{"id":14600,"date":"2018-09-27T18:14:29","date_gmt":"2018-09-27T18:14:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-inverses\/"},"modified":"2025-02-05T05:19:52","modified_gmt":"2025-02-05T05:19:52","slug":"solving-systems-with-inverses","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-inverses\/","title":{"raw":"Solving Systems with Inverses","rendered":"Solving Systems with Inverses"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Find the inverse of a matrix.<\/li>\r\n \t<li>Solve a system of linear equations using an inverse matrix.<\/li>\r\n<\/ul>\r\n<\/div>\r\nNancy 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?\r\n\r\nThere 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.\r\n<h2>Finding the Inverse of a Matrix<\/h2>\r\nWe know that the multiplicative inverse of a real number [latex]a[\/latex] is [latex]{a}^{-1}[\/latex], and [latex]a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1[\/latex]. For example, [latex]{2}^{-1}=\\frac{1}{2}[\/latex] and [latex]\\left(\\frac{1}{2}\\right)2=1[\/latex]. The <strong>multiplicative inverse of a matrix<\/strong> is similar in concept, except that the product of matrix [latex]A[\/latex] and its inverse [latex]{A}^{-1}[\/latex] equals the <strong>identity matrix<\/strong>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by [latex]{I}_{n}[\/latex] where [latex]n[\/latex] represents the dimension of the matrix. The equations below\u00a0are the identity matrices for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a [latex]3\\text{}\\times \\text{}3[\/latex] matrix, respectively.\r\n<div style=\"text-align: center;\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\r\nThe identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A[\/latex].\r\n\r\nA matrix that has a multiplicative inverse has the properties\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/div>\r\nA matrix that has a multiplicative inverse is called an <strong>invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Identity Matrix and Multiplicative Inverse<\/h3>\r\nThe <strong>identity matrix<\/strong>, [latex]{I}_{n}[\/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{I}_{2}=\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right] &amp;&amp;&amp; {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] \\\\ \\text{2 }\\times\\text{ 2} &amp;&amp;&amp; \\text{3 }\\times\\text{ 3}\\end{array}[\/latex]<\/p>\r\nIf [latex]A[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix and [latex]B[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix such that [latex]AB=BA={I}_{n}[\/latex], then [latex]B={A}^{-1}[\/latex], the <strong>multiplicative inverse of a matrix<\/strong> [latex]A[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Showing That the Identity Matrix Acts as a 1<\/h3>\r\nGiven matrix <em>A<\/em>, show that [latex]AI=IA=A[\/latex].\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3&amp; 4\\\\ -2&amp; 5\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"123475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"123475\"]\r\n\r\nUse 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>\r\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right] \\cdot \\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]IA=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two matrices, show that one is the multiplicative inverse of the other.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Given matrix [latex]A[\/latex] of order [latex]n\\times n[\/latex] and matrix [latex]B[\/latex] of order [latex]n\\times n[\/latex] multiply [latex]AB[\/latex].<\/li>\r\n \t<li>If [latex]AB=I[\/latex], then find the product [latex]BA[\/latex]. If [latex]BA=I[\/latex], then [latex]B={A}^{-1}[\/latex] and [latex]A={B}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Showing That Matrix <em>A<\/em> Is the Multiplicative Inverse of Matrix <em>B<\/em><\/h3>\r\nShow that the given matrices are multiplicative inverses of each other.\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"57645\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"57645\"]\r\n\r\nMultiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}AB&amp;=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\\\ &amp;=\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)&amp; \\hfill &amp; \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)&amp; \\hfill &amp; \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ &amp;=\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}BA&amp;=\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\\hfill \\\\ &amp;=\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)&amp; \\hfill &amp; \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)&amp; \\hfill &amp; \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ &amp;=\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{align}[\/latex]<\/p>\r\n[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nShow that the following two matrices are inverses of each other.\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[reveal-answer q=\"250442\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"250442\"]\r\n\r\n[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]\\cdot \\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] \\cdot \\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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding the Multiplicative Inverse Using Matrix Multiplication<\/h2>\r\nWe 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 <strong>matrix multiplication<\/strong>.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\r\nUse matrix multiplication to find the inverse of the given matrix.\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill -3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"502369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"502369\"]\r\n\r\nFor this method, we multiply [latex]A[\/latex] by a matrix containing unknown constants and set it equal to the identity.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill -3\\end{array}\\right]\\cdot\\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]<\/p>\r\nFind the product of the two matrices on the left side of the equal sign.