -2 & -3 & 1 \\\\ -3 & 1 & -19 \\\\ 1 & 0 & 5 \\end{bmatrix} \\begin{bmatrix} 35 \\\\ -26 \\\\ -7 \\end{bmatrix}[\/latex]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]A^{-1}B= \\begin{bmatrix} -70+78-7 \\\\ -105-26+133 \\\\ 35+0-35 \\end{bmatrix} = \\begin{bmatrix} 1 \\\\ 2 \\\\0 \\end{bmatrix}[\/latex]<\/p>\n<p>The solution is [latex](1, 2, 0).[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve the system using the inverse of the coefficient matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} 2x-17y+11z &=& 0 \\\\ -x+11y-7z &=& 8 \\\\ 3y-2z &=& -2 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p id=\"fs-id1165135503565\"><strong>Given a system of equations, solve with matrix inverses using a calculator.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1165135503570\" type=\"1\">\n<li>Save the coefficient matrix and the constant matrix as matrix variables [latex][A][\/latex] and [latex][B].[\/latex]<\/li>\n<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\n<li>If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve the system of equations with matrix inverses using a calculator<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll} 2x+3y+z &=& 32 \\\\ 3x+3y+z &=& -27 \\\\ 2x+4y+z &=& -2 \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>On the matrix page of the calculator, enter the <span id=\"term-00023\" class=\"no-emphasis\" data-type=\"term\">coefficient matrix<\/span> as the matrix variable [latex][A],[\/latex] and enter the constant matrix as the matrix variable [latex][B].[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex][A]= \\begin{bmatrix} 2 & 3 & 1 \\\\ 3 & 3 & 1 \\\\ 2 & 4 & 1 \\end{bmatrix}, [B]= \\begin{bmatrix} 32 \\\\ -27 \\\\ -2 \\end{bmatrix}[\/latex]<\/p>\n<p>On the home screen of the calculator, type in the multiplication to solve for <em>X<\/em>, calling up each matrix variable as needed.<\/p>\n<p style=\"text-align: center;\">[latex][A]^{-1}\\times [B][\/latex]<\/p>\n<p>Evaluate the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{bmatrix} =59 \\\\ -34 \\\\ 252 \\end{bmatrix}[\/latex]<\/p>\n<\/details>\n<section>\n<div><\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Media<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p id=\"fs-id1165137836914\">Access these online resources for additional instruction and practice with solving systems with inverses.<\/p>\n<ul id=\"fs-id1165137836918\">\n<li><a href=\"http:\/\/openstax.org\/l\/identmatrix\" target=\"_blank\" rel=\"noopener nofollow\">The Identity Matrix<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/inversematrix\" target=\"_blank\" rel=\"noopener nofollow\">Determining Inverse Matrices<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/matrixsystem\" target=\"_blank\" rel=\"noopener nofollow\">Using a Matrix Equation to Solve a System of Equations<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"os-eos os-section-exercises-container\" data-uuid-key=\".section-exercises\">\n<h2 data-type=\"document-title\" data-rex-keep=\"true\"><span class=\"os-text\">7.7 Section Exercises<\/span><\/h2>\n<section id=\"fs-id1165135446624\" class=\"section-exercises\" data-depth=\"1\">\n<section id=\"fs-id1165137548736\" data-depth=\"2\">\n<h3 data-type=\"title\">Verbal<\/h3>\n<div id=\"fs-id1165137548741\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137548742\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137548741-solution\">1<\/a><span class=\"os-divider\">. <\/span>In a previous section, we showed that matrix multiplication is not commutative, that is, [latex]AB\\not=BA[\/latex] in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is, [latex]A^{-1}A=AA^{-1}?[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134493452\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134493453\" data-type=\"problem\">\n<p><span class=\"os-number\">2<\/span><span class=\"os-divider\">. <\/span>Does every [latex]2\\times 2[\/latex] matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135547596\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135547597\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135547596-solution\">3<\/a><span class=\"os-divider\">. <\/span>Can you explain whether a [latex]2\\times 2[\/latex] matrix with an entire row of zeros can have an inverse?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133401600\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135702541\" data-type=\"problem\">\n<p><span class=\"os-number\">4<\/span><span class=\"os-divider\">. <\/span>Can a matrix with an entire column of zeros have an inverse? Explain why or why not.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135702546\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135702547\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135702546-solution\">5<\/a><span class=\"os-divider\">. <\/span>Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a [latex]2\\times 2[\/latex] matrix.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<section id=\"fs-id1165133075017\" data-depth=\"2\">\n<h3 data-type=\"title\">Algebraic<\/h3>\n<p id=\"fs-id1165132059611\">In the following exercises, show that matrix <em>A<\/em> is the inverse of matrix <em>B<\/em>.