{"id":181,"date":"2019-08-20T17:03:52","date_gmt":"2019-08-20T21:03:52","guid":{"rendered":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/solving-systems-with-inverses\/"},"modified":"2022-06-01T10:39:37","modified_gmt":"2022-06-01T14:39:37","slug":"solving-systems-with-inverses","status":"publish","type":"chapter","link":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/solving-systems-with-inverses\/","title":{"raw":"Solving Systems with Inverses","rendered":"Solving Systems with Inverses"},"content":{"raw":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\nIn this section, you will:\n<ul>\n \t<li>Find the inverse of a matrix.<\/li>\n \t<li>Solve a system of linear equations using an inverse matrix.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137640111\">Nancy 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 Nancy 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\n<div id=\"fs-id1165135547295\" class=\"bc-section section\">\n<h3>Finding the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165137422539\">We know that the multiplicative inverse of a real number[latex]\\,a\\,[\/latex]is[latex]\\,{a}^{-1},\\,[\/latex]and[latex]\\,a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1.\\,[\/latex]For example,[latex]\\,{2}^{-1}=\\frac{1}{2}\\,[\/latex] and[latex]\\,\\left(\\frac{1}{2}\\right)2=1.\\,[\/latex]The <span class=\"no-emphasis\">multiplicative inverse of a matrix<\/span> is similar in concept, except that the product of matrix[latex]A\\,[\/latex]and its inverse[latex]\\,{A}^{-1}\\,[\/latex]equals the <span class=\"no-emphasis\">identity matrix<\/span>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by[latex]\\,{I}_{n}\\,[\/latex]where[latex]\\,n\\,[\/latex]represents the dimension of the matrix. <a class=\"autogenerated-content\" href=\"#Equation_09_07_01\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Equation_09_07_02\">(Figure)<\/a> are the identity matrices for a[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]matrix and a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]matrix, respectively.<\/p>\n\n<div id=\"Equation_09_07_01\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"Equation_09_07_02\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\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\n<div id=\"fs-id1165137634184\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133354140\">A matrix that has a multiplicative inverse is called an <span class=\"no-emphasis\">invertible matrix<\/span>. Only a square matrix may have a multiplicative inverse, as the reversibility,[latex]\\,A{A}^{-1}={A}^{-1}A=I,\\,[\/latex]is a requirement. Not all square matrices have an inverse, but if[latex]\\,A\\,[\/latex]is invertible, then[latex]\\,{A}^{-1}\\,[\/latex]is unique. We will look at two methods for finding the inverse of a[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]matrix and a third method that can be used on both[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]and[latex]3\\text{}\u00d7\\text{}3\\,[\/latex]matrices.<\/p>\n\n<div id=\"fs-id1165137749300\" class=\"textbox key-takeaways\">\n<h3>The Identity Matrix and Multiplicative Inverse<\/h3>\n<p id=\"fs-id1165134108378\">The identity matrix,[latex]\\,{I}_{n},\\,[\/latex]is a square matrix containing ones down the main diagonal and zeros everywhere else.<\/p>\n\n<div id=\"fs-id1165135264832\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2}=\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}{I}_{3}=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\text{ }2\\,\u00d7\\,2\\text{ 3}\\,\u00d7\\,3\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133097142\">If[latex]\\,A\\,[\/latex]is an[latex]\\,n\\,\u00d7\\,n\\,[\/latex]\nmatrix and[latex]\\,B\\,[\/latex]is an[latex]\\,n\\,\u00d7\\,n\\,[\/latex]\nmatrix such that[latex]\\,AB=BA={I}_{n},\\,[\/latex]then[latex]\\,B={A}^{-1},\\,[\/latex]the multiplicative inverse of a matrix[latex]\\,A.[\/latex]<\/p>\n\n<\/div>\n<div id=\"Example_09_07_01\" class=\"textbox examples\">\n<div id=\"fs-id1165134033268\">\n<div id=\"fs-id1165135244114\">\n<h3>Showing That the Identity Matrix Acts as a 1<\/h3>\n<p id=\"fs-id1165135176569\">Given matrix <em>A<\/em>, show that[latex]\\,AI=IA=A.[\/latex]<\/p>\n\n<div id=\"fs-id1165135207264\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}\\,\\,\\,3&amp; 4\\\\ -2&amp; 5\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137643327\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137643327\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137643327\"]\n<p id=\"fs-id1165134134840\">Use matrix multiplication to show that the product of[latex]\\,A\\,[\/latex]and the identity is equal to the product of the identity and <em>A.<\/em><\/p>\n\n<div id=\"fs-id1165137725097\" class=\"unnumbered aligncenter\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right]\\,\\,\\begin{array}{r}\\hfill \\end{array}\\,\\,\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165137533136\" class=\"unnumbered aligncenter\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\,\\,\\begin{array}{r}\\hfill \\end{array}\\,\\,\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/div>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134372876\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165134073896\"><strong>Given two matrices, show that one is the multiplicative inverse of the other.\n<\/strong><\/p>\n\n<ol id=\"fs-id1165135341279\" type=\"1\">\n \t<li>Given matrix[latex]\\,A\\,[\/latex]of order[latex]\\,n\\,\u00d7\\,n\\,[\/latex]and matrix[latex]\\,B\\,[\/latex]of order[latex]\\,n\\,\u00d7\\,n\\,[\/latex]multiply[latex]\\,AB.[\/latex]<\/li>\n \t<li>If[latex]\\,AB=I,\\,[\/latex]then find the product[latex]\\,BA.\\,[\/latex]If[latex]\\,BA=I,\\,[\/latex]then[latex]\\,B={A}^{-1}\\,[\/latex]and[latex]\\,A={B}^{-1}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_09_07_02\" class=\"textbox examples\">\n<div id=\"fs-id1165135670391\">\n<div id=\"fs-id1165137457026\">\n<h3>Showing That Matrix <em>A<\/em> Is the Multiplicative Inverse of Matrix <em>B<\/em><\/h3>\n<p id=\"fs-id1165137897990\">Show that the given matrices are multiplicative inverses of each other.<\/p>\n\n<div id=\"fs-id1165137701124\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165133015823\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133015823\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133015823\"]\n<p id=\"fs-id1165137888915\">Multiply[latex]\\,AB\\,[\/latex]and[latex]\\,BA.\\,[\/latex]If both products equal the identity, then the two matrices are inverses of each other.<\/p>\n\n<div id=\"fs-id1165132957251\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\u00b7\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)&amp; \\hfill &amp; \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)&amp; \\hfill &amp; \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165137595924\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}BA=\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\u00b7\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)&amp; \\hfill &amp; \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)&amp; \\hfill &amp; \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135255987\">[latex]A\\,[\/latex]and[latex]B[\/latex]are inverses of each other.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827121\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_01\">\n<div id=\"fs-id1165137749877\">\n<p id=\"fs-id1165137749878\">Show that the following two matrices are inverses of each other.<\/p>\n\n<div id=\"fs-id1165135388478\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -1&amp; \\hfill &amp; \\hfill -3\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -3&amp; \\hfill &amp; \\hfill -4\\\\ \\hfill 1&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134122833\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134122833\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134122833\"]\n<div id=\"fs-id1165134122834\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -1&amp; \\hfill &amp; \\hfill -3\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill -3&amp; \\hfill &amp; \\hfill -4\\\\ \\hfill 1&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1\\left(-3\\right)+4\\left(1\\right)&amp; \\hfill &amp; \\hfill 1\\left(-4\\right)+4\\left(1\\right)\\\\ \\hfill -1\\left(-3\\right)+-3\\left(1\\right)&amp; \\hfill &amp; \\hfill -1\\left(-4\\right)+-3\\left(1\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\\\ BA=\\left[\\begin{array}{rrr}\\hfill -3&amp; \\hfill &amp; \\hfill -4\\\\ \\hfill 1&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -1&amp; \\hfill &amp; \\hfill -3\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -3\\left(1\\right)+-4\\left(-1\\right)&amp; \\hfill &amp; \\hfill -3\\left(4\\right)+-4\\left(-3\\right)\\\\ \\hfill 1\\left(1\\right)+1\\left(-1\\right)&amp; \\hfill &amp; \\hfill 1\\left(4\\right)+1\\left(-3\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse Using Matrix Multiplication<\/h4>\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 class=\"no-emphasis\">matrix multiplication<\/span>.<\/p>\n\n<div id=\"Example_09_07_03\" class=\"textbox examples\">\n<div id=\"fs-id1165135347580\">\n<div id=\"fs-id1165135347583\">\n<h3>Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\n<p id=\"fs-id1165133134743\">Use matrix multiplication to find the inverse of the given matrix.<\/p>\n\n<div id=\"fs-id1165133134746\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137642361\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137642361\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137642361\"]\n<p id=\"fs-id1165133067202\">For this method, we multiply[latex]\\,A\\,[\/latex]by a matrix containing unknown constants and set it equal to the identity.<\/p>\n\n<div id=\"fs-id1165134342559\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill a&amp; \\hfill b\\\\ \\hfill c&amp; \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137848625\">Find the product of the two matrices on the left side of the equal sign.<\/p>\n\n<div id=\"fs-id1165137848628\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill a&amp; \\hfill b\\\\ \\hfill c&amp; \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1a-2c&amp; \\hfill 1b-2d\\\\ \\hfill 2a-3c&amp; \\hfill 2b-3d\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135531330\">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\n<div class=\"unnumbered\">[latex]\\begin{array}{c}1a-2c=1\\,\\text{ }{R}_{1}\\\\ 2a-3c=0\\text{ }{R}_{2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137528118\">Using row operations, multiply and add as follows:[latex]\\,\\left(-2\\right){R}_{1}+{R}_{2}\\to {R}_{2}.\\,[\/latex]Add the equations, and solve for[latex]\\,c.[\/latex]<\/p>\n\n<div id=\"fs-id1165134212592\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 1a-2c=1\\,\\,\\,\\,\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135452117\">Back-substitute to solve for[latex]\\,a.[\/latex]<\/p>\n\n<div id=\"fs-id1165135468954\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill a-2\\left(-2\\right)=1\\,\\,\\,\\,\\\\ \\hfill a+4=1\\,\\,\\,\\,\\\\ \\hfill a=-3\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137850118\">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\n<div id=\"fs-id1165134261728\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{rr}\\hfill 1b-2d=0&amp; \\hfill {R}_{1}\\\\ \\hfill 2b-3d=1&amp; \\hfill {R}_{2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134544768\">Using row operations, multiply and add as follows:[latex]\\,\\left(-2\\right){R}_{1}+{R}_{2}={R}_{2}.\\,[\/latex]Add the two equations and solve for[latex]\\,d.[\/latex]<\/p>\n\n<div id=\"fs-id1165135363338\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 1b-2d=0\\\\ \\hfill \\frac{0+1d=1}{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,d=1}\\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165132961982\">Once more, back-substitute and solve for[latex]\\,b.[\/latex]<\/p>\n\n<div id=\"fs-id1165135173163\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill b-2\\left(1\\right)=0\\\\ \\hfill b-2=0\\\\ \\hfill b=2\\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165135195466\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill -3&amp; \\hfill &amp; \\hfill 2\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737676\" class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse by Augmenting with the Identity<\/h4>\n<p id=\"fs-id1165132328759\">Another way to find the <span class=\"no-emphasis\">multiplicative inverse<\/span> is by augmenting with the identity. When matrix[latex]\\,A\\,[\/latex]is transformed into[latex]\\,I,\\,[\/latex]the augmented matrix[latex]\\,I\\,[\/latex]transforms into[latex]\\,{A}^{-1}.[\/latex]<\/p>\n<p id=\"fs-id1165137893432\">For example, given<\/p>\n\n<div id=\"fs-id1165134077301\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 2&amp; \\hfill &amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill &amp; \\hfill 3\\end{array}\\right][\/latex]<\/div>\naugment[latex]\\,A\\,[\/latex]with the identity\n<div id=\"fs-id1165137410710\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 2&amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill 3\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137694941\">Perform <span class=\"no-emphasis\">row operations<\/span> with the goal of turning[latex]\\,A\\,[\/latex]into the identity.<\/p>\n\n<ol id=\"fs-id1165137844094\" type=\"1\">\n \t<li>Switch row 1 and row 2.\n<div id=\"fs-id1165133300766\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 5&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 0&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>Multiply row 2 by[latex]\\,-2\\,[\/latex]and add to row 1.\n<div id=\"fs-id1165131916886\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 2&amp; \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>Multiply row 1 by[latex]\\,-2\\,[\/latex]and add to row 2.\n<div id=\"fs-id1165134055892\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2&amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill -2\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>Add row 2 to row 1.\n<div id=\"fs-id1165137593591\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3&amp; \\hfill -1\\\\ \\hfill 5&amp; \\hfill -2\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>Multiply row 2 by[latex]\\,-1.[\/latex]\n<div id=\"fs-id1165135533052\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3&amp; \\hfill -1\\\\ \\hfill -5&amp; \\hfill 2\\end{array}\\right][\/latex]<\/div><\/li>\n<\/ol>\nThe matrix we have found is[latex]\\,{A}^{-1}.[\/latex]\n<div id=\"fs-id1165134263962\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill -1\\\\ \\hfill -5&amp; \\hfill &amp; \\hfill 2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165132300248\" class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h4>\n<p id=\"fs-id1165135688796\">When we need to find the <span class=\"no-emphasis\">multiplicative inverse<\/span> of a[latex]\\,2\\,\u00d7\\,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[latex]\\,A\\,[\/latex]is a[latex]\\,2\u00d72\\,[\/latex]matrix, such as<\/p>\n\n<div id=\"fs-id1165135512376\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill a&amp; \\hfill &amp; \\hfill b\\\\ \\hfill c&amp; \\hfill &amp; \\hfill d\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137459546\">the multiplicative inverse of[latex]\\,A\\,[\/latex]is given by the formula<\/p>\n\n<div id=\"Equation_09_09_03\">[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{rrr}\\hfill d&amp; \\hfill &amp; \\hfill -b\\\\ \\hfill -c&amp; \\hfill &amp; \\hfill a\\end{array}\\right][\/latex]<\/div>\nwhere[latex]\\,ad-bc\\ne 0.\\,[\/latex]If[latex]\\,ad-bc=0,\\,[\/latex]then[latex]\\,A\\,[\/latex]has no inverse.