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill -3\\end{array}\\right]\\cdot \\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]<\/p>\r\nNext, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}1a - 2c&amp;=1 &amp;&amp; {R}_{1}\\\\ 2a - 3c&amp;=0 &amp;&amp; {R}_{2}\\end{align}[\/latex]<\/p>\r\nUsing 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} -2a + 4c&amp;=-2\\\\ 2a-3c&amp;=0\\\\ \\hline c&amp;=-2\\end{align}[\/latex]<\/p>\r\nBack-substitute to solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} a - 2\\left(-2\\right)=1\\\\ a+4=1\\\\ a=-3\\end{gathered}[\/latex]<\/p>\r\nWrite another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 1b - 2d&amp;=0&amp;&amp; {R}_{1}\\\\ 2b - 3d&amp;=1&amp;&amp; {R}_{2}\\end{align}[\/latex]<\/p>\r\nUsing 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} -2b +4d&amp;=0\\\\ 2b-3d&amp;=1\\\\ \\hline d&amp;=1 \\end{align}[\/latex]<\/p>\r\nOnce more, back-substitute and solve for [latex]b[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} b - 2\\left(1\\right)=0\\\\ b - 2=0\\\\ b=2\\end{gathered}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding the Multiplicative Inverse by Augmenting with the Identity<\/h2>\r\nAnother way to find the <strong>multiplicative inverse<\/strong> 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].\r\n\r\nFor example, given\r\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 2&amp; 1\\\\ \\hfill 5&amp; 3\\end{array}\\right][\/latex]<\/div>\r\naugment [latex]A[\/latex] with the identity\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 2&amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill 3\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\r\nPerform <strong>row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.\r\n<ol>\r\n \t<li>Switch row 1 and row 2.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 5&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill 1\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill 0&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>Multiply row 2 by [latex]-2[\/latex] and add to row 1.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 2&amp; \\hfill 1\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill -2&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>Multiply row 1 by [latex]-2[\/latex] and add to row 2.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill -1\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill -2&amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill -2\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>Add row 2 to row 1.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill -1\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill 3&amp; \\hfill -1\\\\ \\hfill 5&amp; \\hfill -2\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>Multiply row 2 by [latex]-1[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill 3&amp; \\hfill -1\\\\ \\hfill -5&amp; \\hfill 2\\end{array}\\right][\/latex]<\/div><\/li>\r\n<\/ol>\r\nThe matrix we have found is [latex]{A}^{-1}[\/latex].\r\n<div style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3&amp; -1\\\\ \\hfill -5&amp; 2\\end{array}\\right][\/latex]<\/div>\r\n<h2>Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h2>\r\nWhen we need to find the <strong>multiplicative inverse<\/strong> of a [latex]2\\times 2[\/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.\r\n\r\nIf [latex]A[\/latex] is a [latex]2\\times 2[\/latex] matrix, such as\r\n<div style=\"text-align: center;\">[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>\r\nthe multiplicative inverse of [latex]A[\/latex] is given by the formula\r\n<div style=\"text-align: center;\">[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>\r\nwhere [latex]ad-bc\\ne 0[\/latex]. If [latex]ad-bc=0[\/latex], then [latex]A[\/latex] has no inverse.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Using the Formula to Find the Multiplicative Inverse of Matrix <em>A<\/em><\/h3>\r\nUse the formula to find the multiplicative inverse of\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}1&amp; -2\\\\ 2&amp; -3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"458580\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"458580\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{A}^{-1}&amp;=\\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 \\\\ &amp;=\\frac{1}{-3+4}\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\\\ &amp;=\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe 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.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1&amp; -2\\\\ 2&amp; -3\\end{array}\\right\\rvert\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nPerform <strong>row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.\r\n<ol>\r\n \t<li>Multiply row 1 by [latex]-2[\/latex] and add to row 2.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1&amp; -2\\\\ 0&amp; 1\\end{array}\\right\\rvert\\begin{array}{cc}1&amp; 0\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>Multiply row 1 by 2 and add to row 1.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right\\rvert\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/div><\/li>\r\n<\/ol>\r\nSo, we have verified our original solution.\r\n<div style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nUse the formula to find the inverse of matrix [latex]A[\/latex]. Verify your answer by augmenting with the identity matrix.\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}1&amp; -1\\\\ 2&amp; 3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"86995\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"86995\"]\r\n\r\n[latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}&amp;\\hfill \\frac{1}{5}\\\\ -\\frac{2}{5}&amp; \\frac{1}{5}\\end{array}\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Finding the Inverse of the Matrix, If It Exists<\/h3>\r\nFind the inverse, if it exists, of the given matrix.\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3&amp; 6\\\\ 1&amp; 2\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"65547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65547\"]\r\n\r\nWe will use the method of augmenting with the identity.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}3&amp; 6\\\\ 1&amp; 3\\end{array}\\right\\rvert\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/div>\r\n<ol>\r\n \t<li>Switch row 1 and row 2.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1&amp; 3\\\\ 3&amp; 6\\text{ }\\end{array}\\right\\rvert\\begin{array}{cc}0&amp; 1\\\\ 1&amp; 0\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>Multiply row 1 by \u22123 and add it to row 2.\r\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1&amp; 2\\\\ 0&amp; 0\\end{array}\\right\\rvert\\begin{array}{cc}1&amp; 0\\\\ -3&amp; 1\\end{array}\\right][\/latex]<\/div><\/li>\r\n \t<li>There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174704[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h2>\r\nUnfortunately, we do not have a formula similar to the one for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix to find the inverse of a [latex]3\\text{}\\times \\text{}3[\/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use <strong>row operations<\/strong> to obtain the inverse.