<\/p>\n<div id=\"fs-id1165137804955\" class=\"material-set-2\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137804956\" data-type=\"problem\">\n<p><span class=\"os-number\">6<\/span><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix} 1 & 0 \\\\ -1 & 1 \\end{bmatrix}, B= \\begin{bmatrix} 1 & 0 \\\\ 1 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<div id=\"fs-id1165137629353\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137629354\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137629353-solution\">7<\/a><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix}, B= \\begin{bmatrix} -2 & 1 \\\\ \\frac{3}{2} & -\\frac{1}{2} \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137832785\" class=\"material-set-2\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137832786\" data-type=\"problem\">\n<p><span class=\"os-number\">8<\/span><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix} 4 & 5 \\\\ 7 & 0 \\end{bmatrix}, B= \\begin{bmatrix} 0 & \\frac{1}{7} \\\\ \\frac{1}{5} & -\\frac{4}{35} \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137417445\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137417446\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137417445-solution\">9<\/a><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix} -2 & \\frac{1}{2} \\\\ 3 & -1 \\end{bmatrix}, B= \\begin{bmatrix} -2 & 01 \\\\ -6 & -4 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137810179\" class=\"material-set-2\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137810180\" data-type=\"problem\">\n<p><span class=\"os-number\">10<\/span><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix} 1 & 0 & 1 \\\\ 0 & 1 & -1 \\\\ 0 & 1 & 1 \\end{bmatrix}, B= \\frac{1}{2}\\begin{bmatrix} 2 & 1 & -1 \\\\ 0 & 1 & 1 \\\\ 0 & -1 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135481914\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135481915\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135481914-solution\">11<\/a><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix}\u00a0 1 & 2 & 3 \\\\ 4 & 0 & 2 \\\\ 1 & 6 & 9\\end{bmatrix}, B= \\frac{1}{4}\\begin{bmatrix} 6 & 0 & -2 \\\\ 17 & -3 & -5 \\\\ -12 & 2 & 4 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134371186\" class=\"material-set-2\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134371187\" data-type=\"problem\">\n<p><span class=\"os-number\">12<\/span><span class=\"os-divider\">. <\/span> [latex]A= \\begin{bmatrix} 3 & 8 & 2 \\\\ 1 & 1 & 1 \\\\ 5 & 6 & 12 \\end{bmatrix}, B= \\frac{1}{36}\\begin{bmatrix} -6 & 84 & -6 \\\\ 7 & -26 & 1 \\\\ -1 & -22 & 5 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165134282177\">For the following exercises, find the multiplicative inverse of each matrix, if it exists.<\/p>\n<div id=\"fs-id1165134282180\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134282181\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134282180-solution\">13<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 3 & -2 \\\\ 1 & 9 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135445702\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135445703\" data-type=\"problem\">\n<p><span class=\"os-number\">14<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} -2 & 2 \\\\ 3 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137387518\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137387519\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137387518-solution\">15<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} -3 & 7 \\\\ 9 & 2 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134339996\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134339997\" data-type=\"problem\">\n<p><span class=\"os-number\">16<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} -4 & -3 \\\\ -5 & 8 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133103275\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133103276\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165133103275-solution\">17<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 1 \\\\ 2 & 2 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165132957188\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165132957189\" data-type=\"problem\">\n<p><span class=\"os-number\">18<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134185464\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134185465\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134185464-solution\">19<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 0.5 & 1.5 \\\\ 1 & -0.5 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134152537\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134152538\" data-type=\"problem\">\n<p><span class=\"os-number\">20<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 0 & 6 \\\\ 02 & 1 & 7 \\\\ 3 & 0 & 2 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133141439\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135679418\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165133141439-solution\">21<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 