\n<div id=\"Example_09_07_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137780841\">\n<div id=\"fs-id1165137780843\">\n<h3>Using the Formula to Find the Multiplicative Inverse of Matrix <em>A<\/em><\/h3>\n<p id=\"fs-id1165137767546\">Use the formula to find the multiplicative inverse of<\/p>\n\n<div id=\"fs-id1165137846471\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}1&amp; -2\\\\ 2&amp; -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137668153\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137668153\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137668153\"]\n<p id=\"fs-id1165137668155\">Using the formula, we have<\/p>\n\n<div id=\"fs-id1165133296219\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{\\left(1\\right)\\left(-3\\right)-\\left(-2\\right)\\left(2\\right)}\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\frac{1}{-3+4}\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<div id=\"fs-id1165135519186\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137664304\">We can check that our formula works by using one of the other methods to calculate the inverse. Let\u2019s augment[latex]A\\,[\/latex]with the identity.<\/p>\n\n<div id=\"fs-id1165137446409\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1&amp; -2\\\\ 2&amp; -3\\end{array}\\,\\,\\,|\\,\\,\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134365354\">Perform <span class=\"no-emphasis\">row operations<\/span> with the goal of turning[latex]\\,A\\,[\/latex]into the identity.<\/p>\n\n<ol id=\"fs-id1165134550520\" type=\"1\">\n \t<li>Multiply row 1 by[latex]\\,-2\\,[\/latex]and add to row 2.\n<div id=\"fs-id1165137531383\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1&amp; -2\\\\ 0&amp; 1\\end{array}\\,\\,\\,|\\,\\,\\begin{array}{cc}1&amp; 0\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>Multiply row 1 by 2 and add to row 1.\n<div id=\"fs-id1165137463527\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\,\\,\\,|\\,\\,\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/div><\/li>\n<\/ol>\n<p id=\"fs-id1165134190558\">So, we have verified our original solution.<\/p>\n\n<div id=\"fs-id1165137444252\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137914058\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_02\">\n<div>\n\nUse the formula to find the inverse of matrix[latex]\\,A.\\,[\/latex]Verify your answer by augmenting with the identity matrix.\n<div id=\"fs-id1165137836658\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}1&amp; -1\\\\ 2&amp; \\,\\,3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137644723\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137644723\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137644723\"]\n[latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}&amp; \\frac{1}{5}\\\\ -\\frac{2}{5}&amp; \\frac{1}{5}\\end{array}\\right][\/latex][\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_09_07_05\" class=\"textbox examples\">\n<div id=\"fs-id1165133388596\">\n<div id=\"fs-id1165137447551\">\n<h3>Finding the Inverse of the Matrix, If It Exists<\/h3>\n<p id=\"fs-id1165137431926\">Find the inverse, if it exists, of the given matrix.<\/p>\n\n<div id=\"fs-id1165135152090\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}3&amp; 6\\\\ 1&amp; 2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134534268\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134534268\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134534268\"]\n<p id=\"fs-id1165133253462\">We will use the method of augmenting with the identity.<\/p>\n\n<div id=\"fs-id1165135541960\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}3&amp; 6\\\\ 1&amp; 3\\end{array}\\,\\,\\,|\\,\\,\\,\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<ol id=\"fs-id1165134077339\" type=\"1\">\n \t<li>Switch row 1 and row 2.\n<div class=\"unnumbered\">[latex]\\left[\\begin{array}{cc}1&amp; 3\\\\ 3&amp; 6\\,\\text{\u200b}\\end{array}\\text{\u200b}\\,\\,\\text{\u200b}\\text{\u200b}|\\,\\,\\,\\begin{array}{cc}0&amp; 1\\\\ 1&amp; 0\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>Multiply row 1 by \u22123 and add it to row 2.\n<div id=\"fs-id1165137656945\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1&amp; 2\\\\ 0&amp; 0\\end{array}\\,\\,\\,|\\,\\,\\,\\begin{array}{cc}1&amp; 0\\\\ -3&amp; 1\\end{array}\\right][\/latex]<\/div><\/li>\n \t<li>There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.[\/hidden-answer]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137680361\" class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h4>\n<p id=\"fs-id1165134238766\">Unfortunately, we do not have a formula similar to the one for a[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]matrix to find the inverse of a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]matrix. Instead, we will augment the original matrix with the identity matrix and use <span class=\"no-emphasis\">row operations<\/span> to obtain the inverse.<\/p>\n<p id=\"fs-id1165134435528\">Given a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]\nmatrix<\/p>\n\n<div id=\"fs-id1165137882103\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165133394718\">augment[latex]\\,A\\,[\/latex]with the identity matrix<\/p>\n\n<div id=\"fs-id1165133045254\" class=\"unnumbered aligncenter\">[latex]A|I=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\text{ }|\\text{ }\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134357582\">To begin, we write the <span class=\"no-emphasis\">augmented matrix<\/span> with the identity on the right and[latex]\\,A\\,[\/latex]on the left. Performing elementary <span class=\"no-emphasis\">row operations<\/span> so that the <span class=\"no-emphasis\">identity matrix<\/span> appears on the left, we will obtain the <span class=\"no-emphasis\">inverse matrix<\/span> on the right. We will find the inverse of this matrix in the next example.<\/p>\n\n<div id=\"fs-id1165134258652\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135305759\"><strong>Given a[latex]\\,3\\,\u00d7\\,3\\,[\/latex]matrix, find the inverse<\/strong><\/p>\n\n<ol id=\"fs-id1165133015258\" type=\"1\">\n \t<li>Write the original matrix augmented with the identity matrix on the right.<\/li>\n \t<li>Use elementary row operations so that the identity appears on the left.<\/li>\n \t<li>What is obtained on the right is the inverse of the original matrix.<\/li>\n \t<li>Use matrix multiplication to show that[latex]\\,A{A}^{-1}=I\\,[\/latex]and[latex]\\,{A}^{-1}A=I.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_09_07_06\" class=\"textbox examples\">\n<div id=\"fs-id1165135307844\">\n<div id=\"fs-id1165135544452\">\n<h3>Finding the Inverse of a 3 \u00d7 3 Matrix<\/h3>\n<p id=\"fs-id1165135618118\">Given the[latex]\\,3\\,\u00d7\\,3\\,[\/latex]matrix[latex]\\,A,\\,[\/latex]find the inverse.<\/p>\n\n<div id=\"fs-id1165135628520\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134104893\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134104893\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134104893\"]\n<p id=\"fs-id1165133073873\">Augment[latex]\\,A\\,[\/latex]with the identity matrix, and then begin row operations until the identity matrix replaces[latex]\\,A.\\,[\/latex]The matrix on the right will be the inverse of[latex]\\,A.\\,[\/latex]<\/p>\n\n<div id=\"fs-id1165135481050\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\,\\,\\,|\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ \\,\\,0&amp; 0&amp; 1\\end{array}\\right]\\stackrel{\\text{Interchange }{R}_{2}\\,\\text{and }{R}_{1}}{\\to }\\left[\\begin{array}{ccc}3&amp; 3&amp; 1\\\\ 2&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\,\\,\\,|\\begin{array}{ccc}0&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 0\\\\ \\,\\,\\,0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135209090\" class=\"unnumbered aligncenter\">[latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 2&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135434697\" class=\"unnumbered aligncenter\">[latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 2&amp; 3&amp; 1\\\\ 0&amp; 1&amp; 0\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135404828\" class=\"unnumbered aligncenter\">[latex]{R}_{3}\\,\u2194 {R}_{2}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 2&amp; 3&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165137454071\" class=\"unnumbered aligncenter\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 3&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 3&amp; \\hfill -2&amp; \\hfill 0\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135485955\" class=\"unnumbered aligncenter\">[latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 6&amp; \\hfill -2&amp; \\hfill -3\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134485583\">Thus,<\/p>\n\n<div id=\"fs-id1165137871679\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=B=\\left[\\begin{array}{ccc}-1&amp; \\,1&amp; \\,0\\\\ -1&amp; \\,\\,0&amp; \\,\\,1\\\\ \\,\\,6&amp; -2&amp; -3\\end{array}\\,\\right][\/latex][\/hidden-answer]<\/div>\n<\/div>\n<div id=\"fs-id1165137748685\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165135369102\">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]A{A}^{-1}=I\\,[\/latex]and[latex]\\,{A}^{-1}A=I.[\/latex]<\/p>\n\n<div id=\"fs-id1165134409446\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 6&amp; \\hfill -2&amp; \\hfill -3\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}2\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)&amp; \\,\\,2\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)&amp; \\,\\,2\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 3\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)&amp; \\,\\,3\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)&amp; \\,\\,3\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 2\\left(-1\\right)+4\\left(-1\\right)+1\\left(6\\right)&amp; \\,\\,2\\left(1\\right)+4\\left(0\\right)+1\\left(-2\\right)&amp; \\,\\,2\\left(0\\right)+4\\left(1\\right)+1\\left(-3\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"unnumbered\">[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A=\\left[\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 6&amp; \\hfill -2&amp; \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill -1\\left(2\\right)+1\\left(3\\right)+0\\left(2\\right)&amp; \\hfill \\,\\,-1\\left(3\\right)+1\\left(3\\right)+0\\left(4\\right)&amp; \\hfill \\,\\,-1\\left(1\\right)+1\\left(1\\right)+0\\left(1\\right)\\\\ \\hfill -1\\left(2\\right)+0\\left(3\\right)+1\\left(2\\right)&amp; \\hfill \\,\\,-1\\left(3\\right)+0\\left(3\\right)+1\\left(4\\right)&amp; \\hfill \\,\\,-1\\left(1\\right)+0\\left(1\\right)+1\\left(1\\right)\\\\ \\hfill 6\\left(2\\right)+-2\\left(3\\right)+-3\\left(2\\right)&amp; \\hfill \\,\\,6\\left(3\\right)+-2\\left(3\\right)+-3\\left(4\\right)&amp; \\hfill \\,\\,6\\left(1\\right)+-2\\left(1\\right)+-3\\left(1\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137446254\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_03\">\n<div>\n\nFind the inverse of the[latex]\\,3\u00d73\\,[\/latex]matrix.\n<div id=\"fs-id1165137732890\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{ccc}\\,\\,2&amp; -17&amp; 11\\\\ -1&amp; \\,\\,\\,11&amp; -7\\\\ \\,\\,\\,0&amp; \\,\\,\\,\\,\\,3&amp; -2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135478197\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135478197\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135478197\"]\n<p id=\"fs-id1165135478198\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}1&amp; 1&amp; \\,\\,2\\\\ 2&amp; 4&amp; -3\\\\ 3&amp; 6&amp; -5\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137694180\" class=\"bc-section section\">\n<h3>Solving a System of Linear Equations Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165135394318\">Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices:[latex]\\,X\\,[\/latex]is the matrix representing the variables of the system, and[latex]\\,B\\,[\/latex]is the matrix representing the constants. Using <span class=\"no-emphasis\">matrix multiplication<\/span>, we may define a system of equations with the same number of equations as variables as<\/p>\n\n<div id=\"fs-id1165137436526\" class=\"unnumbered aligncenter\">[latex]AX=B[\/latex]<\/div>\n<p id=\"fs-id1165137466371\">To solve a system of linear equations using an <span class=\"no-emphasis\">inverse matrix<\/span>, let[latex]\\,A\\,[\/latex]be the <span class=\"no-emphasis\">coefficient matrix<\/span>, let[latex]\\,X\\,[\/latex]be the variable matrix, and let[latex]\\,B\\,[\/latex]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\n<div id=\"fs-id1165137393188\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137626840\">From this system, the coefficient matrix is<\/p>\n\n<div id=\"fs-id1165137425579\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137628730\">The variable matrix is<\/p>\n\n<div id=\"fs-id1165137696944\" class=\"unnumbered aligncenter\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135670279\">And the constant matrix is<\/p>\n\n<div id=\"fs-id1165135670282\" class=\"unnumbered aligncenter\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134043634\">Then[latex]\\,AX=B\\,[\/latex]looks like<\/p>\n\n<div id=\"fs-id1165137430713\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/div>\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[latex]\\,x,\\,[\/latex]we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of[latex]\\,a.\\,[\/latex]Thus,<\/p>\n\n<div id=\"fs-id1165137406902\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{c}\\text{ }ax=b\\\\ \\text{ }\\left(\\frac{1}{a}\\right)ax=\\left(\\frac{1}{a}\\right)b\\\\ \\left({a}^{-1}\\text{ }\\right)ax=\\left({a}^{-1}\\right)b\\\\ \\left[\\left({a}^{-1}\\right)a\\right]x=\\left({a}^{-1}\\right)b\\\\ \\text{ }1x=\\left({a}^{-1}\\right)b\\\\ \\text{ }x=\\left({a}^{-1}\\right)b\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137843896\">The only difference between a solving a linear equation and a <span class=\"no-emphasis\">system of equations<\/span> 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\\,\u00d7\\,2\\,[\/latex]system and then move on to a[latex]\\,3\\,\u00d7\\,3\\,[\/latex]system.<\/p>\n\n<div id=\"fs-id1165135433007\" class=\"textbox key-takeaways\">\n<h3>Solving a System of Equations Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165131797266\">Given a system of equations, write the coefficient matrix[latex]\\,A,\\,[\/latex]the variable matrix[latex]\\,X,\\,[\/latex]and the constant matrix[latex]\\,B.\\,[\/latex]Then<\/p>\n\n<div id=\"fs-id1165137874489\" class=\"unnumbered aligncenter\">[latex]AX=B[\/latex]<\/div>\n<p id=\"fs-id1165133359429\">Multiply both sides by the inverse of[latex]\\,A\\,[\/latex]to obtain the solution.<\/p>\n\n<div id=\"fs-id1165137805820\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135570400\" class=\"precalculus qa textbox shaded\">\n<p id=\"fs-id1165135149799\"><strong>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>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\n<\/div>\n<div id=\"Example_09_07_07\" class=\"textbox examples\">\n<div id=\"fs-id1165135335932\">\n<div id=\"fs-id1165135335935\">\n<h3>Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165134339834\">Solve the given system of equations using the inverse of a matrix.<\/p>\n\n<div id=\"fs-id1165134339837\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165133361981\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133361981\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133361981\"]\n<p id=\"fs-id1165133361983\">Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.<\/p>\n\n<div id=\"fs-id1165135504917\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right],X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right],B=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135359848\">Then<\/p>\n\n<div class=\"unnumbered\">[latex]\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137759831\">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\n<div id=\"fs-id1165133313245\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d&amp; -b\\\\ -c&amp; a\\end{array}\\right]\\hfill \\\\ \\text{ }=\\frac{1}{3\\left(11\\right)-8\\left(4\\right)}\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; 3\\end{array}\\right]\\hfill \\\\ \\text{ }=\\frac{1}{1}\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; 3\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134256613\">So,<\/p>\n\n<div id=\"fs-id1165137810833\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; \\text{\u200b}\\text{\u200b}\\,\\,3\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137638597\">Now we are ready to solve. Multiply both sides of the equation by[latex]\\,{A}^{-1}.