\r\n\r\nGiven a [latex]3\\text{}\\times \\text{}3[\/latex]\u00a0matrix\r\n<div style=\"text-align: center;\">[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>\r\naugment [latex]A[\/latex] with the identity matrix\r\n<div style=\"text-align: center;\">[latex]A|I=\\left[\\left.\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right\\rvert\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/div>\r\nTo begin, we write the <strong>augmented matrix<\/strong> with the identity on the right and [latex]A[\/latex] on the left. Performing elementary <strong>row operations<\/strong> so that the <strong>identity matrix<\/strong> appears on the left, we will obtain the <strong>inverse matrix<\/strong> on the right. We will find the inverse of this matrix in the next example.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a [latex]3\\times 3[\/latex] matrix, find the inverse<\/h3>\r\n<ol>\r\n \t<li>Write the original matrix augmented with the identity matrix on the right.<\/li>\r\n \t<li>Use elementary row operations so that the identity appears on the left.<\/li>\r\n \t<li>What is obtained on the right is the inverse of the original matrix.<\/li>\r\n \t<li>Use matrix multiplication to show that [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Finding the Inverse of a 3 \u00d7 3 Matrix<\/h3>\r\nGiven the [latex]3\\times 3[\/latex] matrix [latex]A[\/latex], find the inverse.\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[reveal-answer q=\"109384\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"109384\"]\r\n\r\nAugment [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].\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right\\rvert\\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[\\left.\\begin{array}{ccc}3&amp; 3&amp; 1\\\\ 2&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right\\rvert\\begin{array}{ccc}0&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 2&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right\\rvert\\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]<\/p>\r\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 2&amp; 3&amp; 1\\\\ 0&amp; 1&amp; 0\\end{array}\\right\\rvert\\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]<\/p>\r\n<p style=\"text-align: center;\">[latex]{R}_{3}\\leftrightarrow {R}_{2}\\to \\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 2&amp; 3&amp; 1\\end{array}\\right\\rvert\\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]<\/p>\r\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 3&amp; 1\\end{array}\\right\\rvert\\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]<\/p>\r\n<p style=\"text-align: center;\">[latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right\\rvert\\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]<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nTo prove that [latex]B={A}^{-1}[\/latex], let\u2019s multiply the two matrices together to see if the product equals the identity, if [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}&amp;=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right]\\cdot\\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 \\\\ &amp;=\\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 \\\\ &amp;=\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right] \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A&amp;=\\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]\\cdot\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right]\\\\ &amp;=\\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 \\\\ &amp;=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\hfill \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the inverse of the [latex]3\\times 3[\/latex] matrix.\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[reveal-answer q=\"786495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"786495\"]\r\n\r\n[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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]35330[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving a System of Linear Equations Using the Inverse of a Matrix<\/h2>\r\nSolving 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 <strong>matrix multiplication<\/strong>, we may define a system of equations with the same number of equations as variables as\r\n<div style=\"text-align: center;\">[latex]AX=B[\/latex]<\/div>\r\nTo solve a system of linear equations using an <strong>inverse matrix<\/strong>, let [latex]A[\/latex] be the <strong>coefficient matrix<\/strong>, 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.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/div>\r\nFrom this system, the coefficient matrix is\r\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right][\/latex]<\/div>\r\nThe variable matrix is\r\n<div style=\"text-align: center;\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/div>\r\nAnd the constant matrix is\r\n<div style=\"text-align: center;\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/div>\r\nThen [latex]AX=B[\/latex] looks like\r\n<div style=\"text-align: center;\">[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>\r\nRecall 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,\r\n<div style=\"text-align: center;\">[latex]\\begin{align} ax&amp;=b\\\\ \\left(\\frac{1}{a}\\right)ax&amp;=\\left(\\frac{1}{a}\\right)b\\\\ \\left({a}^{-1}\\right)ax&amp;=\\left({a}^{-1}\\right)b\\\\ \\left[\\left({a}^{-1}\\right)a\\right]x&amp;=\\left({a}^{-1}\\right)b\\\\ 1x&amp;=\\left({a}^{-1}\\right)b\\\\ x&amp;=\\left({a}^{-1}\\right)b\\end{align}[\/latex]<\/div>\r\nThe only difference between a solving a linear equation and a <strong>system of equations<\/strong> 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.\r\n\r\nWe will investigate this idea in detail, but it is helpful to begin with a [latex]2\\times 2[\/latex] system and then move on to a [latex]3\\times 3[\/latex] system.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Solving a System of Equations Using the Inverse of a Matrix<\/h3>\r\nGiven 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\r\n<div style=\"text-align: center;\">[latex]AX=B[\/latex]<\/div>\r\nMultiply both sides by the inverse of [latex]A[\/latex] to obtain the solution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\left({A}^{-1}\\right)AX&amp;=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X&amp;=\\left({A}^{-1}\\right)B\\\\ \\hfill IX&amp;=\\left({A}^{-1}\\right)B\\\\ \\hfill X&amp;=\\left({A}^{-1}\\right)B\\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/h3>\r\n<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>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\r\nSolve the given system of equations using the inverse of a matrix.