0 & 1 & -3 \\\\ 4 & 1 & 0 \\\\ 1 & 0 & 5 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134040470\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134040471\" data-type=\"problem\">\n<p><span class=\"os-number\">22<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 2 & -1 \\\\ -3 & 4 & 1 \\\\ -2 & -4 & -5 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134159676\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134159677\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134159676-solution\">23<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 9 & -3 \\\\ 2 & 5 & 6 \\\\ 4 & -2 & 7 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133210779\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133210780\" data-type=\"problem\">\n<p><span class=\"os-number\">24<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & -2 & 3 \\\\ -4 & 8 & -12 \\\\ 1 & 4 & 2 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135397109\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135397110\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135397109-solution\">25<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} \\frac{1}{2} & \\frac{1}{2} & \\frac{1}{2} \\\\ \\frac{1}{3} & \\frac{1}{4} & \\frac{1}{5} \\\\ \\frac{1}{6} & \\frac{1}{7} & \\frac{1}{8} \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165131818536\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165131818537\" data-type=\"problem\">\n<p><span class=\"os-number\">26<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165137675157\">For the following exercises, solve the system using the inverse of a [latex]2\\times 2[\/latex] matrix.<\/p>\n<div id=\"fs-id1165134319666\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134070732\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134319666-solution\">27<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 5x-6y &=& -61 \\\\ 4x+3y &=& -2 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133012293\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133012294\" data-type=\"problem\">\n<p><span class=\"os-number\">28<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 8x+4y &=& -100 \\\\ 3x-4y &=& 1 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165131884588\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165131884589\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165131884588-solution\">29<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 3x-2y &=& 6 \\\\ -x+5y &=& -2 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133045242\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133045243\" data-type=\"problem\">\n<p><span class=\"os-number\">30<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 5x-4y &=& -5 \\\\ 4x+y &=& 2.3 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135205038\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135205039\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135205038-solution\">31<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll}-3x-4y &=& 9 \\\\ 12x+4y &=& -6 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135518212\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135518213\" data-type=\"problem\">\n<p><span class=\"os-number\">32<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} -2x+3y &=& \\frac{3}{10} \\\\ -x+5y &=& \\frac{1}{2} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165131958337\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165131958338\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165131958337-solution\">33<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} \\frac{8}{5}x-\\frac{4}{5}y &=& \\frac{2}{5} \\\\ -\\frac{8}{5}x+\\frac{1}{5}y &=& \\frac{7}{10} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133157406\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135523285\" data-type=\"problem\">\n<p><span class=\"os-number\">34<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} \\frac{1}{2}x+\\frac{1}{5}y &=& -\\frac{1}{4} \\\\ \\frac{1}{2}x-\\frac{3}{5}y &=& -\\frac{9}{4} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165134386500\">For the following exercises, solve a system using the inverse of a [latex]3\\times 3[\/latex] matrix.<\/p>\n<div id=\"fs-id1165132079352\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165132079353\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165132079352-solution\">35<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 3x-2y+5z &=& 21 \\\\ 5x+4y &=& 37 \\\\ x-2y-5z &=& 5 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134177544\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134177545\" data-type=\"problem\">\n<p><span class=\"os-number\">36<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 4x+4y+4z &=& 40 \\\\ 2x-3y+4z &=& -12 \\\\ -x+3y+4z &=& 9 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134226706\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134226707\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134226706-solution\">37<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 6x-5y-z &=& 31 \\\\ -x+2y+z &=& -6 \\\\ 3x+3y+2z &=& 13 