[\/latex]<\/p>\n\n<div id=\"fs-id1165134035963\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\hfill \\\\ \\left[\\begin{array}{rr}\\hfill 11&amp; \\hfill -8\\\\ \\hfill -4&amp; \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 11&amp; \\hfill -8\\\\ \\hfill -4&amp; \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 11\\left(5\\right)+\\left(-8\\right)7\\\\ \\hfill -4\\left(5\\right)+3\\left(7\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill -1\\\\ \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137939473\">The solution is[latex]\\,\\left(-1,1\\right).[\/latex][\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134248791\" class=\"precalculus qa textbox shaded\">\n<p id=\"fs-id1165135661400\"><strong>Can we solve for[latex]\\,X\\,[\/latex]by finding the product[latex]\\,B{A}^{-1}?[\/latex]\n<\/strong><\/p>\n<p id=\"fs-id1165135701407\"><em>No, recall that matrix multiplication is not commutative, so[latex]\\,{A}^{-1}B\\ne B{A}^{-1}.\\,[\/latex]Consider our steps for solving the matrix equation.<\/em><\/p>\n\n<div id=\"fs-id1165134389823\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137770334\"><em>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[latex]\\,A\\,[\/latex]on the left side and to the left of[latex]\\,B\\,[\/latex]on the right side. Because matrix multiplication is not commutative, order matters.<\/em><\/p>\n\n<\/div>\n<div id=\"Example_09_07_08\" class=\"textbox examples\">\n<div id=\"fs-id1165133354131\">\n<div id=\"fs-id1165135169167\">\n<h3>Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165135409796\">Solve the following system using the inverse of a matrix.<\/p>\n\n<div id=\"fs-id1165135189932\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 5x+15y+56z=35\\,\\,\\,\\,\\\\ \\hfill -4x-11y-41z=-26\\\\ \\hfill -x-3y-11z=-7\\,\\,\\,\\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137889814\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137889814\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137889814\"]\n<p id=\"fs-id1165137889816\">Write the equation[latex]\\,AX=B.\\,[\/latex]<\/p>\n\n<div id=\"fs-id1165137473978\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}5&amp; 15&amp; 56\\\\ -4&amp; -11&amp; -41\\\\ -1&amp; -3&amp; -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135496349\">First, we will find the inverse of[latex]\\,A\\,[\/latex]by augmenting with the identity.<\/p>\n\n<div id=\"fs-id1165134547371\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rrr}\\hfill 5&amp; \\hfill 15&amp; \\hfill 56\\\\ \\hfill -4&amp; \\hfill -11&amp; \\hfill -41\\\\ \\hfill -1&amp; \\hfill -3&amp; \\hfill -11\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165132324521\">Multiply row 1 by[latex]\\,\\frac{1}{5}.[\/latex]<\/p>\n\n<div id=\"fs-id1165137870885\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ -4&amp; -11&amp; -41\\\\ -1&amp; -3&amp; -11\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135525873\">Multiply row 1 by 4 and add to row 2.<\/p>\n\n<div id=\"fs-id1165135525876\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ -1&amp; -3&amp; -11\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165133309272\">Add row 1 to row 3.<\/p>\n\n<div id=\"fs-id1165133309275\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; \\frac{1}{5}\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ \\frac{1}{5}&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165133141472\">Multiply row 2 by \u22123 and add to row 1.<\/p>\n\n<div id=\"fs-id1165133141476\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; -\\frac{1}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; \\frac{1}{5}\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-\\frac{11}{5}&amp; -3&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ \\frac{1}{5}&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134183817\">Multiply row 3 by 5.<\/p>\n\n<div id=\"fs-id1165134183820\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; -\\frac{1}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-\\frac{11}{5}&amp; -3&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135238352\">Multiply row 3 by[latex]\\,\\frac{1}{5}\\,[\/latex]and add to row 1.<\/p>\n\n<div id=\"fs-id1165134042101\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134584807\">Multiply row 3 by[latex]\\,-\\frac{19}{5}\\,[\/latex]and add to row 2.<\/p>\n\n<div id=\"fs-id1165135622432\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ -3&amp; 1&amp; -19\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135388626\">So,<\/p>\n\n<div id=\"fs-id1165135388629\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ -3&amp; 1&amp; -19\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/div>\nMultiply both sides of the equation by[latex]\\,{A}^{-1}.\\,[\/latex]We want[latex]\\,{A}^{-1}AX={A}^{-1}B:[\/latex]\n<div id=\"fs-id1165134043976\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rrr}\\hfill -2&amp; \\hfill -3&amp; \\hfill 1\\\\ \\hfill -3&amp; \\hfill 1&amp; \\hfill -19\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill 5&amp; \\hfill 15&amp; \\hfill 56\\\\ \\hfill -4&amp; \\hfill -11&amp; \\hfill -41\\\\ \\hfill -1&amp; \\hfill -3&amp; \\hfill -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -2&amp; \\hfill -3&amp; \\hfill 1\\\\ \\hfill -3&amp; \\hfill 1&amp; \\hfill -19\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135319442\">Thus,<\/p>\n\n<div id=\"fs-id1165135401791\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}B=\\left[\\begin{array}{r}\\hfill -70+78-7\\\\ \\hfill -105-26+133\\\\ \\hfill 35+0-35\\end{array}\\right]=\\left[\\begin{array}{c}1\\\\ 2\\\\ 0\\end{array}\\right][\/latex]<\/div>\nThe solution is[latex]\\,\\left(1,2,0\\right).[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134323584\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_04\">\n<div id=\"fs-id1165133155808\">\n<p id=\"fs-id1165133155809\">Solve the system using the inverse of the coefficient matrix.<\/p>\n\n<div id=\"fs-id1165133155812\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\text{ }2x-17y+11z=0\\hfill \\\\ \\text{ }-x+11y-7z=8\\hfill \\\\ \\text{ }3y-2z=-2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137446595\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137446595\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137446595\"]\n<p id=\"fs-id1165137446596\">[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135407360\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135503565\"><strong>Given a system of equations, solve with matrix inverses using a calculator.\n<\/strong><\/p>\n\n<ol id=\"fs-id1165135503570\" type=\"1\">\n \t<li>Save the coefficient matrix and the constant matrix as matrix variables[latex]\\,\\left[A\\right]\\,[\/latex]and[latex]\\,\\left[B\\right].[\/latex]<\/li>\n \t<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\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>\n<\/ol>\n<\/div>\n<div id=\"Example_09_07_09\" class=\"textbox examples\">\n<div id=\"fs-id1165131997848\">\n<div id=\"fs-id1165137611786\">\n<h3>Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\n<p id=\"fs-id1165137611791\">Solve the system of equations with matrix inverses using a calculator<\/p>\n\n<div id=\"fs-id1165137811747\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}2x+3y+z=32\\hfill \\\\ 3x+3y+z=-27\\hfill \\\\ 2x+4y+z=-2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134394498\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134394498\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134394498\"]\n<p id=\"fs-id1165134394500\">On the matrix page of the calculator, enter the <span class=\"no-emphasis\">coefficient matrix<\/span> as the matrix variable[latex]\\,\\left[A\\right],\\,[\/latex]and enter the constant matrix as the matrix variable[latex]\\,\\left[B\\right].[\/latex]<\/p>\n\n<div id=\"fs-id1165134117288\" class=\"unnumbered aligncenter\">[latex]\\left[A\\right]=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right],\\text{\u2003}\\left[B\\right]=\\left[\\begin{array}{c}32\\\\ -27\\\\ -2\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134089389\">On the home screen of the calculator, type in the multiplication to solve for[latex]\\,X,\\,[\/latex]calling up each matrix variable as needed.<\/p>\n\n<div class=\"unnumbered\">[latex]{\\left[A\\right]}^{-1}\u00d7\\left[B\\right][\/latex]<\/div>\n<p id=\"fs-id1165135481265\">Evaluate the expression.<\/p>\n\n<div id=\"fs-id1165137863372\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131884558\" class=\"precalculus media\">\n<p id=\"fs-id1165137836914\">Access these online resources for additional instruction and practice with solving systems with inverses.<\/p>\n\n<ul id=\"fs-id1165137836918\">\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/identmatrix\">The Identity Matrix<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/inversematrix\">Determining Inverse Matrices<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/matrixsystem\">Using a Matrix Equation to Solve a System of Equations<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135500881\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"eip-id1165137848559\" summary=\"..\"><colgroup> <col> <col><\/colgroup>\n<tbody>\n<tr valign=\"middle\">\n<td>Identity matrix for a[latex]2\\text{}\u00d7\\text{}2[\/latex]matrix<\/td>\n<td>[latex]{I}_{2}=\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Identity matrix for a[latex]\\text{3}\\text{}\u00d7\\text{}3[\/latex]matrix<\/td>\n<td>[latex]{I}_{3}=\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Multiplicative inverse of a[latex]2\\text{}\u00d7\\text{}2[\/latex]matrix<\/td>\n<td>[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d&amp; -b\\\\ -c&amp; a\\end{array}\\right],\\text{ where }ad-bc\\ne 0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165134225617\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165134225620\">\n \t<li>An identity matrix has the property[latex]\\,AI=IA=A.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_09_07_01\">(Figure)<\/a>.<\/li>\n \t<li>An invertible matrix has the property[latex]\\,A{A}^{-1}={A}^{-1}A=I.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_09_07_02\">(Figure)<\/a>.<\/li>\n \t<li>Use matrix multiplication and the identity to find the inverse of a[latex]\\,2\u00d72\\,[\/latex]matrix. See <a class=\"autogenerated-content\" href=\"#Example_09_07_03\">(Figure)<\/a>.<\/li>\n \t<li>The multiplicative inverse can be found using a formula. See <a class=\"autogenerated-content\" href=\"#Example_09_07_04\">(Figure)<\/a>.<\/li>\n \t<li>Another method of finding the inverse is by augmenting with the identity. See <a class=\"autogenerated-content\" href=\"#Example_09_07_05\">(Figure)<\/a>.<\/li>\n \t<li>We can augment a[latex]\\,3\u00d73\\,[\/latex]matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See <a class=\"autogenerated-content\" href=\"#Example_09_07_06\">(Figure)<\/a>.<\/li>\n \t<li>Write the system of equations as[latex]\\,AX=B,\\,[\/latex]and multiply both sides by the inverse of[latex]\\,A:{A}^{-1}AX={A}^{-1}B.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_09_07_07\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_09_07_08\">(Figure)<\/a>.<\/li>\n \t<li>We can also use a calculator to solve a system of equations with matrix inverses. See <a class=\"autogenerated-content\" href=\"#Example_09_07_09\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165135446624\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id1165137548736\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165137548741\">\n<div id=\"fs-id1165137548742\">\n<p id=\"fs-id1165134540056\">In a previous section, we showed that matrix multiplication is not commutative, that is,[latex]\\,AB\\ne BA\\,[\/latex]in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is,[latex]\\,{A}^{-1}A=A{A}^{-1}?[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134494231\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134494231\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134494231\"]If[latex]\\,{A}^{-1}\\,[\/latex]is the inverse of[latex]\\,A,\\,[\/latex]then[latex]\\,A{A}^{-1}=I,\\,[\/latex]the identity matrix. Since[latex]\\,A\\,[\/latex]is also the inverse of[latex]\\,{A}^{-1},{A}^{-1}A=I.\\,[\/latex]You can also check by proving this for a[latex]\\,2\u00d72\\,[\/latex]matrix.\n\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134493452\">\n<div id=\"fs-id1165134493453\">\n<p id=\"fs-id1165134493454\">Does every[latex]\\,2\u00d72\\,[\/latex]matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135547596\">\n<div id=\"fs-id1165135547597\">\n<p id=\"fs-id1165135547598\">Can you explain whether a[latex]\\,2\u00d72\\,[\/latex]matrix with an entire row of zeros can have an inverse?<\/p>\n\n<\/div>\n<div id=\"fs-id1165133436150\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133436150\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133436150\"]\n<p id=\"fs-id1165133436152\">No, because[latex]\\,ad\\,[\/latex]and[latex]\\,bc\\,[\/latex]are both 0, so[latex]\\,ad-bc=0,\\,[\/latex]which requires us to divide by 0 in the formula.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133401600\">\n<div>\n<p id=\"fs-id1165135702542\">Can a matrix with an entire column of zeros have an inverse? Explain why or why not.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135702546\">\n<div id=\"fs-id1165135702547\">\n<p id=\"fs-id1165135702548\">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\u00d72\\,[\/latex]matrix.<\/p>\n\n<\/div>\n<div id=\"fs-id1165134138529\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134138529\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134138529\"]\n<p id=\"fs-id1165134138531\">Yes. Consider the matrix[latex]\\,\\left[\\begin{array}{cc}0&amp; 1\\\\ 1&amp; 0\\end{array}\\right].\\,[\/latex]The inverse is found with the following calculation:[latex]\\,{A}^{-1}=\\frac{1}{0\\left(0\\right)-1\\left(1\\right)}\\left[\\begin{array}{cc}0&amp; -1\\\\ -1&amp; 0\\end{array}\\right]=\\left[\\begin{array}{cc}0&amp; 1\\\\ 1&amp; 0\\end{array}\\right].[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133075017\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\nIn the following exercises, show that matrix[latex]\\,A\\,[\/latex]is the inverse of matrix[latex]\\,B.[\/latex]\n<div id=\"fs-id1165137804955\">\n<div id=\"fs-id1165137804956\">\n<p id=\"fs-id1165137804958\">[latex]A=\\left[\\begin{array}{cc}1&amp; 0\\\\ -1&amp; 1\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}1&amp; 0\\\\ 1&amp; 1\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137629353\">\n<div id=\"fs-id1165137629354\">\n<p id=\"fs-id1165135611304\">[latex]A=\\left[\\begin{array}{cc}1&amp; 2\\\\ 3&amp; 4\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}-2&amp; 1\\\\ \\frac{3}{2}&amp; -\\frac{1}{2}\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165132949834\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165132949834\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165132949834\"]\n<p id=\"fs-id1165132949836\">[latex]AB=BA=\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right]=I[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div>\n<p id=\"fs-id1165137832787\">[latex]A=\\left[\\begin{array}{cc}4&amp; 5\\\\ 7&amp; 0\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}0&amp; \\frac{1}{7}\\\\ \\frac{1}{5}&amp; -\\frac{4}{35}\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137417445\">\n<div id=\"fs-id1165137417446\">\n<p id=\"fs-id1165137417447\">[latex]A=\\left[\\begin{array}{cc}-2&amp; \\frac{1}{2}\\\\ 3&amp; -1\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}-2&amp; -1\\\\ -6&amp; -4\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165135186218\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135186218\"]\n<p id=\"fs-id1165135186218\">[latex]AB=BA=\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right]=I[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137810179\">\n<div id=\"fs-id1165137810180\">\n<p id=\"fs-id1165137810181\">[latex]A=\\left[\\begin{array}{ccc}1&amp; 0&amp; 1\\\\ 0&amp; 1&amp; -1\\\\ 0&amp; 1&amp; 1\\end{array}\\right],\\,B=\\frac{1}{2}\\left[\\begin{array}{ccc}2&amp; 1&amp; -1\\\\ 0&amp; 1&amp; 1\\\\ 0&amp; -1&amp; 1\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div>\n<div>\n<p id=\"fs-id1165135481916\">[latex]A=\\left[\\begin{array}{ccc}1&amp; 2&amp; 3\\\\ 4&amp; 0&amp; 2\\\\ 1&amp; 6&amp; 9\\end{array}\\right],\\,B=\\frac{1}{4}\\left[\\begin{array}{ccc}6&amp; 0&amp; -2\\\\ 17&amp; -3&amp; -5\\\\ -12&amp; 2&amp; 4\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135513422\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135513422\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135513422\"]\n<p id=\"fs-id1165135513424\">[latex]AB=BA=\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right]=I[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165134371187\">\n<p id=\"fs-id1165134371188\">[latex]A=\\left[\\begin{array}{ccc}3&amp; 8&amp; 2\\\\ 1&amp; 1&amp; 1\\\\ 5&amp; 6&amp; 12\\end{array}\\right],\\,B=\\frac{1}{36}\\left[\\begin{array}{ccc}-6&amp; 84&amp; -6\\\\ 7&amp; -26&amp; 1\\\\ -1&amp; -22&amp; 5\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\nFor the following exercises, find the multiplicative inverse of each matrix, if it exists.