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"999483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"999483\"]\r\n\r\nWrite the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThen\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right]\\cdot\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/p>\r\nFirst, we need to calculate [latex]{A}^{-1}[\/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{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]<\/p>\r\nSo,\r\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; \\text{ }\\text{ }3\\end{array}\\right][\/latex]<\/p>\r\nNow we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{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]<\/p>\r\nThe solution is [latex]\\left(-1,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Can we solve for [latex]X[\/latex] by finding the product [latex]B{A}^{-1}?[\/latex]<\/h3>\r\n<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>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\left({A}^{-1}\\right)AX&amp;=\\left({A}^{-1}\\right)B\\\\ \\left[\\left({A}^{-1}\\right)A\\right]X&amp;=\\left({A}^{-1}\\right)B\\\\ IX&amp;=\\left({A}^{-1}\\right)B\\\\ X&amp;=\\left({A}^{-1}\\right)B\\end{align}[\/latex]<\/p>\r\n<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>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\r\nSolve the following system using the inverse of a matrix.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} 5x+15y+56z=35\\\\ -4x - 11y - 41z=-26\\\\ -x - 3y - 11z=-7\\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"342836\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342836\"]\r\n\r\nWrite the equation [latex]AX=B[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc}5&amp; 15&amp; 56\\\\ -4&amp; -11&amp; -41\\\\ -1&amp; -3&amp; -11\\end{array}\\right]\\cdot\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\r\nFirst, we will find the inverse of [latex]A[\/latex] by augmenting with the identity.\r\n<p style=\"text-align: center;\">[latex]\\left[\\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\\rvert\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 1 by [latex]\\frac{1}{5}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ -4&amp; -11&amp; -41\\\\ \\hfill-1&amp; -3&amp; -11\\end{array}\\right\\rvert\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 1 by 4 and add to row 2.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ \\hfill-1&amp; -3&amp; -11\\end{array}\\right\\rvert\\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]<\/p>\r\nAdd row 1 to row 3.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ \\hfill0&amp; 0&amp; \\frac{1}{5}\\end{array}\\right\\rvert\\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]<\/p>\r\nMultiply row 2 by \u22123 and add to row 1.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; -\\frac{1}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ \\hfill0&amp; 0&amp; \\frac{1}{5}\\end{array}\\right\\rvert\\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]<\/p>\r\nMultiply row 3 by 5.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; -\\frac{1}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ \\hfill0&amp; 0&amp; 1\\end{array}\\right\\rvert\\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]<\/p>\r\nMultiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ \\hfill0&amp; 0&amp; 1\\end{array}\\right\\rvert\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\\\hfill 0&amp; 0&amp; 1\\end{array}\\right\\rvert\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ -3&amp; 1&amp; -19\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\r\nSo,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nMultiply both sides of the equation by [latex]{A}^{-1}[\/latex]. We want [latex]{A}^{-1}AX={A}^{-1}B:[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\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]\\cdot\\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]\\cdot\\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]\\cdot\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThe solution is [latex]\\left(1,2,0\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve the system using the inverse of the coefficient matrix.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 17y+11z&amp;=0 \\\\ -x+11y - 7z&amp;=8 \\\\ 3y - 2z&amp;=-2 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"145598\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145598\"]\r\n\r\n[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve with matrix inverses using a calculator.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165135503570\">\r\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>\r\n \t<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\r\n \t<li>If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\r\nSolve the system of equations with matrix inverses using a calculator\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+3y+z=32\\hfill \\\\ 3x+3y+z=-27\\hfill \\\\ 2x+4y+z=-2\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"332326\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"332326\"]\r\n\r\nOn the matrix page of the calculator, enter the <strong>coefficient matrix<\/strong> as the matrix variable [latex]\\left[A\\right][\/latex], and enter the constant matrix as the matrix variable [latex]\\left[B\\right][\/latex].\r\n<p style=\"text-align: center;\">[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{ }\\left[B\\right]=\\left[\\begin{array}{c}32\\\\ -27\\\\ -2\\end{array}\\right][\/latex]<\/p>\r\nOn the home screen of the calculator, type in the multiplication to solve for [latex]X[\/latex], calling up each matrix variable as needed.\r\n<p style=\"text-align: center;\">[latex]{\\left[A\\right]}^{-1}\\times \\left[B\\right][\/latex]<\/p>\r\nEvaluate the expression.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table id=\"eip-id1165137848559\" summary=\"..\"><colgroup> <col \/> <col \/> <\/colgroup>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td>Identity matrix for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\r\n<td>[latex]{I}_{2}=\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>Identity matrix for a [latex]\\text{3}\\text{}\\times \\text{}3[\/latex] matrix<\/td>\r\n<td>[latex]{I}_{3}=\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>Multiplicative inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\r\n<td>[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d&amp; -b\\\\ -c&amp; a\\end{array}\\right],\\text{ where }ad-bc\\ne 0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>An identity matrix has the property [latex]AI=IA=A[\/latex].<\/li>\r\n \t<li>An invertible matrix has the property [latex]A{A}^{-1}={A}^{-1}A=I[\/latex].<\/li>\r\n \t<li>Use matrix multiplication and the identity to find the inverse of a [latex]2\\times 2[\/latex] matrix.<\/li>\r\n \t<li>The multiplicative inverse can be found using a formula.