{"id":2291,"date":"2016-11-03T20:01:47","date_gmt":"2016-11-03T20:01:47","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2291"},"modified":"2017-04-04T19:55:03","modified_gmt":"2017-04-04T19:55:03","slug":"find-the-inverse-of-a-matrix","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/find-the-inverse-of-a-matrix\/","title":{"raw":"Find the Inverse of a Matrix","rendered":"Find the Inverse of a Matrix"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Verify that multiplying a matrix by it's inverse results in 1<\/li>\r\n \t<li>Use matrix multiplication to find the inverse of a matrix<\/li>\r\n \t<li>Finding an inverse by augmenting with an identity matrix<\/li>\r\n<\/ul>\r\n<\/div>\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<p style=\"text-align: center;\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nThe identity matrix acts as a 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<p style=\"text-align: center;\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/p>\r\nA matrix that has a multiplicative inverse is called an <strong>invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.\r\n<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}{l}\\hfill\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2}=\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}{I}_{3}=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\text{ }2\\times 2\\text{ 3}\\times 3\\hfill \\end{array}\\hfill \\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 exercises\">\r\n<h3>Example: 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[latex]A=\\left[\\begin{array}{cc}3&amp; 4\\\\ -2&amp; 5\\end{array}\\right][\/latex]\r\n\r\n[reveal-answer q=\"38185\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"38185\"]\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]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0&amp; \\hfill &amp; \\hfill &amp; \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)&amp; \\hfill &amp; \\hfill &amp; \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/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 exercises\">\r\n<h3>Example: 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=\"812081\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"812081\"]\r\n\r\nMultiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}AB=\\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 \\\\ =\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)&amp; \\hfill &amp; \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)&amp; \\hfill &amp; \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}BA=\\left[\\begin{array}{rrr}\\hfill -9&amp; \\hfill &amp; \\hfill -5\\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 5\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill -9\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)&amp; \\hfill &amp; \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)&amp; \\hfill &amp; \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1&amp; &amp; 0\\\\ 0&amp; &amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/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=\"textbox key-takeaways\">\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=\"159815\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"159815\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -1&amp; \\hfill &amp; \\hfill -3\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill -3&amp; \\hfill &amp; \\hfill -4\\\\ \\hfill 1&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1\\left(-3\\right)+4\\left(1\\right)&amp; \\hfill &amp; \\hfill 1\\left(-4\\right)+4\\left(1\\right)\\\\ \\hfill -1\\left(-3\\right)+-3\\left(1\\right)&amp; \\hfill &amp; \\hfill -1\\left(-4\\right)+-3\\left(1\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\\\ BA=\\left[\\begin{array}{rrr}\\hfill -3&amp; \\hfill &amp; \\hfill -4\\\\ \\hfill 1&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 4\\\\ \\hfill -1&amp; \\hfill &amp; \\hfill -3\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -3\\left(1\\right)+-4\\left(-1\\right)&amp; \\hfill &amp; \\hfill -3\\left(4\\right)+-4\\left(-3\\right)\\\\ \\hfill 1\\left(1\\right)+1\\left(-1\\right)&amp; \\hfill &amp; \\hfill 1\\left(4\\right)+1\\left(-3\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill &amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\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 exercises\">\r\n<h3>Example: 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=\"31546\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"31546\"]\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]\\text{ }\\left[\\begin{array}{rr}\\hfill a&amp; \\hfill b\\\\ \\hfill c&amp; \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/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]\\text{ }\\left[\\begin{array}{rr}\\hfill a&amp; \\hfill b\\\\ \\hfill c&amp; \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1a - 2c&amp; \\hfill 1b - 2d\\\\ \\hfill 2a - 3c&amp; \\hfill 2b - 3d\\end{array}\\right][\/latex]<\/p>\r\nNext, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}1a - 2c=1\\text{ }{R}_{1}\\\\ 2a - 3c=0\\text{ }{R}_{2}\\end{array}[\/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{array}{r}\\hfill 