\n<div id=\"fs-id1165134282180\">\n<div id=\"fs-id1165134282181\">\n<p id=\"fs-id1165134282182\">[latex]\\left[\\begin{array}{cc}3&amp; -2\\\\ 1&amp; 9\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165135363094\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135363094\"]\n<p id=\"fs-id1165135363094\">[latex]\\frac{1}{29}\\left[\\begin{array}{cc}9&amp; 2\\\\ -1&amp; 3\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div>[latex]\\left[\\begin{array}{cc}-2&amp; 2\\\\ 3&amp; 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137387518\">\n<div id=\"fs-id1165137387519\">\n<p id=\"fs-id1165137387520\">[latex]\\left[\\begin{array}{cc}-3&amp; 7\\\\ 9&amp; 2\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135192366\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135192366\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135192366\"]\n<p id=\"fs-id1165134493412\">[latex]\\frac{1}{69}\\left[\\begin{array}{cc}-2&amp; 7\\\\ 9&amp; 3\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134339996\">\n<div id=\"fs-id1165134339997\">\n<p id=\"fs-id1165134339998\">[latex]\\left[\\begin{array}{cc}-4&amp; -3\\\\ -5&amp; 8\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133103275\">\n<div id=\"fs-id1165133103276\">[latex]\\left[\\begin{array}{cc}1&amp; 1\\\\ 2&amp; 2\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165132957181\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165132957181\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165132957181\"]There is no inverse\n\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165132957189\">\n<p id=\"fs-id1165132957190\">[latex]\\left[\\begin{array}{cc}0&amp; 1\\\\ 1&amp; 0\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134185464\">\n<div id=\"fs-id1165134185465\">\n<p id=\"fs-id1165134185466\">[latex]\\left[\\begin{array}{cc}0.5&amp; 1.5\\\\ 1&amp; -0.5\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134237283\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134237283\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134237283\"]\n<p id=\"fs-id1165134237285\">[latex]\\frac{4}{7}\\left[\\begin{array}{cc}0.5&amp; 1.5\\\\ 1&amp; -0.5\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134152537\">\n<div id=\"fs-id1165134152538\">\n<p id=\"fs-id1165134135289\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; 6\\\\ -2&amp; 1&amp; 7\\\\ 3&amp; 0&amp; 2\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133141439\">\n<div id=\"fs-id1165135679418\">\n<p id=\"fs-id1165135679419\">[latex]\\left[\\begin{array}{ccc}0&amp; 1&amp; -3\\\\ 4&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135684091\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135684091\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135684091\"]\n<p id=\"fs-id1165135684094\">[latex]\\frac{1}{17}\\left[\\begin{array}{ccc}-5&amp; 5&amp; -3\\\\ 20&amp; -3&amp; 12\\\\ 1&amp; -1&amp; 4\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134040470\">\n<div id=\"fs-id1165134040471\">\n<p id=\"fs-id1165134040472\">[latex]\\left[\\begin{array}{ccc}1&amp; 2&amp; -1\\\\ -3&amp; 4&amp; 1\\\\ -2&amp; -4&amp; -5\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134159676\">\n<div id=\"fs-id1165134159677\">[latex]\\left[\\begin{array}{ccc}1&amp; 9&amp; -3\\\\ 2&amp; 5&amp; 6\\\\ 4&amp; -2&amp; 7\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165134086077\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134086077\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134086077\"]\n[latex]\\frac{1}{209}\\left[\\begin{array}{ccc}47&amp; -57&amp; 69\\\\ 10&amp; 19&amp; -12\\\\ -24&amp; 38&amp; -13\\end{array}\\right][\/latex][\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133210779\">\n<div id=\"fs-id1165133210780\">\n<p id=\"fs-id1165133210782\">[latex]\\left[\\begin{array}{ccc}1&amp; -2&amp; 3\\\\ -4&amp; 8&amp; -12\\\\ 1&amp; 4&amp; 2\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135397109\">\n<div id=\"fs-id1165135397110\">\n<p id=\"fs-id1165135397111\">[latex]\\left[\\begin{array}{ccc}\\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{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134389849\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134389849\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134389849\"]\n[latex]\\left[\\begin{array}{ccc}18&amp; 60&amp; -168\\\\ -56&amp; -140&amp; 448\\\\ 40&amp; 80&amp; -280\\end{array}\\right][\/latex][\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165131818536\">\n<div id=\"fs-id1165131818537\">\n<p id=\"fs-id1165131818538\">[latex]\\left[\\begin{array}{ccc}1&amp; 2&amp; 3\\\\ 4&amp; 5&amp; 6\\\\ 7&amp; 8&amp; 9\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165137675157\">For the following exercises, solve the system using the inverse of a[latex]\\,2\\,\u00d7\\,2\\,[\/latex]matrix.<\/p>\n\n<div id=\"fs-id1165134319666\">\n<div id=\"fs-id1165134070732\">\n<p id=\"fs-id1165134070733\">[latex]\\begin{array}{l}\\text{ }5x-6y=-61\\hfill \\\\ 4x+3y=-2\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165133176692\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133176692\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133176692\"]\n<p id=\"fs-id1165133176694\">[latex]\\left(-5,6\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133012293\">\n<div id=\"fs-id1165133012294\">\n<p id=\"fs-id1165133409845\">[latex]\\begin{array}{l}8x+4y=-100\\\\ 3x-4y=1\\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165131884588\">\n<div id=\"fs-id1165131884589\">\n<p id=\"fs-id1165135169381\">[latex]\\begin{array}{l}\\,3x-2y=6\\hfill \\\\ -x+5y=-2\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135649520\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135649520\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135649520\"]\n<p id=\"fs-id1165132938289\">[latex]\\left(2,0\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165133045243\">\n<p id=\"fs-id1165133045244\">[latex]\\begin{array}{l}5x-4y=-5\\hfill \\\\ \\,\\,\\,4x+y=2.3\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135205038\">\n<div id=\"fs-id1165135205039\">\n<p id=\"fs-id1165135205040\">[latex]\\begin{array}{l}-3x-4y=9\\hfill \\\\ \\,12x+4y=-6\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135358856\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135358856\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135358856\"]\n<p id=\"fs-id1165135358859\">[latex]\\left(\\frac{1}{3},-\\frac{5}{2}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div>[latex]\\begin{array}{l}-2x+3y=\\frac{3}{10}\\hfill \\\\ \\,\\,-x+5y=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div>\n<div>[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\frac{8}{5}x-\\frac{4}{5}y=\\frac{2}{5}\\hfill \\\\ -\\frac{8}{5}x+\\frac{1}{5}y=\\frac{7}{10}\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165135328736\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135328736\"]\n<p id=\"fs-id1165135328736\">[latex]\\left(-\\frac{2}{3},-\\frac{11}{6}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133157406\">\n<div id=\"fs-id1165135523285\">\n<p id=\"fs-id1165135523286\">[latex]\\begin{array}{l}\\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\n<\/div>\n<\/div>\n<p id=\"fs-id1165134386500\">For the following exercises, solve a system using the inverse of a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]\nmatrix.<\/p>\n\n<div id=\"fs-id1165132079352\">\n<div id=\"fs-id1165132079353\">\n<p id=\"fs-id1165132079354\">[latex]\\begin{array}{l}3x-2y+5z=21\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5x+4y=37\\hfill \\\\ \\,\\,\\,x-2y-5z=5\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137612251\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137612251\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137612251\"]\n<p id=\"fs-id1165137612253\">[latex]\\left(7,\\frac{1}{2},\\frac{1}{5}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165134177545\">[latex]\\begin{array}{l}\\text{ }4x+4y+4z=40\\hfill \\\\ \\text{ }2x-3y+4z=-12\\hfill \\\\ \\text{ }-x+3y+4z=9\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134226706\">\n<div id=\"fs-id1165134226707\">\n<p id=\"fs-id1165134226708\">[latex]\\begin{array}{l}\\text{ }6x-5y-z=31\\hfill \\\\ \\text{ }-x+2y+z=-6\\hfill \\\\ \\text{ }3x+3y+2z=13\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134129992\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134129992\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134129992\"]\n<p id=\"fs-id1165134129994\">[latex]\\left(5,0,-1\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165134224076\">\n<p id=\"fs-id1165134583960\">[latex]\\begin{array}{l}6x-5y+2z=-4\\hfill \\\\ \\,\\,2x+5y-z=12\\hfill \\\\ \\,\\,2x+5y+z=12\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134185366\">\n<div id=\"fs-id1165134185367\">\n<p id=\"fs-id1165134185368\">[latex]\\begin{array}{l}4x-2y+3z=-12\\hfill \\\\ 2x+2y-9z=33\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6y-4z=1\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137654966\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137654966\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137654966\"]\n<p id=\"fs-id1165137654968\">[latex]\\frac{1}{34}\\left(-35,-97,-154\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135207327\">\n<div id=\"fs-id1165135207328\">\n<p id=\"fs-id1165134419566\">[latex]\\begin{array}{l}\\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\n<\/div>\n<\/div>\n<div id=\"fs-id1165135421457\">\n<div>\n<p id=\"fs-id1165134342520\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\frac{1}{2}x-\\frac{1}{5}y+\\frac{1}{5}z=\\frac{31}{100}\\hfill \\\\ -\\frac{3}{4}x-\\frac{1}{4}y+\\frac{1}{2}z=\\frac{7}{40}\\hfill \\\\ -\\frac{4}{5}x-\\frac{1}{2}y+\\frac{3}{2}z=\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134203408\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134203408\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134203408\"]\n<p id=\"fs-id1165134203410\">[latex]\\frac{1}{690}\\left(65,-1136,-229\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165132944845\">\n<div id=\"fs-id1165132944846\">\n<p id=\"fs-id1165132944848\">[latex]\\begin{array}{l}0.1x+0.2y+0.3z=-1.4\\hfill \\\\ 0.1x-0.2y+0.3z=0.6\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0.4y+0.9z=-2\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137900020\" class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1165134356163\">For the following exercises, use a calculator to solve the system of equations with matrix inverses.<\/p>\n\n<div id=\"fs-id1165134356166\">\n<div id=\"fs-id1165134356167\">\n<p id=\"fs-id1165134356168\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,2x-y=-3\\hfill \\\\ -x+2y=2.3\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137938316\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137938316\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137938316\"]\n<p id=\"fs-id1165137938319\">[latex]\\left(-\\frac{37}{30},\\frac{8}{15}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135186095\">\n<div id=\"fs-id1165135186096\">\n<p id=\"fs-id1165135186097\">[latex]\\begin{array}{l}-\\frac{1}{2}x-\\frac{3}{2}y=-\\frac{43}{20}\\hfill \\\\ \\,\\,\\frac{5}{2}x+\\frac{11}{5}y=\\frac{31}{4}\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135652325\">\n<div id=\"fs-id1165135652326\">\n<p id=\"fs-id1165135652327\">[latex]\\begin{array}{l}12.3x-2y-2.5z=2\\hfill \\\\ 36.9x+7y-7.5z=-7\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8y-5z=-10\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135407474\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135407474\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135407474\"]\n<p id=\"fs-id1165135369163\">[latex]\\left(\\frac{10}{123},-1,\\frac{2}{5}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134467729\">\n<div id=\"fs-id1165134467730\">\n<p id=\"fs-id1165134467731\">[latex]\\begin{array}{l}0.5x-3y+6z=-0.8\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0.7x-2y=-0.06\\hfill \\\\ 0.5x+4y+5z=0\\hfill \\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134537745\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<p id=\"fs-id1165135339476\">For the following exercises, find the inverse of the given matrix.<\/p>\n\n<div id=\"fs-id1165135339479\">\n<div id=\"fs-id1165135339480\">\n<p id=\"fs-id1165135339481\">[latex]\\left[\\begin{array}{cccc}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{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div>\n<div class=\"textbox shaded\">[reveal-answer q=\"313213\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"313213\"][latex]\\frac{1}{2}\\left[\\begin{array}{rrrr}\\hfill 2&amp; \\hfill 1&amp; \\hfill -1&amp; \\hfill -1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill -1\\\\ \\hfill 0&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -1&amp; \\hfill 1\\end{array}\\right][\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<div>\n<div>\n<p id=\"fs-id1165135258899\">[latex]\\left[\\begin{array}{rrrr}\\hfill -1&amp; \\hfill 0&amp; \\hfill 2&amp; \\hfill 5\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 2\\\\ \\hfill 0&amp; \\hfill 2&amp; \\hfill -1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill -3&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191959\">\n<div id=\"fs-id1165135191960\">\n<p id=\"fs-id1165135191962\">[latex]\\left[\\begin{array}{rrrr}\\hfill 1&amp; \\hfill -2&amp; \\hfill 3&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 2\\\\ \\hfill 1&amp; \\hfill 4&amp; \\hfill -2&amp; \\hfill 3\\\\ \\hfill -5&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165133409785\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133409785\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133409785\"]\n[latex]\\frac{1}{39}\\left[\\begin{array}{rrrr}\\hfill 3&amp; \\hfill 2&amp; \\hfill 1&amp; \\hfill -7\\\\ \\hfill 18&amp; \\hfill -53&amp; \\hfill 32&amp; \\hfill 10\\\\ \\hfill 24&amp; \\hfill -36&amp; \\hfill 21&amp; \\hfill 9\\\\ \\hfill -9&amp; \\hfill 46&amp; \\hfill -16&amp; \\hfill -5\\end{array}\\right][\/latex][\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134234210\">\n<div id=\"fs-id1165134234211\">\n<p id=\"fs-id1165134234212\">[latex]\\left[\\begin{array}{rrrrr}\\hfill 1&amp; \\hfill 2&amp; \\hfill 0&amp; \\hfill 2&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 2&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 3&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 2&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 2&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134385542\">\n<div id=\"fs-id1165134385543\">\n<p id=\"fs-id1165134385544\">[latex]\\left[\\begin{array}{rrrrrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134248766\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134248766\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134248766\"]\n<p id=\"fs-id1165134248768\">[latex]\\left[\\begin{array}{rrrrrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill -1&amp; \\hfill -1&amp; \\hfill -1&amp; \\hfill -1&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133094400\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\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\n<div id=\"fs-id1165133260340\">\n<div id=\"fs-id1165133260342\">\n<p id=\"fs-id1165133260343\">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\n<\/div>\n<\/div>\n<div id=\"fs-id1165133260348\">\n<div id=\"fs-id1165133260349\">\n<p id=\"fs-id1165133260350\">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\n<\/div>\n<div id=\"fs-id1165133111141\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133111141\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133111141\"]\n<p id=\"fs-id1165133111143\">Infinite solutions.