<\/li>\r\n \t<li>Another method of finding the inverse is by augmenting with the identity.<\/li>\r\n \t<li>We can augment a [latex]3\\times 3[\/latex] matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse.<\/li>\r\n \t<li>Write the system of equations as [latex]AX=B[\/latex], and multiply both sides by the inverse of [latex]A:{A}^{-1}AX={A}^{-1}B[\/latex].<\/li>\r\n \t<li>We can also use a calculator to solve a system of equations with matrix inverses.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165134179616\" class=\"definition\">\r\n \t<dt>identity matrix<\/dt>\r\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>\r\n<\/dl>\r\n<dl id=\"fs-id1165134179626\" class=\"definition\">\r\n \t<dt>multiplicative inverse of a matrix<\/dt>\r\n \t<dd id=\"fs-id1165135528440\">a matrix that, when multiplied by the original, equals the identity matrix<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/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>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>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<h2>Finding the Inverse of a Matrix<\/h2>\n<p>We know that the multiplicative inverse of a real number [latex]a[\/latex] is [latex]{a}^{-1}[\/latex], and [latex]a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1[\/latex]. For example, [latex]{2}^{-1}=\\frac{1}{2}[\/latex] and [latex]\\left(\\frac{1}{2}\\right)2=1[\/latex]. The <strong>multiplicative inverse of a matrix<\/strong> is similar in concept, except that the product of matrix [latex]A[\/latex] and its inverse [latex]{A}^{-1}[\/latex] equals the <strong>identity matrix<\/strong>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by [latex]{I}_{n}[\/latex] where [latex]n[\/latex] represents the dimension of the matrix. The equations below\u00a0are the identity matrices for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a [latex]3\\text{}\\times \\text{}3[\/latex] matrix, respectively.<\/p>\n<div style=\"text-align: center;\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div><\/div>\n<div style=\"text-align: center;\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 0& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<p>The identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A[\/latex].<\/p>\n<p>A matrix that has a multiplicative inverse has the properties<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/div>\n<p>A matrix that has a multiplicative inverse is called an <strong>invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Identity Matrix and Multiplicative Inverse<\/h3>\n<p>The <strong>identity matrix<\/strong>, [latex]{I}_{n}[\/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{I}_{2}=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right] &&& {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] \\\\ \\text{2 }\\times\\text{ 2} &&& \\text{3 }\\times\\text{ 3}\\end{array}[\/latex]<\/p>\n<p>If [latex]A[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix and [latex]B[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix such that [latex]AB=BA={I}_{n}[\/latex], then [latex]B={A}^{-1}[\/latex], the <strong>multiplicative inverse of a matrix<\/strong> [latex]A[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Showing That the Identity Matrix Acts as a 1<\/h3>\n<p>Given matrix <em>A<\/em>, show that [latex]AI=IA=A[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3& 4\\\\ -2& 5\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q123475\">Show Solution<\/span><\/p>\n<div id=\"q123475\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right] \\cdot \\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0& \\hfill & \\hfill & \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0& \\hfill & \\hfill & \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]IA=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two matrices, show that one is the multiplicative inverse of the other.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Given matrix [latex]A[\/latex] of order [latex]n\\times n[\/latex] and matrix [latex]B[\/latex] of order [latex]n\\times n[\/latex] multiply [latex]AB[\/latex].<\/li>\n<li>If [latex]AB=I[\/latex], then find the product [latex]BA[\/latex]. If [latex]BA=I[\/latex], then [latex]B={A}^{-1}[\/latex] and [latex]A={B}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Showing That Matrix <em>A<\/em> Is the Multiplicative Inverse of Matrix <em>B<\/em><\/h3>\n<p>Show that the given matrices are multiplicative inverses of each other.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q57645\">Show Solution<\/span><\/p>\n<div id=\"q57645\" class=\"hidden-answer\" style=\"display: none\">\n<p>Multiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}AB&=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ &=\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)& \\hfill & \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)& \\hfill & \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ &=\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}BA&=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\hfill \\\\ &=\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)& \\hfill & \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)& \\hfill & \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ &=\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{align}[\/latex]<\/p>\n<p>[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Show that the following two matrices are inverses of each other.<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q250442\">Show Solution<\/span><\/p>\n<div id=\"q250442\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]\\cdot \\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] \\cdot \\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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding the Multiplicative Inverse Using Matrix Multiplication<\/h2>\n<p>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 <strong>matrix multiplication<\/strong>.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 3: Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\n<p>Use matrix multiplication to find the inverse of the given matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill -2\\\\ \\hfill 2& \\hfill & \\hfill -3\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q502369\">Show Solution<\/span><\/p>\n<div id=\"q502369\" class=\"hidden-answer\" style=\"display: none\">\n<p>For this method, we multiply [latex]A[\/latex] by a matrix containing unknown constants and set it equal to the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -3\\end{array}\\right]\\cdot\\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]<\/p>\n<p>Find the product of the two matrices on the left side of the equal sign.