1a - 2c=1\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/p>\r\nBack-substitute to solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill a - 2\\left(-2\\right)=1\\\\ \\hfill a+4=1\\\\ \\hfill a=-3\\end{array}[\/latex]<\/p>\r\nWrite another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rr}\\hfill 1b - 2d=0&amp; \\hfill {R}_{1}\\\\ \\hfill 2b - 3d=1&amp; \\hfill {R}_{2}\\end{array}[\/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{array}{r}\\hfill 1b - 2d=0\\\\ \\hfill \\frac{0+1d=1}{d=1}\\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nOnce more, back-substitute and solve for [latex]b[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill b - 2\\left(1\\right)=0\\\\ \\hfill b - 2=0\\\\ \\hfill b=2\\end{array}[\/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<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 2&amp; \\hfill &amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill &amp; \\hfill 3\\end{array}\\right][\/latex]<\/p>\r\naugment [latex]A[\/latex] with the identity\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr}\\hfill 2&amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill 3\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/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>Switch row 1 and row 2.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 5&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 0&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/li>\r\n \t<li>Multiply row 2 by [latex]-2[\/latex] and add to row 1.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 2&amp; \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/li>\r\n \t<li>Multiply row 1 by [latex]-2[\/latex] and add to row 2.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2&amp; \\hfill 1\\\\ \\hfill 5&amp; \\hfill -2\\end{array}\\right][\/latex]<\/li>\r\n \t<li>Add row 2 to row 1.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3&amp; \\hfill -1\\\\ \\hfill 5&amp; \\hfill -2\\end{array}\\right][\/latex]<\/li>\r\n \t<li>Multiply row 2 by [latex]-1[\/latex].\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3&amp; \\hfill -1\\\\ \\hfill -5&amp; \\hfill 2\\end{array}\\right][\/latex]<\/li>\r\n<\/ol>\r\nThe matrix we have found is [latex]{A}^{-1}[\/latex].\r\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill &amp; \\hfill -1\\\\ \\hfill -5&amp; \\hfill &amp; \\hfill 2\\end{array}\\right][\/latex]<\/p>\r\n\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<p 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]<\/p>\r\nthe multiplicative inverse of [latex]A[\/latex] is given by the formula\r\n<p 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]<\/p>\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 exercises\">\r\n<h3>Example: 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=\"11636\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"11636\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{\\left(1\\right)\\left(-3\\right)-\\left(-2\\right)\\left(2\\right)}\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\\\ =\\frac{1}{-3+4}\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/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[\\begin{array}{cc}1&amp; -2\\\\ 2&amp; -3\\end{array}|\\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[latex]\\left[\\begin{array}{cc}1&amp; -2\\\\ 0&amp; 1\\end{array}|\\begin{array}{cc}1&amp; 0\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/li>\r\n \t<li>Multiply row 1 by 2 and add to row 1.\r\n[latex]\\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}|\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/li>\r\n<\/ol>\r\nSo, we have verified our original solution.\r\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}-3&amp; 2\\\\ -2&amp; 1\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\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=\"161972\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"161972\"][latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}&amp; \\frac{1}{5}\\\\ -\\frac{2}{5}&amp; \\frac{1}{5}\\end{array}\\right][\/latex][\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6363&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"300\"><\/iframe>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: 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=\"930665\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"930665\"]\r\n\r\nWe will use the method of augmenting with the identity.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3&amp; 6\\\\ 1&amp; 3\\end{array}|\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\n\r\n<ol>\r\n \t<li>Switch row 1 and row 2.\r\n[latex]\\left[\\begin{array}{cc}1&amp; 3\\\\ 3&amp; 6\\text{ }\\end{array}\\text{ }\\text{ }\\text{ }|\\begin{array}{cc}0&amp; 1\\\\ 1&amp; 0\\end{array}\\right][\/latex]<\/li>\r\n \t<li>Multiply row 1 by \u22123 and add it to row 2.