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133111147\">\n<div id=\"fs-id1165133111148\">\n<p id=\"fs-id1165133111149\">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\n<\/div>\n<\/div>\n<div id=\"fs-id1165134081432\">\n<div id=\"fs-id1165134081433\">\n<p id=\"fs-id1165134081434\">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\n<\/div>\n<div id=\"fs-id1165134081439\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134081439\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134081439\"]\n<p id=\"fs-id1165134081441\">50% oranges, 25% bananas, 20% apples<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135705936\">\n<div id=\"fs-id1165135705937\">\n<p id=\"fs-id1165135705938\">A sorority 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\n<\/div>\n<\/div>\n<div id=\"fs-id1165135705944\">\n<div id=\"fs-id1165135705945\">\n<p id=\"fs-id1165135705946\">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\n<\/div>\n<div id=\"fs-id1165133353997\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133353997\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133353997\"]\n<p id=\"fs-id1165133353999\">10 straw hats, 50 beanies, 40 cowboy hats<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133354003\">\n<div id=\"fs-id1165133354004\">\n<p id=\"fs-id1165133354006\">Anna, Ashley, and Andrea weigh a combined 370 lb. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137501545\">\n<div id=\"fs-id1165137501546\">\n<p id=\"fs-id1165137501547\">Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137501553\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137501553\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137501553\"]\n<p id=\"fs-id1165135547272\">Tom ate 6, Joe ate 3, and Albert ate 3.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165135547278\">\n<p id=\"fs-id1165135547279\">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{\\text{ ft}}^{2}.\\,[\/latex]He used[latex]\\,{\\text{3 ft}}^{2}\\,[\/latex]more chicken wire than plywood. How much of each material in did the farmer use?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134531903\">\n<div id=\"fs-id1165134531904\">\n<p id=\"fs-id1165134531906\">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\n<\/div>\n<div id=\"fs-id1165133045226\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133045226\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133045226\"]\n<p id=\"fs-id1165133045228\">124 oranges, 10 lemons, 8 pomegranates<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl>\n \t<dt>identity matrix<\/dt>\n \t<dd id=\"fs-id1165134179622\">a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134179626\">\n \t<dt>multiplicative inverse of a matrix<\/dt>\n \t<dd id=\"fs-id1165135528440\">a matrix that, when multiplied by the original, equals the identity matrix<\/dd>\n<\/dl>\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Find the inverse of a matrix.<\/li>\n<li>Solve a system of linear equations using an inverse matrix.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137640111\">Nancy 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 Nancy 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<div id=\"fs-id1165135547295\" class=\"bc-section section\">\n<h3>Finding the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165137422539\">We know that the multiplicative inverse of a real number[latex]\\,a\\,[\/latex]is[latex]\\,{a}^{-1},\\,[\/latex]and[latex]\\,a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1.\\,[\/latex]For example,[latex]\\,{2}^{-1}=\\frac{1}{2}\\,[\/latex] and[latex]\\,\\left(\\frac{1}{2}\\right)2=1.\\,[\/latex]The <span class=\"no-emphasis\">multiplicative inverse of a matrix<\/span> is similar in concept, except that the product of matrix[latex]A\\,[\/latex]and its inverse[latex]\\,{A}^{-1}\\,[\/latex]equals the <span class=\"no-emphasis\">identity matrix<\/span>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by[latex]\\,{I}_{n}\\,[\/latex]where[latex]\\,n\\,[\/latex]represents the dimension of the matrix. <a class=\"autogenerated-content\" href=\"#Equation_09_07_01\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Equation_09_07_02\">(Figure)<\/a> are the identity matrices for a[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]matrix and a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]matrix, respectively.<\/p>\n<div id=\"Equation_09_07_01\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"Equation_09_07_02\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 0& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\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<div id=\"fs-id1165137634184\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133354140\">A matrix that has a multiplicative inverse is called an <span class=\"no-emphasis\">invertible matrix<\/span>. Only a square matrix may have a multiplicative inverse, as the reversibility,[latex]\\,A{A}^{-1}={A}^{-1}A=I,\\,[\/latex]is a requirement. Not all square matrices have an inverse, but if[latex]\\,A\\,[\/latex]is invertible, then[latex]\\,{A}^{-1}\\,[\/latex]is unique. We will look at two methods for finding the inverse of a[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]matrix and a third method that can be used on both[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]and[latex]3\\text{}\u00d7\\text{}3\\,[\/latex]matrices.<\/p>\n<div id=\"fs-id1165137749300\" class=\"textbox key-takeaways\">\n<h3>The Identity Matrix and Multiplicative Inverse<\/h3>\n<p id=\"fs-id1165134108378\">The identity matrix,[latex]\\,{I}_{n},\\,[\/latex]is a square matrix containing ones down the main diagonal and zeros everywhere else.<\/p>\n<div id=\"fs-id1165135264832\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2}=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right]\\begin{array}{cccc}& & & \\end{array}{I}_{3}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\text{ }2\\,\u00d7\\,2\\text{ 3}\\,\u00d7\\,3\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133097142\">If[latex]\\,A\\,[\/latex]is an[latex]\\,n\\,\u00d7\\,n\\,[\/latex]<br \/>\nmatrix and[latex]\\,B\\,[\/latex]is an[latex]\\,n\\,\u00d7\\,n\\,[\/latex]<br \/>\nmatrix such that[latex]\\,AB=BA={I}_{n},\\,[\/latex]then[latex]\\,B={A}^{-1},\\,[\/latex]the multiplicative inverse of a matrix[latex]\\,A.[\/latex]<\/p>\n<\/div>\n<div id=\"Example_09_07_01\" class=\"textbox examples\">\n<div id=\"fs-id1165134033268\">\n<div id=\"fs-id1165135244114\">\n<h3>Showing That the Identity Matrix Acts as a 1<\/h3>\n<p id=\"fs-id1165135176569\">Given matrix <em>A<\/em>, show that[latex]\\,AI=IA=A.[\/latex]<\/p>\n<div id=\"fs-id1165135207264\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}\\,\\,\\,3& 4\\\\ -2& 5\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137643327\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134134840\">Use matrix multiplication to show that the product of[latex]\\,A\\,[\/latex]and the identity is equal to the product of the identity and <em>A.<\/em><\/p>\n<div id=\"fs-id1165137725097\" class=\"unnumbered aligncenter\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]\\,\\,\\begin{array}{r}\\hfill \\end{array}\\,\\,\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0& \\hfill & \\hfill & \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0& \\hfill & \\hfill & \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165137533136\" class=\"unnumbered aligncenter\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\,\\,\\begin{array}{r}\\hfill \\end{array}\\,\\,\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/div>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134372876\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\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[latex]\\,A\\,[\/latex]of order[latex]\\,n\\,\u00d7\\,n\\,[\/latex]and matrix[latex]\\,B\\,[\/latex]of order[latex]\\,n\\,\u00d7\\,n\\,[\/latex]multiply[latex]\\,AB.[\/latex]<\/li>\n<li>If[latex]\\,AB=I,\\,[\/latex]then find the product[latex]\\,BA.\\,[\/latex]If[latex]\\,BA=I,\\,[\/latex]then[latex]\\,B={A}^{-1}\\,[\/latex]and[latex]\\,A={B}^{-1}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_09_07_02\" class=\"textbox examples\">\n<div id=\"fs-id1165135670391\">\n<div id=\"fs-id1165137457026\">\n<h3>Showing That Matrix <em>A<\/em> Is the Multiplicative Inverse of Matrix <em>B<\/em><\/h3>\n<p id=\"fs-id1165137897990\">Show that the given matrices are multiplicative inverses of each other.<\/p>\n<div id=\"fs-id1165137701124\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165133015823\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137888915\">Multiply[latex]\\,AB\\,[\/latex]and[latex]\\,BA.\\,[\/latex]If both products equal the identity, then the two matrices are inverses of each other.<\/p>\n<div id=\"fs-id1165132957251\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\u00b7\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)& \\hfill & \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)& \\hfill & \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165137595924\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}BA=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\u00b7\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)& \\hfill & \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)& \\hfill & \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135255987\">[latex]A\\,[\/latex]and[latex]B[\/latex]are inverses of each other.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827121\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_01\">\n<div id=\"fs-id1165137749877\">\n<p id=\"fs-id1165137749878\">Show that the following two matrices are inverses of each other.<\/p>\n<div id=\"fs-id1165135388478\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134122833\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<div id=\"fs-id1165134122834\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1\\left(-3\\right)+4\\left(1\\right)& \\hfill & \\hfill 1\\left(-4\\right)+4\\left(1\\right)\\\\ \\hfill -1\\left(-3\\right)+-3\\left(1\\right)& \\hfill & \\hfill -1\\left(-4\\right)+-3\\left(1\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ BA=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -3\\left(1\\right)+-4\\left(-1\\right)& \\hfill & \\hfill -3\\left(4\\right)+-4\\left(-3\\right)\\\\ \\hfill 1\\left(1\\right)+1\\left(-1\\right)& \\hfill & \\hfill 1\\left(4\\right)+1\\left(-3\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse Using Matrix Multiplication<\/h4>\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 class=\"no-emphasis\">matrix multiplication<\/span>.<\/p>\n<div id=\"Example_09_07_03\" class=\"textbox examples\">\n<div id=\"fs-id1165135347580\">\n<div id=\"fs-id1165135347583\">\n<h3>Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\n<p id=\"fs-id1165133134743\">Use matrix multiplication to find the inverse of the given matrix.<\/p>\n<div id=\"fs-id1165133134746\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill -2\\\\ \\hfill 2& \\hfill & \\hfill -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137642361\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133067202\">For this method, we multiply[latex]\\,A\\,[\/latex]by a matrix containing unknown constants and set it equal to the identity.<\/p>\n<div id=\"fs-id1165134342559\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill a& \\hfill b\\\\ \\hfill c& \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137848625\">Find the product of the two matrices on the left side of the equal sign.<\/p>\n<div id=\"fs-id1165137848628\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill a& \\hfill b\\\\ \\hfill c& \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1a-2c& \\hfill 1b-2d\\\\ \\hfill 2a-3c& \\hfill 2b-3d\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135531330\">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<div class=\"unnumbered\">[latex]\\begin{array}{c}1a-2c=1\\,\\text{ }{R}_{1}\\\\ 2a-3c=0\\text{ }{R}_{2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137528118\">Using row operations, multiply and add as follows:[latex]\\,\\left(-2\\right){R}_{1}+{R}_{2}\\to {R}_{2}.\\,[\/latex]Add the equations, and solve for[latex]\\,c.[\/latex]<\/p>\n<div id=\"fs-id1165134212592\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 1a-2c=1\\,\\,\\,\\,\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135452117\">Back-substitute to solve for[latex]\\,a.[\/latex]<\/p>\n<div id=\"fs-id1165135468954\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill a-2\\left(-2\\right)=1\\,\\,\\,\\,\\\\ \\hfill a+4=1\\,\\,\\,\\,\\\\ \\hfill a=-3\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137850118\">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<div id=\"fs-id1165134261728\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{rr}\\hfill 1b-2d=0& \\hfill {R}_{1}\\\\ \\hfill 2b-3d=1& \\hfill {R}_{2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134544768\">Using row operations, multiply and add as follows:[latex]\\,\\left(-2\\right){R}_{1}+{R}_{2}={R}_{2}.\\,[\/latex]Add the two equations and solve for[latex]\\,d.[\/latex]<\/p>\n<div id=\"fs-id1165135363338\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 1b-2d=0\\\\ \\hfill \\frac{0+1d=1}{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,d=1}\\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165132961982\">Once more, back-substitute and solve for[latex]\\,b.[\/latex]<\/p>\n<div id=\"fs-id1165135173163\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill b-2\\left(1\\right)=0\\\\ \\hfill b-2=0\\\\ \\hfill b=2\\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165135195466\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill 2\\\\ \\hfill -2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737676\" class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse by Augmenting with the Identity<\/h4>\n<p id=\"fs-id1165132328759\">Another way to find the <span class=\"no-emphasis\">multiplicative inverse<\/span> is by augmenting with the identity. When matrix[latex]\\,A\\,[\/latex]is transformed into[latex]\\,I,\\,[\/latex]the augmented matrix[latex]\\,I\\,[\/latex]transforms into[latex]\\,{A}^{-1}.[\/latex]<\/p>\n<p id=\"fs-id1165137893432\">For example, given<\/p>\n<div id=\"fs-id1165134077301\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 2& \\hfill & \\hfill 1\\\\ \\hfill 5& \\hfill & \\hfill 3\\end{array}\\right][\/latex]<\/div>\n<p>augment[latex]\\,A\\,[\/latex]with the identity<\/p>\n<div id=\"fs-id1165137410710\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 2& \\hfill 1\\\\ \\hfill 5& \\hfill 3\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137694941\">Perform <span class=\"no-emphasis\">row operations<\/span> with the goal of turning[latex]\\,A\\,[\/latex]into the identity.<\/p>\n<ol id=\"fs-id1165137844094\" type=\"1\">\n<li>Switch row 1 and row 2.\n<div id=\"fs-id1165133300766\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 5& \\hfill 3\\\\ \\hfill 2& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>Multiply row 2 by[latex]\\,-2\\,[\/latex]and add to row 1.\n<div id=\"fs-id1165131916886\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 2& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2& \\hfill 1\\\\ \\hfill 1& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>Multiply row 1 by[latex]\\,-2\\,[\/latex]and add to row 2.\n<div id=\"fs-id1165134055892\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2& \\hfill 1\\\\ \\hfill 5& \\hfill -2\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>Add row 2 to row 1.\n<div id=\"fs-id1165137593591\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3& \\hfill -1\\\\ \\hfill 5& \\hfill -2\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>Multiply row 2 by[latex]\\,-1.