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134224075\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134224076\" data-type=\"problem\">\n<p><span class=\"os-number\">38<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 6x-5y+2z &=& -4 \\\\ 2x+5y-z &=& 12 \\\\ 2x+5y+z &=& 12 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134185366\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134185367\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134185366-solution\">39<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 4x-2y+3z &=& -12 \\\\ 2x+2y-9z &=& 33 \\\\ 6y-4z &=& 1 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135207327\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135207328\" data-type=\"problem\">\n<p><span class=\"os-number\">40<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} \\frac{1}{10}x-\\frac{1}{5}y+4z &=& \\frac{-41}{2} \\\\ \\frac{1}{5}x-20y+\\frac{2}{5}z &=& -101 \\\\ \\frac{3}{10}x+4y-\\frac{3}{10}z &=& 23 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135421457\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135421458\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135421457-solution\">41<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} \\frac{1}{2}x-\\frac{1}{5}y+\\frac{1}{5}z &=& \\frac{31}{100} \\\\ -\\frac{3}{4}x-\\frac{1}{4}y+\\frac{1}{2}z &=& \\frac{7}{40} \\\\ -\\frac{4}{5}x-\\frac{1}{2}y+\\frac{3}{2}z &=& \\frac{1}{4} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165132944845\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165132944846\" data-type=\"problem\">\n<p><span class=\"os-number\">42<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} -.1x+0.2y+0.3z &=& -1.4 \\\\ 0.1x-0.2y+0.3z &=& -.6 \\\\ 0.4y+0.9z &=& -2 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<section id=\"fs-id1165137900020\" data-depth=\"2\">\n<h3 data-type=\"title\">Technology<\/h3>\n<p id=\"fs-id1165134356163\">For the following exercises, use a calculator to solve the system of equations with matrix inverses.<\/p>\n<div id=\"fs-id1165134356166\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134356167\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134356166-solution\">43<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 2x-y &=& -3 \\\\ -x+2y &=& 2.3 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<div id=\"fs-id1165135186095\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135186096\" data-type=\"problem\">\n<p><span class=\"os-number\">44<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} -\\frac{1}{2}x-\\frac{3}{2}y &=& -\\frac{43}{20} \\\\ \\frac{5}{2}x+\\frac{11}{5}y &=& \\frac{31}{4} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135652325\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135652326\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135652325-solution\">45<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 12.3x-2y-2.5z &=& 2 \\\\ 36.9x+7y-7.5z &=& -7 \\\\ 8y-5z &=& -10 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134467729\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134467730\" data-type=\"problem\">\n<p><span class=\"os-number\">46<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{array}{rll} 0.5x-3y+6z &=& -0.8 \\\\ 0.7x-2y &=& -0.06 \\\\ 0.5x+4y+5z &=& 0 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<section id=\"fs-id1165134537745\" data-depth=\"2\">\n<h3 data-type=\"title\">Extensions<\/h3>\n<p id=\"fs-id1165135339476\">For the following exercises, find the inverse of the given matrix.<\/p>\n<div id=\"fs-id1165135339479\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135339480\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135339479-solution\">47<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 1 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<div id=\"fs-id1165135258897\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135258898\" data-type=\"problem\">\n<p><span class=\"os-number\">48<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} -1 & 0 & 2 & 5 \\\\ 0 & 0 & 0 & 2 \\\\ 0 & 2 & -1 & 0 \\\\ 1 & -3 & 0 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135191959\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135191960\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135191959-solution\">49<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & -2 & 3 & 0 \\\\ 0 & 1 & 0 & 2 \\\\ 1 & 4 & -2 & 3 \\\\ -5 & 0 & 1 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134234210\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134234211\" data-type=\"problem\">\n<p><span class=\"os-number\">50<\/span><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 2 & 0 & 2 & 3 \\\\ 0 & 2 & 1 & 0 & 0 \\\\ 0 & 0 & 3 & 0 & 1 \\\\ 0 & 2 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 2 & 0 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134385542\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134385543\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134385542-solution\">51<\/a><span class=\"os-divider\">. <\/span> [latex]\\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 0 & ) \\\\ 0 & 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 1 & 0 \\\\ 1 & 1 & 1 & 1 & 1 & 1 \\end{bmatrix}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<section id=\"fs-id1165133094400\" data-depth=\"2\">\n<h3 data-type=\"title\">Real-World Applications<\/h3>\n<p id=\"fs-id1165133094406\">For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix.<\/p>\n<div id=\"fs-id1165133260340\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133260342\" data-type=\"problem\">\n<p><span class=\"os-number\">52<\/span><span class=\"os-divider\">. <\/span>2,400 tickets were sold for a basketball game. If the prices for floor 1 and floor 2 were different, and the total amount of money brought in is $64,000, how much was the price of each ticket?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133260348\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133260349\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165133260348-solution\">53<\/a><span class=\"os-divider\">. <\/span>In the previous exercise, if you were told there were 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133111147\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133111148\" data-type=\"problem\">\n<p><span class=\"os-number\">54<\/span><span class=\"os-divider\">. <\/span>A food drive collected two different types of canned goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134081432\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134081433\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134081432-solution\">55<\/a><span class=\"os-divider\">. <\/span>Students were asked to bring their favorite fruit to class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135705936\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135705937\" data-type=\"problem\">\n<p><span class=\"os-number\">56<\/span><span class=\"os-divider\">. <\/span>The nursing club held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at $0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135705944\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135705945\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165135705944-solution\">57<\/a><span class=\"os-divider\">. <\/span>A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at $13.99, the beanie at $7.99, and the cowboy hat at $14.49. If 100 hats were sold this past quarter, $1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133354003\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133354004\" data-type=\"problem\">\n<p><span class=\"os-number\">58<\/span><span class=\"os-divider\">. <\/span>Anna, Percy, and Morgan weigh a combined 370 lb. If Morgan weighs 20 lb more than Percy, and Anna weighs 1.5 times as much as Percy, how much does each person weigh?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137501545\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137501546\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165137501545-solution\">59<\/a><span class=\"os-divider\">. <\/span>Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Micah ate twice as many ice cream bars as Joe, and Albert ate three less than Micah, how many ice cream bars did each roommate eat?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135547276\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135547278\" data-type=\"problem\">\n<p><span class=\"os-number\">60<\/span><span class=\"os-divider\">. <\/span>A farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost $2 per square foot, the wood $10 per square foot, and the plywood $5 per square foot. The farmer spent a total of $51, and the total amount of materials used was [latex]14 ft^2.[\/latex] He used [latex]3 ft^2[\/latex] more chicken wire than plywood. How much of each material in did the farmer use?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134531903\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134531904\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-7\" data-page-slug=\"chapter-7\" data-page-uuid=\"f0a43272-7860-5760-ac27-5047923a0f08\" data-page-fragment=\"fs-id1165134531903-solution\">61<\/a><span class=\"os-divider\">. <\/span>Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n","protected":false},"author":158,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-218","chapter","type-chapter","status-publish","hentry"],"part":162,"_links":{"self":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/218","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/users\/158"}],"version-history":[{"count":16,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/218\/revisions"}],"predecessor-version":[{"id":1776,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/218\/revisions\/1776"}],"part":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/parts\/162"}],"metadata":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/218\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/media?parent=218"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=218"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/contributor?post=218"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/license?post=218"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}