\r\n[latex]\\left[\\begin{array}{cc}1&amp; 2\\\\ 0&amp; 0\\end{array}|\\begin{array}{cc}1&amp; 0\\\\ -3&amp; 1\\end{array}\\right][\/latex]<\/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<h3>Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h3>\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<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]\r\naugment [latex]A[\/latex] with the identity matrix\r\n[latex]A|I=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\text{ }|\\text{ }\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\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 exercises\">\r\n<h3>Example: 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=\"144003\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"144003\"]\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[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}|\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right]\\stackrel{\\text{Interchange }{R}_{2}\\text{and }{R}_{1}}{\\to }\\left[\\begin{array}{ccc}3&amp; 3&amp; 1\\\\ 2&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}|\\begin{array}{ccc}0&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 2&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}|\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 2&amp; 3&amp; 1\\\\ 0&amp; 1&amp; 0\\end{array}|\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{R}_{3}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 2&amp; 3&amp; 1\\end{array}|\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 3&amp; 1\\end{array}|\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 3&amp; \\hfill -2&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}|\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 6&amp; \\hfill -2&amp; \\hfill -3\\end{array}\\right][\/latex]<\/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{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 6&amp; \\hfill -2&amp; \\hfill -3\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{ccc}2\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)&amp; 2\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)&amp; 2\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 3\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)&amp; 3\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)&amp; 3\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 2\\left(-1\\right)+4\\left(-1\\right)+1\\left(6\\right)&amp; 2\\left(1\\right)+4\\left(0\\right)+1\\left(-2\\right)&amp; 2\\left(0\\right)+4\\left(1\\right)+1\\left(-3\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right]\\hfill \\end{array}[\/latex]\r\n[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A=\\left[\\begin{array}{rrr}\\hfill -1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill -1&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 6&amp; \\hfill -2&amp; \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -1\\left(2\\right)+1\\left(3\\right)+0\\left(2\\right)&amp; \\hfill -1\\left(3\\right)+1\\left(3\\right)+0\\left(4\\right)&amp; \\hfill -1\\left(1\\right)+1\\left(1\\right)+0\\left(1\\right)\\\\ \\hfill -1\\left(2\\right)+0\\left(3\\right)+1\\left(2\\right)&amp; \\hfill -1\\left(3\\right)+0\\left(3\\right)+1\\left(4\\right)&amp; \\hfill -1\\left(1\\right)+0\\left(1\\right)+1\\left(1\\right)\\\\ \\hfill 6\\left(2\\right)+-2\\left(3\\right)+-3\\left(2\\right)&amp; \\hfill 6\\left(3\\right)+-2\\left(3\\right)+-3\\left(4\\right)&amp; \\hfill 6\\left(1\\right)+-2\\left(1\\right)+-3\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\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=\"202597\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"202597\"][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][\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=127903&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"450\"><\/iframe>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Verify that multiplying a matrix by it&#8217;s inverse results in 1<\/li>\n<li>Use matrix multiplication to find the inverse of a matrix<\/li>\n<li>Finding an inverse by augmenting with an identity matrix<\/li>\n<\/ul>\n<\/div>\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<p style=\"text-align: center;\">[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 0& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>The identity matrix acts as a 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<p style=\"text-align: center;\">[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/p>\n<p>A matrix that has a multiplicative inverse is called an <strong>invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.<\/p>\n<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}{l}\\hfill\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2}=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right]\\begin{array}{cccc}& & & \\end{array}{I}_{3}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\text{ }2\\times 2\\text{ 3}\\times 3\\hfill \\end{array}\\hfill \\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 exercises\">\n<h3>Example: Showing That the Identity Matrix Acts as a 1<\/h3>\n<p>Given matrix <em>A<\/em>, show that [latex]AI=IA=A[\/latex].