[\/latex]\n<div id=\"fs-id1165135533052\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3& \\hfill -1\\\\ \\hfill -5& \\hfill 2\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>The matrix we have found is[latex]\\,{A}^{-1}.[\/latex]<\/p>\n<div id=\"fs-id1165134263962\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill -1\\\\ \\hfill -5& \\hfill & \\hfill 2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165132300248\" class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h4>\n<p id=\"fs-id1165135688796\">When we need to find the <span class=\"no-emphasis\">multiplicative inverse<\/span> of a[latex]\\,2\\,\u00d7\\,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[latex]\\,A\\,[\/latex]is a[latex]\\,2\u00d72\\,[\/latex]matrix, such as<\/p>\n<div id=\"fs-id1165135512376\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{rrr}\\hfill a& \\hfill & \\hfill b\\\\ \\hfill c& \\hfill & \\hfill d\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137459546\">the multiplicative inverse of[latex]\\,A\\,[\/latex]is given by the formula<\/p>\n<div id=\"Equation_09_09_03\">[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{rrr}\\hfill d& \\hfill & \\hfill -b\\\\ \\hfill -c& \\hfill & \\hfill a\\end{array}\\right][\/latex]<\/div>\n<p>where[latex]\\,ad-bc\\ne 0.\\,[\/latex]If[latex]\\,ad-bc=0,\\,[\/latex]then[latex]\\,A\\,[\/latex]has no inverse.<\/p>\n<div id=\"Example_09_07_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137780841\">\n<div id=\"fs-id1165137780843\">\n<h3>Using the Formula to Find the Multiplicative Inverse of Matrix <em>A<\/em><\/h3>\n<p id=\"fs-id1165137767546\">Use the formula to find the multiplicative inverse of<\/p>\n<div id=\"fs-id1165137846471\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137668153\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137668155\">Using the formula, we have<\/p>\n<div id=\"fs-id1165133296219\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{\\left(1\\right)\\left(-3\\right)-\\left(-2\\right)\\left(2\\right)}\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\frac{1}{-3+4}\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135519186\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137664304\">We can check that our formula works by using one of the other methods to calculate the inverse. Let\u2019s augment[latex]A\\,[\/latex]with the identity.<\/p>\n<div id=\"fs-id1165137446409\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}\\,\\,\\,|\\,\\,\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134365354\">Perform <span class=\"no-emphasis\">row operations<\/span> with the goal of turning[latex]\\,A\\,[\/latex]into the identity.<\/p>\n<ol id=\"fs-id1165134550520\" type=\"1\">\n<li>Multiply row 1 by[latex]\\,-2\\,[\/latex]and add to row 2.\n<div id=\"fs-id1165137531383\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1& -2\\\\ 0& 1\\end{array}\\,\\,\\,|\\,\\,\\begin{array}{cc}1& 0\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>Multiply row 1 by 2 and add to row 1.\n<div id=\"fs-id1165137463527\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\,\\,\\,|\\,\\,\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-id1165134190558\">So, we have verified our original solution.<\/p>\n<div id=\"fs-id1165137444252\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137914058\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_02\">\n<div>\n<p>Use the formula to find the inverse of matrix[latex]\\,A.\\,[\/latex]Verify your answer by augmenting with the identity matrix.<\/p>\n<div id=\"fs-id1165137836658\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}1& -1\\\\ 2& \\,\\,3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137644723\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}& \\frac{1}{5}\\\\ -\\frac{2}{5}& \\frac{1}{5}\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_09_07_05\" class=\"textbox examples\">\n<div id=\"fs-id1165133388596\">\n<div id=\"fs-id1165137447551\">\n<h3>Finding the Inverse of the Matrix, If It Exists<\/h3>\n<p id=\"fs-id1165137431926\">Find the inverse, if it exists, of the given matrix.<\/p>\n<div id=\"fs-id1165135152090\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}3& 6\\\\ 1& 2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134534268\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133253462\">We will use the method of augmenting with the identity.<\/p>\n<div id=\"fs-id1165135541960\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}3& 6\\\\ 1& 3\\end{array}\\,\\,\\,|\\,\\,\\,\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\n<ol id=\"fs-id1165134077339\" type=\"1\">\n<li>Switch row 1 and row 2.\n<div class=\"unnumbered\">[latex]\\left[\\begin{array}{cc}1& 3\\\\ 3& 6\\,\\text{\u200b}\\end{array}\\text{\u200b}\\,\\,\\text{\u200b}\\text{\u200b}|\\,\\,\\,\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>Multiply row 1 by \u22123 and add it to row 2.\n<div id=\"fs-id1165137656945\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}1& 2\\\\ 0& 0\\end{array}\\,\\,\\,|\\,\\,\\,\\begin{array}{cc}1& 0\\\\ -3& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<li>There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/details>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137680361\" class=\"bc-section section\">\n<h4>Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h4>\n<p id=\"fs-id1165134238766\">Unfortunately, we do not have a formula similar to the one for a[latex]\\,2\\text{}\u00d7\\text{}2\\,[\/latex]matrix to find the inverse of a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]matrix. Instead, we will augment the original matrix with the identity matrix and use <span class=\"no-emphasis\">row operations<\/span> to obtain the inverse.<\/p>\n<p id=\"fs-id1165134435528\">Given a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]<br \/>\nmatrix<\/p>\n<div id=\"fs-id1165137882103\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165133394718\">augment[latex]\\,A\\,[\/latex]with the identity matrix<\/p>\n<div id=\"fs-id1165133045254\" class=\"unnumbered aligncenter\">[latex]A|I=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\text{ }|\\text{ }\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134357582\">To begin, we write the <span class=\"no-emphasis\">augmented matrix<\/span> with the identity on the right and[latex]\\,A\\,[\/latex]on the left. Performing elementary <span class=\"no-emphasis\">row operations<\/span> so that the <span class=\"no-emphasis\">identity matrix<\/span> appears on the left, we will obtain the <span class=\"no-emphasis\">inverse matrix<\/span> on the right. We will find the inverse of this matrix in the next example.<\/p>\n<div id=\"fs-id1165134258652\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135305759\"><strong>Given a[latex]\\,3\\,\u00d7\\,3\\,[\/latex]matrix, find the inverse<\/strong><\/p>\n<ol id=\"fs-id1165133015258\" type=\"1\">\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]\\,A{A}^{-1}=I\\,[\/latex]and[latex]\\,{A}^{-1}A=I.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_09_07_06\" class=\"textbox examples\">\n<div id=\"fs-id1165135307844\">\n<div id=\"fs-id1165135544452\">\n<h3>Finding the Inverse of a 3 \u00d7 3 Matrix<\/h3>\n<p id=\"fs-id1165135618118\">Given the[latex]\\,3\\,\u00d7\\,3\\,[\/latex]matrix[latex]\\,A,\\,[\/latex]find the inverse.<\/p>\n<div id=\"fs-id1165135628520\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134104893\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133073873\">Augment[latex]\\,A\\,[\/latex]with the identity matrix, and then begin row operations until the identity matrix replaces[latex]\\,A.\\,[\/latex]The matrix on the right will be the inverse of[latex]\\,A.\\,[\/latex]<\/p>\n<div id=\"fs-id1165135481050\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\,\\,\\,|\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ \\,\\,0& 0& 1\\end{array}\\right]\\stackrel{\\text{Interchange }{R}_{2}\\,\\text{and }{R}_{1}}{\\to }\\left[\\begin{array}{ccc}3& 3& 1\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}\\,\\,\\,|\\begin{array}{ccc}0& 1& 0\\\\ 1& 0& 0\\\\ \\,\\,\\,0& 0& 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135209090\" class=\"unnumbered aligncenter\">[latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135434697\" class=\"unnumbered aligncenter\">[latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 0& 1& 0\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135404828\" class=\"unnumbered aligncenter\">[latex]{R}_{3}\\,\u2194 {R}_{2}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 2& 3& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165137454071\" class=\"unnumbered aligncenter\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 3& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 3& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165135485955\" class=\"unnumbered aligncenter\">[latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134485583\">Thus,<\/p>\n<div id=\"fs-id1165137871679\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=B=\\left[\\begin{array}{ccc}-1& \\,1& \\,0\\\\ -1& \\,\\,0& \\,\\,1\\\\ \\,\\,6& -2& -3\\end{array}\\,\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137748685\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165135369102\">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]A{A}^{-1}=I\\,[\/latex]and[latex]\\,{A}^{-1}A=I.[\/latex]<\/p>\n<div id=\"fs-id1165134409446\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}2\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& \\,\\,2\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& \\,\\,2\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 3\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& \\,\\,3\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& \\,\\,3\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 2\\left(-1\\right)+4\\left(-1\\right)+1\\left(6\\right)& \\,\\,2\\left(1\\right)+4\\left(0\\right)+1\\left(-2\\right)& \\,\\,2\\left(0\\right)+4\\left(1\\right)+1\\left(-3\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"unnumbered\">[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill -1\\left(2\\right)+1\\left(3\\right)+0\\left(2\\right)& \\hfill \\,\\,-1\\left(3\\right)+1\\left(3\\right)+0\\left(4\\right)& \\hfill \\,\\,-1\\left(1\\right)+1\\left(1\\right)+0\\left(1\\right)\\\\ \\hfill -1\\left(2\\right)+0\\left(3\\right)+1\\left(2\\right)& \\hfill \\,\\,-1\\left(3\\right)+0\\left(3\\right)+1\\left(4\\right)& \\hfill \\,\\,-1\\left(1\\right)+0\\left(1\\right)+1\\left(1\\right)\\\\ \\hfill 6\\left(2\\right)+-2\\left(3\\right)+-3\\left(2\\right)& \\hfill \\,\\,6\\left(3\\right)+-2\\left(3\\right)+-3\\left(4\\right)& \\hfill \\,\\,6\\left(1\\right)+-2\\left(1\\right)+-3\\left(1\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137446254\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_03\">\n<div>\n<p>Find the inverse of the[latex]\\,3\u00d73\\,[\/latex]matrix.<\/p>\n<div id=\"fs-id1165137732890\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{ccc}\\,\\,2& -17& 11\\\\ -1& \\,\\,\\,11& -7\\\\ \\,\\,\\,0& \\,\\,\\,\\,\\,3& -2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135478197\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135478198\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}1& 1& \\,\\,2\\\\ 2& 4& -3\\\\ 3& 6& -5\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137694180\" class=\"bc-section section\">\n<h3>Solving a System of Linear Equations Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165135394318\">Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices:[latex]\\,X\\,[\/latex]is the matrix representing the variables of the system, and[latex]\\,B\\,[\/latex]is the matrix representing the constants. Using <span class=\"no-emphasis\">matrix multiplication<\/span>, we may define a system of equations with the same number of equations as variables as<\/p>\n<div id=\"fs-id1165137436526\" class=\"unnumbered aligncenter\">[latex]AX=B[\/latex]<\/div>\n<p id=\"fs-id1165137466371\">To solve a system of linear equations using an <span class=\"no-emphasis\">inverse matrix<\/span>, let[latex]\\,A\\,[\/latex]be the <span class=\"no-emphasis\">coefficient matrix<\/span>, let[latex]\\,X\\,[\/latex]be the variable matrix, and let[latex]\\,B\\,[\/latex]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<div id=\"fs-id1165137393188\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137626840\">From this system, the coefficient matrix is<\/p>\n<div id=\"fs-id1165137425579\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137628730\">The variable matrix is<\/p>\n<div id=\"fs-id1165137696944\" class=\"unnumbered aligncenter\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135670279\">And the constant matrix is<\/p>\n<div id=\"fs-id1165135670282\" class=\"unnumbered aligncenter\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134043634\">Then[latex]\\,AX=B\\,[\/latex]looks like<\/p>\n<div id=\"fs-id1165137430713\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/div>\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[latex]\\,x,\\,[\/latex]we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of[latex]\\,a.\\,[\/latex]Thus,<\/p>\n<div id=\"fs-id1165137406902\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{c}\\text{ }ax=b\\\\ \\text{ }\\left(\\frac{1}{a}\\right)ax=\\left(\\frac{1}{a}\\right)b\\\\ \\left({a}^{-1}\\text{ }\\right)ax=\\left({a}^{-1}\\right)b\\\\ \\left[\\left({a}^{-1}\\right)a\\right]x=\\left({a}^{-1}\\right)b\\\\ \\text{ }1x=\\left({a}^{-1}\\right)b\\\\ \\text{ }x=\\left({a}^{-1}\\right)b\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137843896\">The only difference between a solving a linear equation and a <span class=\"no-emphasis\">system of equations<\/span> 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\\,\u00d7\\,2\\,[\/latex]system and then move on to a[latex]\\,3\\,\u00d7\\,3\\,[\/latex]system.<\/p>\n<div id=\"fs-id1165135433007\" class=\"textbox key-takeaways\">\n<h3>Solving a System of Equations Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165131797266\">Given a system of equations, write the coefficient matrix[latex]\\,A,\\,[\/latex]the variable matrix[latex]\\,X,\\,[\/latex]and the constant matrix[latex]\\,B.\\,[\/latex]Then<\/p>\n<div id=\"fs-id1165137874489\" class=\"unnumbered aligncenter\">[latex]AX=B[\/latex]<\/div>\n<p id=\"fs-id1165133359429\">Multiply both sides by the inverse of[latex]\\,A\\,[\/latex]to obtain the solution.<\/p>\n<div id=\"fs-id1165137805820\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135570400\" class=\"precalculus qa textbox shaded\">\n<p id=\"fs-id1165135149799\"><strong>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>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 id=\"Example_09_07_07\" class=\"textbox examples\">\n<div id=\"fs-id1165135335932\">\n<div id=\"fs-id1165135335935\">\n<h3>Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165134339834\">Solve the given system of equations using the inverse of a matrix.<\/p>\n<div id=\"fs-id1165134339837\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165133361981\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133361983\">Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.<\/p>\n<div id=\"fs-id1165135504917\" class=\"unnumbered aligncenter\">[latex]A=\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right],X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right],B=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135359848\">Then<\/p>\n<div class=\"unnumbered\">[latex]\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137759831\">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<div id=\"fs-id1165133313245\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d& -b\\\\ -c& a\\end{array}\\right]\\hfill \\\\ \\text{ }=\\frac{1}{3\\left(11\\right)-8\\left(4\\right)}\\left[\\begin{array}{cc}11& -8\\\\ -4& 3\\end{array}\\right]\\hfill \\\\ \\text{ }=\\frac{1}{1}\\left[\\begin{array}{cc}11& -8\\\\ -4& 3\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134256613\">So,<\/p>\n<div id=\"fs-id1165137810833\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}11& -8\\\\ -4& \\text{\u200b}\\text{\u200b}\\,\\,3\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165137638597\">Now we are ready to solve. Multiply both sides of the equation by[latex]\\,{A}^{-1}.