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -3\\end{array}\\right]\\cdot \\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]<\/p>\n<p>Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}1a - 2c&=1 && {R}_{1}\\\\ 2a - 3c&=0 && {R}_{2}\\end{align}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{align} -2a + 4c&=-2\\\\ 2a-3c&=0\\\\ \\hline c&=-2\\end{align}[\/latex]<\/p>\n<p>Back-substitute to solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} a - 2\\left(-2\\right)=1\\\\ a+4=1\\\\ a=-3\\end{gathered}[\/latex]<\/p>\n<p>Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 1b - 2d&=0&& {R}_{1}\\\\ 2b - 3d&=1&& {R}_{2}\\end{align}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{align} -2b +4d&=0\\\\ 2b-3d&=1\\\\ \\hline d&=1 \\end{align}[\/latex]<\/p>\n<p>Once more, back-substitute and solve for [latex]b[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} b - 2\\left(1\\right)=0\\\\ b - 2=0\\\\ b=2\\end{gathered}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill 2\\\\ \\hfill -2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding the Multiplicative Inverse by Augmenting with the Identity<\/h2>\n<p>Another way to find the <strong>multiplicative inverse<\/strong> 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>For example, given<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 2& 1\\\\ \\hfill 5& 3\\end{array}\\right][\/latex]<\/div>\n<p>augment [latex]A[\/latex] with the identity<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 2& \\hfill 1\\\\ \\hfill 5& \\hfill 3\\end{array}\\right\\rvert\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<p>Perform <strong>row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.<\/p>\n<ol>\n<li>Switch row 1 and row 2.\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 5& \\hfill 3\\\\ \\hfill 2& \\hfill 1\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 2& \\hfill 1\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill -1\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill -1\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3& -1\\\\ \\hfill -5& 2\\end{array}\\right][\/latex]<\/div>\n<h2>Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h2>\n<p>When we need to find the <strong>multiplicative inverse<\/strong> of a [latex]2\\times 2[\/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.<\/p>\n<p>If [latex]A[\/latex] is a [latex]2\\times 2[\/latex] matrix, such as<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill a& \\hfill & \\hfill b\\\\ \\hfill c& \\hfill & \\hfill d\\end{array}\\right][\/latex]<\/div>\n<p>the multiplicative inverse of [latex]A[\/latex] is given by the formula<\/p>\n<div style=\"text-align: center;\">[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 class=\"textbox shaded\">\n<h3>Example 4: Using the Formula to Find the Multiplicative Inverse of Matrix <em>A<\/em><\/h3>\n<p>Use the formula to find the multiplicative inverse of<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q458580\">Show Solution<\/span><\/p>\n<div id=\"q458580\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{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{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>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<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}\\right\\rvert\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/p>\n<p>Perform <strong>row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.<\/p>\n<ol>\n<li>Multiply row 1 by [latex]-2[\/latex] and add to row 2.\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1& -2\\\\ 0& 1\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right\\rvert\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>So, we have verified our original solution.<\/p>\n<div style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\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<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}1& -1\\\\ 2& 3\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q86995\">Show Solution<\/span><\/p>\n<div id=\"q86995\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}&\\hfill \\frac{1}{5}\\\\ -\\frac{2}{5}& \\frac{1}{5}\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding the Inverse of the Matrix, If It Exists<\/h3>\n<p>Find the inverse, if it exists, of the given matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3& 6\\\\ 1& 2\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q65547\">Show Solution<\/span><\/p>\n<div id=\"q65547\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will use the method of augmenting with the identity.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}3& 6\\\\ 1& 3\\end{array}\\right\\rvert\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\n<ol>\n<li>Switch row 1 and row 2.\n<div style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1& 3\\\\ 3& 6\\text{ }\\end{array}\\right\\rvert\\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 style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{cc}1& 2\\\\ 0& 0\\end{array}\\right\\rvert\\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.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174704\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174704&theme=oea&iframe_resize_id=ohm174704\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h2>\n<p>Unfortunately, we do not have a formula similar to the one for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix to find the inverse of a [latex]3\\text{}\\times \\text{}3[\/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use <strong>row operations<\/strong> to obtain the inverse.<\/p>\n<p>Given a [latex]3\\text{}\\times \\text{}3[\/latex]\u00a0matrix<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\n<p>augment [latex]A[\/latex] with the identity matrix<\/p>\n<div style=\"text-align: center;\">[latex]A|I=\\left[\\left.\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right\\rvert\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n<p>To begin, we write the <strong>augmented matrix<\/strong> with the identity on the right and [latex]A[\/latex] on the left. Performing elementary <strong>row operations<\/strong> so that the <strong>identity matrix<\/strong> appears on the left, we will obtain the <strong>inverse matrix<\/strong> on the right. We will find the inverse of this matrix in the next example.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a [latex]3\\times 3[\/latex] matrix, find the inverse<\/h3>\n<ol>\n<li>Write the original matrix augmented with the identity matrix on the right.<\/li>\n<li>Use elementary row operations so that the identity appears on the left.<\/li>\n<li>What is obtained on the right is the inverse of the original matrix.