<br \/>\n[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=\"q38185\">Solution<\/span><\/p>\n<div id=\"q38185\" 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]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0& \\hfill & \\hfill & \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0& \\hfill & \\hfill & \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/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 exercises\">\n<h3>Example: 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=\"q812081\">Solution<\/span><\/p>\n<div id=\"q812081\" class=\"hidden-answer\" style=\"display: none\">\n<p>Multiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)& \\hfill & \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)& \\hfill & \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}BA=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)& \\hfill & \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)& \\hfill & \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p>[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\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=\"q159815\">Solution<\/span><\/p>\n<div id=\"q159815\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1\\left(-3\\right)+4\\left(1\\right)& \\hfill & \\hfill 1\\left(-4\\right)+4\\left(1\\right)\\\\ \\hfill -1\\left(-3\\right)+-3\\left(1\\right)& \\hfill & \\hfill -1\\left(-4\\right)+-3\\left(1\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ BA=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -3\\left(1\\right)+-4\\left(-1\\right)& \\hfill & \\hfill -3\\left(4\\right)+-4\\left(-3\\right)\\\\ \\hfill 1\\left(1\\right)+1\\left(-1\\right)& \\hfill & \\hfill 1\\left(4\\right)+1\\left(-3\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/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 exercises\">\n<h3>Example: 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=\"q31546\">Solution<\/span><\/p>\n<div id=\"q31546\" 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]\\text{ }\\left[\\begin{array}{rr}\\hfill a& \\hfill b\\\\ \\hfill c& \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/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]\\text{ }\\left[\\begin{array}{rr}\\hfill a& \\hfill b\\\\ \\hfill c& \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1a - 2c& \\hfill 1b - 2d\\\\ \\hfill 2a - 3c& \\hfill 2b - 3d\\end{array}\\right][\/latex]<\/p>\n<p>Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}1a - 2c=1\\text{ }{R}_{1}\\\\ 2a - 3c=0\\text{ }{R}_{2}\\end{array}[\/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{array}{r}\\hfill 1a - 2c=1\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/p>\n<p>Back-substitute to solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill a - 2\\left(-2\\right)=1\\\\ \\hfill a+4=1\\\\ \\hfill a=-3\\end{array}[\/latex]<\/p>\n<p>Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rr}\\hfill 1b - 2d=0& \\hfill {R}_{1}\\\\ \\hfill 2b - 3d=1& \\hfill {R}_{2}\\end{array}[\/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{array}{r}\\hfill 1b - 2d=0\\\\ \\hfill \\frac{0+1d=1}{d=1}\\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>Once more, back-substitute and solve for [latex]b[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill b - 2\\left(1\\right)=0\\\\ \\hfill b - 2=0\\\\ \\hfill b=2\\end{array}[\/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<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill 2& \\hfill & \\hfill 1\\\\ \\hfill 5& \\hfill & \\hfill 3\\end{array}\\right][\/latex]<\/p>\n<p>augment [latex]A[\/latex] with the identity<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rr}\\hfill 2& \\hfill 1\\\\ \\hfill 5& \\hfill 3\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\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.<br \/>\n[latex]\\left[\\begin{array}{rr}\\hfill 5& \\hfill 3\\\\ \\hfill 2& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0\\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 2 by [latex]-2[\/latex] and add to row 1.<br \/>\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 2& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2& \\hfill 1\\\\ \\hfill 1& \\hfill 0\\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 1 by [latex]-2[\/latex] and add to row 2.<br \/>\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2& \\hfill 1\\\\ \\hfill 5& \\hfill -2\\end{array}\\right][\/latex]<\/li>\n<li>Add row 2 to row 1.<br \/>\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3& \\hfill -1\\\\ \\hfill 5& \\hfill -2\\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 2 by [latex]-1[\/latex].<br \/>\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3& \\hfill -1\\\\ \\hfill -5& \\hfill 2\\end{array}\\right][\/latex]<\/li>\n<\/ol>\n<p>The matrix we have found is [latex]{A}^{-1}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill -1\\\\ \\hfill -5& \\hfill & \\hfill 2\\end{array}\\right][\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill a& \\hfill & \\hfill b\\\\ \\hfill c& \\hfill & \\hfill d\\end{array}\\right][\/latex]<\/p>\n<p>the multiplicative inverse of [latex]A[\/latex] is given by the formula<\/p>\n<p 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]<\/p>\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 exercises\">\n<h3>Example: 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=\"q11636\">Solution<\/span><\/p>\n<div id=\"q11636\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{\\left(1\\right)\\left(-3\\right)-\\left(-2\\right)\\left(2\\right)}\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\\\ =\\frac{1}{-3+4}\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/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[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}|\\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.