[\/latex]<\/p>\n<div id=\"fs-id1165134035963\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\hfill \\\\ \\left[\\begin{array}{rr}\\hfill 11& \\hfill -8\\\\ \\hfill -4& \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 11& \\hfill -8\\\\ \\hfill -4& \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 11\\left(5\\right)+\\left(-8\\right)7\\\\ \\hfill -4\\left(5\\right)+3\\left(7\\right)\\end{array}\\right]\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill -1\\\\ \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137939473\">The solution is[latex]\\,\\left(-1,1\\right).[\/latex]<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134248791\" class=\"precalculus qa textbox shaded\">\n<p id=\"fs-id1165135661400\"><strong>Can we solve for[latex]\\,X\\,[\/latex]by finding the product[latex]\\,B{A}^{-1}?[\/latex]<br \/>\n<\/strong><\/p>\n<p id=\"fs-id1165135701407\"><em>No, recall that matrix multiplication is not commutative, so[latex]\\,{A}^{-1}B\\ne B{A}^{-1}.\\,[\/latex]Consider our steps for solving the matrix equation.<\/em><\/p>\n<div id=\"fs-id1165134389823\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137770334\"><em>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[latex]\\,A\\,[\/latex]on the left side and to the left of[latex]\\,B\\,[\/latex]on the right side. Because matrix multiplication is not commutative, order matters.<\/em><\/p>\n<\/div>\n<div id=\"Example_09_07_08\" class=\"textbox examples\">\n<div id=\"fs-id1165133354131\">\n<div id=\"fs-id1165135169167\">\n<h3>Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\n<p id=\"fs-id1165135409796\">Solve the following system using the inverse of a matrix.<\/p>\n<div id=\"fs-id1165135189932\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{r}\\hfill 5x+15y+56z=35\\,\\,\\,\\,\\\\ \\hfill -4x-11y-41z=-26\\\\ \\hfill -x-3y-11z=-7\\,\\,\\,\\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137889814\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137889816\">Write the equation[latex]\\,AX=B.\\,[\/latex]<\/p>\n<div id=\"fs-id1165137473978\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}5& 15& 56\\\\ -4& -11& -41\\\\ -1& -3& -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135496349\">First, we will find the inverse of[latex]\\,A\\,[\/latex]by augmenting with the identity.<\/p>\n<div id=\"fs-id1165134547371\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rrr}\\hfill 5& \\hfill 15& \\hfill 56\\\\ \\hfill -4& \\hfill -11& \\hfill -41\\\\ \\hfill -1& \\hfill -3& \\hfill -11\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165132324521\">Multiply row 1 by[latex]\\,\\frac{1}{5}.[\/latex]<\/p>\n<div id=\"fs-id1165137870885\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 3& \\frac{56}{5}\\\\ -4& -11& -41\\\\ -1& -3& -11\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135525873\">Multiply row 1 by 4 and add to row 2.<\/p>\n<div id=\"fs-id1165135525876\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 3& \\frac{56}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ -1& -3& -11\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ \\frac{4}{5}& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165133309272\">Add row 1 to row 3.<\/p>\n<div id=\"fs-id1165133309275\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 3& \\frac{56}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& \\frac{1}{5}\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ \\frac{4}{5}& 1& 0\\\\ \\frac{1}{5}& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165133141472\">Multiply row 2 by \u22123 and add to row 1.<\/p>\n<div id=\"fs-id1165133141476\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 0& -\\frac{1}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& \\frac{1}{5}\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-\\frac{11}{5}& -3& 0\\\\ \\frac{4}{5}& 1& 0\\\\ \\frac{1}{5}& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134183817\">Multiply row 3 by 5.<\/p>\n<div id=\"fs-id1165134183820\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 0& -\\frac{1}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-\\frac{11}{5}& -3& 0\\\\ \\frac{4}{5}& 1& 0\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135238352\">Multiply row 3 by[latex]\\,\\frac{1}{5}\\,[\/latex]and add to row 1.<\/p>\n<div id=\"fs-id1165134042101\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-2& -3& 1\\\\ \\frac{4}{5}& 1& 0\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134584807\">Multiply row 3 by[latex]\\,-\\frac{19}{5}\\,[\/latex]and add to row 2.<\/p>\n<div id=\"fs-id1165135622432\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\,\\,|\\,\\,\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135388626\">So,<\/p>\n<div id=\"fs-id1165135388629\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/div>\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<div id=\"fs-id1165134043976\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{rrr}\\hfill -2& \\hfill -3& \\hfill 1\\\\ \\hfill -3& \\hfill 1& \\hfill -19\\\\ \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill 5& \\hfill 15& \\hfill 56\\\\ \\hfill -4& \\hfill -11& \\hfill -41\\\\ \\hfill -1& \\hfill -3& \\hfill -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -2& \\hfill -3& \\hfill 1\\\\ \\hfill -3& \\hfill 1& \\hfill -19\\\\ \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165135319442\">Thus,<\/p>\n<div id=\"fs-id1165135401791\" class=\"unnumbered aligncenter\">[latex]{A}^{-1}B=\\left[\\begin{array}{r}\\hfill -70+78-7\\\\ \\hfill -105-26+133\\\\ \\hfill 35+0-35\\end{array}\\right]=\\left[\\begin{array}{c}1\\\\ 2\\\\ 0\\end{array}\\right][\/latex]<\/div>\n<p>The solution is[latex]\\,\\left(1,2,0\\right).[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134323584\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_09_07_04\">\n<div id=\"fs-id1165133155808\">\n<p id=\"fs-id1165133155809\">Solve the system using the inverse of the coefficient matrix.<\/p>\n<div id=\"fs-id1165133155812\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\text{ }2x-17y+11z=0\\hfill \\\\ \\text{ }-x+11y-7z=8\\hfill \\\\ \\text{ }3y-2z=-2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137446595\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137446596\">[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135407360\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\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]\\,\\left[A\\right]\\,[\/latex]and[latex]\\,\\left[B\\right].[\/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 id=\"Example_09_07_09\" class=\"textbox examples\">\n<div id=\"fs-id1165131997848\">\n<div id=\"fs-id1165137611786\">\n<h3>Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\n<p id=\"fs-id1165137611791\">Solve the system of equations with matrix inverses using a calculator<\/p>\n<div id=\"fs-id1165137811747\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}2x+3y+z=32\\hfill \\\\ 3x+3y+z=-27\\hfill \\\\ 2x+4y+z=-2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134394498\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134394500\">On the matrix page of the calculator, enter the <span class=\"no-emphasis\">coefficient matrix<\/span> as the matrix variable[latex]\\,\\left[A\\right],\\,[\/latex]and enter the constant matrix as the matrix variable[latex]\\,\\left[B\\right].[\/latex]<\/p>\n<div id=\"fs-id1165134117288\" class=\"unnumbered aligncenter\">[latex]\\left[A\\right]=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right],\\text{\u2003}\\left[B\\right]=\\left[\\begin{array}{c}32\\\\ -27\\\\ -2\\end{array}\\right][\/latex]<\/div>\n<p id=\"fs-id1165134089389\">On the home screen of the calculator, type in the multiplication to solve for[latex]\\,X,\\,[\/latex]calling up each matrix variable as needed.<\/p>\n<div class=\"unnumbered\">[latex]{\\left[A\\right]}^{-1}\u00d7\\left[B\\right][\/latex]<\/div>\n<p id=\"fs-id1165135481265\">Evaluate the expression.<\/p>\n<div id=\"fs-id1165137863372\" class=\"unnumbered aligncenter\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131884558\" class=\"precalculus media\">\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:\/\/openstaxcollege.org\/l\/identmatrix\">The Identity Matrix<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/inversematrix\">Determining Inverse Matrices<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/matrixsystem\">Using a Matrix Equation to Solve a System of Equations<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135500881\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"eip-id1165137848559\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr valign=\"middle\">\n<td>Identity matrix for a[latex]2\\text{}\u00d7\\text{}2[\/latex]matrix<\/td>\n<td>[latex]{I}_{2}=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Identity matrix for a[latex]\\text{3}\\text{}\u00d7\\text{}3[\/latex]matrix<\/td>\n<td>[latex]{I}_{3}=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Multiplicative inverse of a[latex]2\\text{}\u00d7\\text{}2[\/latex]matrix<\/td>\n<td>[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d& -b\\\\ -c& a\\end{array}\\right],\\text{ where }ad-bc\\ne 0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165134225617\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165134225620\">\n<li>An identity matrix has the property[latex]\\,AI=IA=A.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_09_07_01\">(Figure)<\/a>.<\/li>\n<li>An invertible matrix has the property[latex]\\,A{A}^{-1}={A}^{-1}A=I.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_09_07_02\">(Figure)<\/a>.<\/li>\n<li>Use matrix multiplication and the identity to find the inverse of a[latex]\\,2\u00d72\\,[\/latex]matrix. See <a class=\"autogenerated-content\" href=\"#Example_09_07_03\">(Figure)<\/a>.<\/li>\n<li>The multiplicative inverse can be found using a formula. See <a class=\"autogenerated-content\" href=\"#Example_09_07_04\">(Figure)<\/a>.<\/li>\n<li>Another method of finding the inverse is by augmenting with the identity. See <a class=\"autogenerated-content\" href=\"#Example_09_07_05\">(Figure)<\/a>.<\/li>\n<li>We can augment a[latex]\\,3\u00d73\\,[\/latex]matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See <a class=\"autogenerated-content\" href=\"#Example_09_07_06\">(Figure)<\/a>.<\/li>\n<li>Write the system of equations as[latex]\\,AX=B,\\,[\/latex]and multiply both sides by the inverse of[latex]\\,A:{A}^{-1}AX={A}^{-1}B.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_09_07_07\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_09_07_08\">(Figure)<\/a>.<\/li>\n<li>We can also use a calculator to solve a system of equations with matrix inverses. See <a class=\"autogenerated-content\" href=\"#Example_09_07_09\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165135446624\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id1165137548736\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165137548741\">\n<div id=\"fs-id1165137548742\">\n<p id=\"fs-id1165134540056\">In a previous section, we showed that matrix multiplication is not commutative, that is,[latex]\\,AB\\ne BA\\,[\/latex]in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is,[latex]\\,{A}^{-1}A=A{A}^{-1}?[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134494231\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>If[latex]\\,{A}^{-1}\\,[\/latex]is the inverse of[latex]\\,A,\\,[\/latex]then[latex]\\,A{A}^{-1}=I,\\,[\/latex]the identity matrix. Since[latex]\\,A\\,[\/latex]is also the inverse of[latex]\\,{A}^{-1},{A}^{-1}A=I.\\,[\/latex]You can also check by proving this for a[latex]\\,2\u00d72\\,[\/latex]matrix.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134493452\">\n<div id=\"fs-id1165134493453\">\n<p id=\"fs-id1165134493454\">Does every[latex]\\,2\u00d72\\,[\/latex]matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135547596\">\n<div id=\"fs-id1165135547597\">\n<p id=\"fs-id1165135547598\">Can you explain whether a[latex]\\,2\u00d72\\,[\/latex]matrix with an entire row of zeros can have an inverse?<\/p>\n<\/div>\n<div id=\"fs-id1165133436150\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133436152\">No, because[latex]\\,ad\\,[\/latex]and[latex]\\,bc\\,[\/latex]are both 0, so[latex]\\,ad-bc=0,\\,[\/latex]which requires us to divide by 0 in the formula.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133401600\">\n<div>\n<p id=\"fs-id1165135702542\">Can a matrix with an entire column of zeros have an inverse? Explain why or why not.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135702546\">\n<div id=\"fs-id1165135702547\">\n<p id=\"fs-id1165135702548\">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\u00d72\\,[\/latex]matrix.<\/p>\n<\/div>\n<div id=\"fs-id1165134138529\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134138531\">Yes. Consider the matrix[latex]\\,\\left[\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right].\\,[\/latex]The inverse is found with the following calculation:[latex]\\,{A}^{-1}=\\frac{1}{0\\left(0\\right)-1\\left(1\\right)}\\left[\\begin{array}{cc}0& -1\\\\ -1& 0\\end{array}\\right]=\\left[\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right].[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133075017\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p>In the following exercises, show that matrix[latex]\\,A\\,[\/latex]is the inverse of matrix[latex]\\,B.[\/latex]<\/p>\n<div id=\"fs-id1165137804955\">\n<div id=\"fs-id1165137804956\">\n<p id=\"fs-id1165137804958\">[latex]A=\\left[\\begin{array}{cc}1& 0\\\\ -1& 1\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}1& 0\\\\ 1& 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137629353\">\n<div id=\"fs-id1165137629354\">\n<p id=\"fs-id1165135611304\">[latex]A=\\left[\\begin{array}{cc}1& 2\\\\ 3& 4\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}-2& 1\\\\ \\frac{3}{2}& -\\frac{1}{2}\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165132949834\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165132949836\">[latex]AB=BA=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]=I[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div>\n<p id=\"fs-id1165137832787\">[latex]A=\\left[\\begin{array}{cc}4& 5\\\\ 7& 0\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}0& \\frac{1}{7}\\\\ \\frac{1}{5}& -\\frac{4}{35}\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137417445\">\n<div id=\"fs-id1165137417446\">\n<p id=\"fs-id1165137417447\">[latex]A=\\left[\\begin{array}{cc}-2& \\frac{1}{2}\\\\ 3& -1\\end{array}\\right],\\,B=\\left[\\begin{array}{cc}-2& -1\\\\ -6& -4\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135186218\">[latex]AB=BA=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]=I[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137810179\">\n<div id=\"fs-id1165137810180\">\n<p id=\"fs-id1165137810181\">[latex]A=\\left[\\begin{array}{ccc}1& 0& 1\\\\ 0& 1& -1\\\\ 0& 1& 1\\end{array}\\right],\\,B=\\frac{1}{2}\\left[\\begin{array}{ccc}2& 1& -1\\\\ 0& 1& 1\\\\ 0& -1& 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p id=\"fs-id1165135481916\">[latex]A=\\left[\\begin{array}{ccc}1& 2& 3\\\\ 4& 0& 2\\\\ 1& 6& 9\\end{array}\\right],\\,B=\\frac{1}{4}\\left[\\begin{array}{ccc}6& 0& -2\\\\ 17& -3& -5\\\\ -12& 2& 4\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135513422\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135513424\">[latex]AB=BA=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]=I[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165134371187\">\n<p id=\"fs-id1165134371188\">[latex]A=\\left[\\begin{array}{ccc}3& 8& 2\\\\ 1& 1& 1\\\\ 5& 6& 12\\end{array}\\right],\\,B=\\frac{1}{36}\\left[\\begin{array}{ccc}-6& 84& -6\\\\ 7& -26& 1\\\\ -1& -22& 5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<p>For the following exercises, find the multiplicative inverse of each matrix, if it exists.