<\/li>\n<li>Use matrix multiplication to show that [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Finding the Inverse of a 3 \u00d7 3 Matrix<\/h3>\n<p>Given the [latex]3\\times 3[\/latex] matrix [latex]A[\/latex], find the inverse.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q109384\">Show Solution<\/span><\/p>\n<div id=\"q109384\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right\\rvert\\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[\\left.\\begin{array}{ccc}3& 3& 1\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}\\right\\rvert\\begin{array}{ccc}0& 1& 0\\\\ 1& 0& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}\\right\\rvert\\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]<\/p>\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 0& 1& 0\\end{array}\\right\\rvert\\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]<\/p>\n<p style=\"text-align: center;\">[latex]{R}_{3}\\leftrightarrow {R}_{2}\\to \\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 2& 3& 1\\end{array}\\right\\rvert\\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]<\/p>\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 3& 1\\end{array}\\right\\rvert\\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]<\/p>\n<p style=\"text-align: center;\">[latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right\\rvert\\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]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=B=\\left[\\begin{array}{ccc}-1& 1& 0\\\\ -1& 0& 1\\\\ 6& -2& -3\\end{array}\\right][\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>To prove that [latex]B={A}^{-1}[\/latex], let\u2019s multiply the two matrices together to see if the product equals the identity, if [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\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]\\cdot\\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 \\\\ &=\\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] \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\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]\\cdot\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\\\ &=\\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{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the inverse of the [latex]3\\times 3[\/latex] matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{ccc}2& -17& 11\\\\ -1& 11& -7\\\\ 0& 3& -2\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786495\">Show Solution<\/span><\/p>\n<div id=\"q786495\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{A}^{-1}=\\left[\\begin{array}{ccc}1& 1& 2\\\\ 2& 4& -3\\\\ 3& 6& -5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm35330\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35330&theme=oea&iframe_resize_id=ohm35330\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving a System of Linear Equations Using the Inverse of a Matrix<\/h2>\n<p>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 <strong>matrix multiplication<\/strong>, we may define a system of equations with the same number of equations as variables as<\/p>\n<div style=\"text-align: center;\">[latex]AX=B[\/latex]<\/div>\n<p>To solve a system of linear equations using an <strong>inverse matrix<\/strong>, let [latex]A[\/latex] be the <strong>coefficient matrix<\/strong>, 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 style=\"text-align: center;\">[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>From this system, the coefficient matrix is<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right][\/latex]<\/div>\n<p>The variable matrix is<\/p>\n<div style=\"text-align: center;\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/div>\n<p>And the constant matrix is<\/p>\n<div style=\"text-align: center;\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/div>\n<p>Then [latex]AX=B[\/latex] looks like<\/p>\n<div style=\"text-align: center;\">[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>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 style=\"text-align: center;\">[latex]\\begin{align} ax&=b\\\\ \\left(\\frac{1}{a}\\right)ax&=\\left(\\frac{1}{a}\\right)b\\\\ \\left({a}^{-1}\\right)ax&=\\left({a}^{-1}\\right)b\\\\ \\left[\\left({a}^{-1}\\right)a\\right]x&=\\left({a}^{-1}\\right)b\\\\ 1x&=\\left({a}^{-1}\\right)b\\\\ x&=\\left({a}^{-1}\\right)b\\end{align}[\/latex]<\/div>\n<p>The only difference between a solving a linear equation and a <strong>system of equations<\/strong> 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>We will investigate this idea in detail, but it is helpful to begin with a [latex]2\\times 2[\/latex] system and then move on to a [latex]3\\times 3[\/latex] system.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Solving a System of Equations Using the Inverse of a Matrix<\/h3>\n<p>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 style=\"text-align: center;\">[latex]AX=B[\/latex]<\/div>\n<p>Multiply both sides by the inverse of [latex]A[\/latex] to obtain the solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\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{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/h3>\n<p><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 class=\"textbox shaded\">\n<h3>Example 7: Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\n<p>Solve the given system of equations using the inverse of a matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q999483\">Show Solution<\/span><\/p>\n<div id=\"q999483\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\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]<\/p>\n<p>Then<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right]\\cdot\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/p>\n<p>First, we need to calculate [latex]{A}^{-1}[\/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{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]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}11& -8\\\\ -4& \\text{ }\\text{ }3\\end{array}\\right][\/latex]<\/p>\n<p>Now we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{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]<\/p>\n<p>The solution is [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Can we solve for [latex]X[\/latex] by finding the product [latex]B{A}^{-1}?