<br \/>\n[latex]\\left[\\begin{array}{cc}1& -2\\\\ 0& 1\\end{array}|\\begin{array}{cc}1& 0\\\\ -2& 1\\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 1 by 2 and add to row 1.<br \/>\n[latex]\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}|\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/li>\n<\/ol>\n<p>So, we have verified our original solution.<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\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=\"q161972\">Solution<\/span><\/p>\n<div id=\"q161972\" class=\"hidden-answer\" style=\"display: none\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}& \\frac{1}{5}\\\\ -\\frac{2}{5}& \\frac{1}{5}\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6363&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"300\"><\/iframe>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: 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=\"q930665\">Solution<\/span><\/p>\n<div id=\"q930665\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will use the method of augmenting with the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3& 6\\\\ 1& 3\\end{array}|\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/p>\n<ol>\n<li>Switch row 1 and row 2.<br \/>\n[latex]\\left[\\begin{array}{cc}1& 3\\\\ 3& 6\\text{ }\\end{array}\\text{ }\\text{ }\\text{ }|\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right][\/latex]<\/li>\n<li>Multiply row 1 by \u22123 and add it to row 2.<br \/>\n[latex]\\left[\\begin{array}{cc}1& 2\\\\ 0& 0\\end{array}|\\begin{array}{cc}1& 0\\\\ -3& 1\\end{array}\\right][\/latex]<\/li>\n<li>There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3>Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h3>\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<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<br \/>\naugment [latex]A[\/latex] with the identity matrix<br \/>\n[latex]A|I=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\text{ }|\\text{ }\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/p>\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 exercises\">\n<h3>Example: 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=\"q144003\">Solution<\/span><\/p>\n<div id=\"q144003\" 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[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\stackrel{\\text{Interchange }{R}_{2}\\text{and }{R}_{1}}{\\to }\\left[\\begin{array}{ccc}3& 3& 1\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{ccc}0& 1& 0\\\\ 1& 0& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 0& 1& 0\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{R}_{3}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 2& 3& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 3& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 3& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right][\/latex]<\/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{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{ccc}2\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& 2\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& 2\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 3\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& 3\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& 3\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 2\\left(-1\\right)+4\\left(-1\\right)+1\\left(6\\right)& 2\\left(1\\right)+4\\left(0\\right)+1\\left(-2\\right)& 2\\left(0\\right)+4\\left(1\\right)+1\\left(-3\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<br \/>\n[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -1\\left(2\\right)+1\\left(3\\right)+0\\left(2\\right)& \\hfill -1\\left(3\\right)+1\\left(3\\right)+0\\left(4\\right)& \\hfill -1\\left(1\\right)+1\\left(1\\right)+0\\left(1\\right)\\\\ \\hfill -1\\left(2\\right)+0\\left(3\\right)+1\\left(2\\right)& \\hfill -1\\left(3\\right)+0\\left(3\\right)+1\\left(4\\right)& \\hfill -1\\left(1\\right)+0\\left(1\\right)+1\\left(1\\right)\\\\ \\hfill 6\\left(2\\right)+-2\\left(3\\right)+-3\\left(2\\right)& \\hfill 6\\left(3\\right)+-2\\left(3\\right)+-3\\left(4\\right)& \\hfill 6\\left(1\\right)+-2\\left(1\\right)+-3\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\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=\"q202597\">Solution<\/span><\/p>\n<div id=\"q202597\" class=\"hidden-answer\" style=\"display: none\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}1& 1& 2\\\\ 2& 4& -3\\\\ 3& 6& -5\\end{array}\\right][\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=127903&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"450\"><\/iframe>\n<\/div>\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-2291\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul 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href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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