<\/p>\n<div id=\"fs-id1165134282180\">\n<div id=\"fs-id1165134282181\">\n<p id=\"fs-id1165134282182\">[latex]\\left[\\begin{array}{cc}3& -2\\\\ 1& 9\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135363094\">[latex]\\frac{1}{29}\\left[\\begin{array}{cc}9& 2\\\\ -1& 3\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div>[latex]\\left[\\begin{array}{cc}-2& 2\\\\ 3& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137387518\">\n<div id=\"fs-id1165137387519\">\n<p id=\"fs-id1165137387520\">[latex]\\left[\\begin{array}{cc}-3& 7\\\\ 9& 2\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135192366\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134493412\">[latex]\\frac{1}{69}\\left[\\begin{array}{cc}-2& 7\\\\ 9& 3\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134339996\">\n<div id=\"fs-id1165134339997\">\n<p id=\"fs-id1165134339998\">[latex]\\left[\\begin{array}{cc}-4& -3\\\\ -5& 8\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133103275\">\n<div id=\"fs-id1165133103276\">[latex]\\left[\\begin{array}{cc}1& 1\\\\ 2& 2\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165132957181\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>There is no inverse<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165132957189\">\n<p id=\"fs-id1165132957190\">[latex]\\left[\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134185464\">\n<div id=\"fs-id1165134185465\">\n<p id=\"fs-id1165134185466\">[latex]\\left[\\begin{array}{cc}0.5& 1.5\\\\ 1& -0.5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134237283\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134237285\">[latex]\\frac{4}{7}\\left[\\begin{array}{cc}0.5& 1.5\\\\ 1& -0.5\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134152537\">\n<div id=\"fs-id1165134152538\">\n<p id=\"fs-id1165134135289\">[latex]\\left[\\begin{array}{ccc}1& 0& 6\\\\ -2& 1& 7\\\\ 3& 0& 2\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133141439\">\n<div id=\"fs-id1165135679418\">\n<p id=\"fs-id1165135679419\">[latex]\\left[\\begin{array}{ccc}0& 1& -3\\\\ 4& 1& 0\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135684091\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135684094\">[latex]\\frac{1}{17}\\left[\\begin{array}{ccc}-5& 5& -3\\\\ 20& -3& 12\\\\ 1& -1& 4\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134040470\">\n<div id=\"fs-id1165134040471\">\n<p id=\"fs-id1165134040472\">[latex]\\left[\\begin{array}{ccc}1& 2& -1\\\\ -3& 4& 1\\\\ -2& -4& -5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134159676\">\n<div id=\"fs-id1165134159677\">[latex]\\left[\\begin{array}{ccc}1& 9& -3\\\\ 2& 5& 6\\\\ 4& -2& 7\\end{array}\\right][\/latex]<\/div>\n<div id=\"fs-id1165134086077\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]\\frac{1}{209}\\left[\\begin{array}{ccc}47& -57& 69\\\\ 10& 19& -12\\\\ -24& 38& -13\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133210779\">\n<div id=\"fs-id1165133210780\">\n<p id=\"fs-id1165133210782\">[latex]\\left[\\begin{array}{ccc}1& -2& 3\\\\ -4& 8& -12\\\\ 1& 4& 2\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135397109\">\n<div id=\"fs-id1165135397110\">\n<p id=\"fs-id1165135397111\">[latex]\\left[\\begin{array}{ccc}\\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{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134389849\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]\\left[\\begin{array}{ccc}18& 60& -168\\\\ -56& -140& 448\\\\ 40& 80& -280\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131818536\">\n<div id=\"fs-id1165131818537\">\n<p id=\"fs-id1165131818538\">[latex]\\left[\\begin{array}{ccc}1& 2& 3\\\\ 4& 5& 6\\\\ 7& 8& 9\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137675157\">For the following exercises, solve the system using the inverse of a[latex]\\,2\\,\u00d7\\,2\\,[\/latex]matrix.<\/p>\n<div id=\"fs-id1165134319666\">\n<div id=\"fs-id1165134070732\">\n<p id=\"fs-id1165134070733\">[latex]\\begin{array}{l}\\text{ }5x-6y=-61\\hfill \\\\ 4x+3y=-2\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165133176692\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133176694\">[latex]\\left(-5,6\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133012293\">\n<div id=\"fs-id1165133012294\">\n<p id=\"fs-id1165133409845\">[latex]\\begin{array}{l}8x+4y=-100\\\\ 3x-4y=1\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131884588\">\n<div id=\"fs-id1165131884589\">\n<p id=\"fs-id1165135169381\">[latex]\\begin{array}{l}\\,3x-2y=6\\hfill \\\\ -x+5y=-2\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135649520\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165132938289\">[latex]\\left(2,0\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165133045243\">\n<p id=\"fs-id1165133045244\">[latex]\\begin{array}{l}5x-4y=-5\\hfill \\\\ \\,\\,\\,4x+y=2.3\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135205038\">\n<div id=\"fs-id1165135205039\">\n<p id=\"fs-id1165135205040\">[latex]\\begin{array}{l}-3x-4y=9\\hfill \\\\ \\,12x+4y=-6\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135358856\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135358859\">[latex]\\left(\\frac{1}{3},-\\frac{5}{2}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div>[latex]\\begin{array}{l}-2x+3y=\\frac{3}{10}\\hfill \\\\ \\,\\,-x+5y=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div>\n<div>[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\frac{8}{5}x-\\frac{4}{5}y=\\frac{2}{5}\\hfill \\\\ -\\frac{8}{5}x+\\frac{1}{5}y=\\frac{7}{10}\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135328736\">[latex]\\left(-\\frac{2}{3},-\\frac{11}{6}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133157406\">\n<div id=\"fs-id1165135523285\">\n<p id=\"fs-id1165135523286\">[latex]\\begin{array}{l}\\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<\/div>\n<p id=\"fs-id1165134386500\">For the following exercises, solve a system using the inverse of a[latex]\\,3\\text{}\u00d7\\text{}3\\,[\/latex]<br \/>\nmatrix.<\/p>\n<div id=\"fs-id1165132079352\">\n<div id=\"fs-id1165132079353\">\n<p id=\"fs-id1165132079354\">[latex]\\begin{array}{l}3x-2y+5z=21\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5x+4y=37\\hfill \\\\ \\,\\,\\,x-2y-5z=5\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137612251\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137612253\">[latex]\\left(7,\\frac{1}{2},\\frac{1}{5}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165134177545\">[latex]\\begin{array}{l}\\text{ }4x+4y+4z=40\\hfill \\\\ \\text{ }2x-3y+4z=-12\\hfill \\\\ \\text{ }-x+3y+4z=9\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134226706\">\n<div id=\"fs-id1165134226707\">\n<p id=\"fs-id1165134226708\">[latex]\\begin{array}{l}\\text{ }6x-5y-z=31\\hfill \\\\ \\text{ }-x+2y+z=-6\\hfill \\\\ \\text{ }3x+3y+2z=13\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134129992\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134129994\">[latex]\\left(5,0,-1\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165134224076\">\n<p id=\"fs-id1165134583960\">[latex]\\begin{array}{l}6x-5y+2z=-4\\hfill \\\\ \\,\\,2x+5y-z=12\\hfill \\\\ \\,\\,2x+5y+z=12\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134185366\">\n<div id=\"fs-id1165134185367\">\n<p id=\"fs-id1165134185368\">[latex]\\begin{array}{l}4x-2y+3z=-12\\hfill \\\\ 2x+2y-9z=33\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6y-4z=1\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137654966\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137654968\">[latex]\\frac{1}{34}\\left(-35,-97,-154\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135207327\">\n<div id=\"fs-id1165135207328\">\n<p id=\"fs-id1165134419566\">[latex]\\begin{array}{l}\\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<\/div>\n<div id=\"fs-id1165135421457\">\n<div>\n<p id=\"fs-id1165134342520\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\frac{1}{2}x-\\frac{1}{5}y+\\frac{1}{5}z=\\frac{31}{100}\\hfill \\\\ -\\frac{3}{4}x-\\frac{1}{4}y+\\frac{1}{2}z=\\frac{7}{40}\\hfill \\\\ -\\frac{4}{5}x-\\frac{1}{2}y+\\frac{3}{2}z=\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134203408\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134203410\">[latex]\\frac{1}{690}\\left(65,-1136,-229\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165132944845\">\n<div id=\"fs-id1165132944846\">\n<p id=\"fs-id1165132944848\">[latex]\\begin{array}{l}0.1x+0.2y+0.3z=-1.4\\hfill \\\\ 0.1x-0.2y+0.3z=0.6\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0.4y+0.9z=-2\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137900020\" class=\"bc-section section\">\n<h4>Technology<\/h4>\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\">\n<div id=\"fs-id1165134356167\">\n<p id=\"fs-id1165134356168\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,2x-y=-3\\hfill \\\\ -x+2y=2.3\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137938316\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137938319\">[latex]\\left(-\\frac{37}{30},\\frac{8}{15}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135186095\">\n<div id=\"fs-id1165135186096\">\n<p id=\"fs-id1165135186097\">[latex]\\begin{array}{l}-\\frac{1}{2}x-\\frac{3}{2}y=-\\frac{43}{20}\\hfill \\\\ \\,\\,\\frac{5}{2}x+\\frac{11}{5}y=\\frac{31}{4}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135652325\">\n<div id=\"fs-id1165135652326\">\n<p id=\"fs-id1165135652327\">[latex]\\begin{array}{l}12.3x-2y-2.5z=2\\hfill \\\\ 36.9x+7y-7.5z=-7\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8y-5z=-10\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135407474\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135369163\">[latex]\\left(\\frac{10}{123},-1,\\frac{2}{5}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134467729\">\n<div id=\"fs-id1165134467730\">\n<p id=\"fs-id1165134467731\">[latex]\\begin{array}{l}0.5x-3y+6z=-0.8\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0.7x-2y=-0.06\\hfill \\\\ 0.5x+4y+5z=0\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134537745\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<p id=\"fs-id1165135339476\">For the following exercises, find the inverse of the given matrix.<\/p>\n<div id=\"fs-id1165135339479\">\n<div id=\"fs-id1165135339480\">\n<p id=\"fs-id1165135339481\">[latex]\\left[\\begin{array}{cccc}1& 0& 1& 0\\\\ 0& 1& 0& 1\\\\ 0& 1& 1& 0\\\\ 0& 0& 1& 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]\\frac{1}{2}\\left[\\begin{array}{rrrr}\\hfill 2& \\hfill 1& \\hfill -1& \\hfill -1\\\\ \\hfill 0& \\hfill 1& \\hfill 1& \\hfill -1\\\\ \\hfill 0& \\hfill -1& \\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -1& \\hfill 1\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div>\n<p id=\"fs-id1165135258899\">[latex]\\left[\\begin{array}{rrrr}\\hfill -1& \\hfill 0& \\hfill 2& \\hfill 5\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 2\\\\ \\hfill 0& \\hfill 2& \\hfill -1& \\hfill 0\\\\ \\hfill 1& \\hfill -3& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191959\">\n<div id=\"fs-id1165135191960\">\n<p id=\"fs-id1165135191962\">[latex]\\left[\\begin{array}{rrrr}\\hfill 1& \\hfill -2& \\hfill 3& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 2\\\\ \\hfill 1& \\hfill 4& \\hfill -2& \\hfill 3\\\\ \\hfill -5& \\hfill 0& \\hfill 1& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165133409785\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]\\frac{1}{39}\\left[\\begin{array}{rrrr}\\hfill 3& \\hfill 2& \\hfill 1& \\hfill -7\\\\ \\hfill 18& \\hfill -53& \\hfill 32& \\hfill 10\\\\ \\hfill 24& \\hfill -36& \\hfill 21& \\hfill 9\\\\ \\hfill -9& \\hfill 46& \\hfill -16& \\hfill -5\\end{array}\\right][\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134234210\">\n<div id=\"fs-id1165134234211\">\n<p id=\"fs-id1165134234212\">[latex]\\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill 2& \\hfill 0& \\hfill 2& \\hfill 3\\\\ \\hfill 0& \\hfill 2& \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 3& \\hfill 0& \\hfill 1\\\\ \\hfill 0& \\hfill 2& \\hfill 0& \\hfill 0& \\hfill 1\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134385542\">\n<div id=\"fs-id1165134385543\">\n<p id=\"fs-id1165134385544\">[latex]\\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 1& \\hfill 1& \\hfill 1& \\hfill 1& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134248766\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134248768\">[latex]\\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill -1& \\hfill -1& \\hfill -1& \\hfill -1& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133094400\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\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\">\n<div id=\"fs-id1165133260342\">\n<p id=\"fs-id1165133260343\">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<\/div>\n<div id=\"fs-id1165133260348\">\n<div id=\"fs-id1165133260349\">\n<p id=\"fs-id1165133260350\">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<div id=\"fs-id1165133111141\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133111143\">Infinite solutions.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133111147\">\n<div id=\"fs-id1165133111148\">\n<p id=\"fs-id1165133111149\">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<\/div>\n<div id=\"fs-id1165134081432\">\n<div id=\"fs-id1165134081433\">\n<p id=\"fs-id1165134081434\">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<div id=\"fs-id1165134081439\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134081441\">50% oranges, 25% bananas, 20% apples<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135705936\">\n<div id=\"fs-id1165135705937\">\n<p id=\"fs-id1165135705938\">A sorority 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<\/div>\n<div id=\"fs-id1165135705944\">\n<div id=\"fs-id1165135705945\">\n<p id=\"fs-id1165135705946\">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<div id=\"fs-id1165133353997\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133353999\">10 straw hats, 50 beanies, 40 cowboy hats<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133354003\">\n<div id=\"fs-id1165133354004\">\n<p id=\"fs-id1165133354006\">Anna, Ashley, and Andrea weigh a combined 370 lb. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137501545\">\n<div id=\"fs-id1165137501546\">\n<p id=\"fs-id1165137501547\">Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat?<\/p>\n<\/div>\n<div id=\"fs-id1165137501553\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135547272\">Tom ate 6, Joe ate 3, and Albert ate 3.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165135547278\">\n<p id=\"fs-id1165135547279\">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{\\text{ ft}}^{2}.\\,[\/latex]He used[latex]\\,{\\text{3 ft}}^{2}\\,[\/latex]more chicken wire than plywood. How much of each material in did the farmer use?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134531903\">\n<div id=\"fs-id1165134531904\">\n<p id=\"fs-id1165134531906\">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<div id=\"fs-id1165133045226\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133045228\">124 oranges, 10 lemons, 8 pomegranates<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl>\n<dt>identity matrix<\/dt>\n<dd id=\"fs-id1165134179622\">a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134179626\">\n<dt>multiplicative inverse of a matrix<\/dt>\n<dd id=\"fs-id1165135528440\">a matrix that, when multiplied by the original, equals the identity matrix<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":291,"menu_order":8,"template":"","meta":{"pb_show_title":null,"pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-181","chapter","type-chapter","status-publish","hentry"],"part":166,"_links":{"self":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/181","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/users\/291"}],"version-history":[{"count":1,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/181\/revisions"}],"predecessor-version":[{"id":182,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/181\/revisions\/182"}],"part":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/parts\/166"}],"metadata":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/181\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/media?parent=181"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapter-type?post=181"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/contributor?post=181"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/license?post=181"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}