[\/latex]<\/h3>\n<p><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<p style=\"text-align: center;\">[latex]\\begin{align} \\left({A}^{-1}\\right)AX&=\\left({A}^{-1}\\right)B\\\\ \\left[\\left({A}^{-1}\\right)A\\right]X&=\\left({A}^{-1}\\right)B\\\\ IX&=\\left({A}^{-1}\\right)B\\\\ X&=\\left({A}^{-1}\\right)B\\end{align}[\/latex]<\/p>\n<p><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 class=\"textbox shaded\">\n<h3>Example 8: Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\n<p>Solve the following system using the inverse of a matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} 5x+15y+56z=35\\\\ -4x - 11y - 41z=-26\\\\ -x - 3y - 11z=-7\\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342836\">Show Solution<\/span><\/p>\n<div id=\"q342836\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the equation [latex]AX=B[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc}5& 15& 56\\\\ -4& -11& -41\\\\ -1& -3& -11\\end{array}\\right]\\cdot\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\n<p>First, we will find the inverse of [latex]A[\/latex] by augmenting with the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\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\\rvert\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 1 by [latex]\\frac{1}{5}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 3& \\frac{56}{5}\\\\ -4& -11& -41\\\\ \\hfill-1& -3& -11\\end{array}\\right\\rvert\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 1 by 4 and add to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 3& \\frac{56}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ \\hfill-1& -3& -11\\end{array}\\right\\rvert\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ \\frac{4}{5}& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p>Add row 1 to row 3.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 3& \\frac{56}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ \\hfill0& 0& \\frac{1}{5}\\end{array}\\right\\rvert\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ \\frac{4}{5}& 1& 0\\\\ \\frac{1}{5}& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 2 by \u22123 and add to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 0& -\\frac{1}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ \\hfill0& 0& \\frac{1}{5}\\end{array}\\right\\rvert\\begin{array}{ccc}-\\frac{11}{5}& -3& 0\\\\ \\frac{4}{5}& 1& 0\\\\ \\frac{1}{5}& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 3 by 5.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 0& -\\frac{1}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ \\hfill0& 0& 1\\end{array}\\right\\rvert\\begin{array}{ccc}-\\frac{11}{5}& -3& 0\\\\ \\frac{4}{5}& 1& 0\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& \\frac{19}{5}\\\\ \\hfill0& 0& 1\\end{array}\\right\\rvert\\begin{array}{ccc}-2& -3& 1\\\\ \\frac{4}{5}& 1& 0\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\\\hfill 0& 0& 1\\end{array}\\right\\rvert\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\n<p>Multiply both sides of the equation by [latex]{A}^{-1}[\/latex]. We want [latex]{A}^{-1}AX={A}^{-1}B:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\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]\\cdot\\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]\\cdot\\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]\\cdot\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>The solution is [latex]\\left(1,2,0\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve the system using the inverse of the coefficient matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 17y+11z&=0 \\\\ -x+11y - 7z&=8 \\\\ 3y - 2z&=-2 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q145598\">Show Solution<\/span><\/p>\n<div id=\"q145598\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve with matrix inverses using a calculator.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165135503570\">\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 class=\"textbox shaded\">\n<h3>Example 9: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\n<p>Solve the system of equations with matrix inverses using a calculator<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+3y+z=32\\hfill \\\\ 3x+3y+z=-27\\hfill \\\\ 2x+4y+z=-2\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q332326\">Show Solution<\/span><\/p>\n<div id=\"q332326\" class=\"hidden-answer\" style=\"display: none\">\n<p>On the matrix page of the calculator, enter the <strong>coefficient matrix<\/strong> 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<p style=\"text-align: center;\">[latex]\\left[A\\right]=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right],\\text{ }\\left[B\\right]=\\left[\\begin{array}{c}32\\\\ -27\\\\ -2\\end{array}\\right][\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]{\\left[A\\right]}^{-1}\\times \\left[B\\right][\/latex]<\/p>\n<p>Evaluate the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Equations<\/h2>\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{}\\times \\text{}2[\/latex] matrix<\/td>\n<td>[latex]{I}_{2}=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Identity matrix for a [latex]\\text{3}\\text{}\\times \\text{}3[\/latex] matrix<\/td>\n<td>[latex]{I}_{3}=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Multiplicative inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix<\/td>\n<td>[latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d& -b\\\\ -c& a\\end{array}\\right],\\text{ where }ad-bc\\ne 0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>An identity matrix has the property [latex]AI=IA=A[\/latex].<\/li>\n<li>An invertible matrix has the property [latex]A{A}^{-1}={A}^{-1}A=I[\/latex].<\/li>\n<li>Use matrix multiplication and the identity to find the inverse of a [latex]2\\times 2[\/latex] matrix.<\/li>\n<li>The multiplicative inverse can be found using a formula.<\/li>\n<li>Another method of finding the inverse is by augmenting with the identity.<\/li>\n<li>We can augment a [latex]3\\times 3[\/latex] matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse.<\/li>\n<li>Write the system of equations as [latex]AX=B[\/latex], and multiply both sides by the inverse of [latex]A:{A}^{-1}AX={A}^{-1}B[\/latex].<\/li>\n<li>We can also use a calculator to solve a system of equations with matrix inverses.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165134179616\" class=\"definition\">\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\" class=\"definition\">\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\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-14600\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14600","chapter","type-chapter","status-publish","hentry"],"part":14549,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14600","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14600\/revisions"}],"predecessor-version":[{"id":15853,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14600\/revisions\/15853"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14549"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14600\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=14600"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14600"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=14600"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=14600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}