{"id":333,"date":"2023-06-21T13:22:57","date_gmt":"2023-06-21T13:22:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/row-operations-and-the-augmented-matrix\/"},"modified":"2023-08-10T23:47:34","modified_gmt":"2023-08-10T23:47:34","slug":"row-operations-and-the-augmented-matrix","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/row-operations-and-the-augmented-matrix\/","title":{"raw":"\u25aa   Row Operations and Augmented Matrices","rendered":"\u25aa   Row Operations and Augmented Matrices"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the augmented matrix for a system of equations.<\/li>\r\n \t<li>Perform row operations on an augmented matrix.<\/li>\r\n<\/ul>\r\n<\/div>\r\nA <strong>matrix<\/strong> can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an <strong>augmented matrix<\/strong>.\r\n\r\nFor example, consider the following [latex]2\\times 2[\/latex] system of equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x+4y=7\\\\ 4x - 2y=5\\end{array}[\/latex]<\/div>\r\nWe can write this system as an augmented matrix:\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 3&amp; \\hfill 4&amp; \\hfill 7\\\\ \\hfill 4&amp; \\hfill -2&amp; \\hfill 5\\\\ \\end{array}\\right][\/latex]<\/div>\r\nWe can also write a matrix containing just the coefficients. This is called the <strong>coefficient matrix<\/strong>.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3&amp; 4\\\\ 4&amp; -2\\end{array}\\right][\/latex]<\/div>\r\nA three-by-three <strong>system of equations<\/strong> such as\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x-y-z=0\\hfill \\\\ \\text{ }x+y=5\\hfill \\\\ \\text{ }2x - 3z=2\\hfill \\end{array}[\/latex]<\/div>\r\nhas a coefficient matrix\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{rrr}\\hfill 3&amp; \\hfill -1&amp; \\hfill -1\\\\ \\hfill 1&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -3\\end{array}\\right][\/latex]<\/div>\r\nand is represented by the augmented matrix\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 3&amp; \\hfill -1&amp; \\hfill -1&amp; \\hfill 0\\\\ \\hfill 1&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -3&amp; \\hfill 2\\\\ \\end{array}\\right][\/latex]<\/p>\r\nNotice that the matrix is written so that the variables line up in their own columns: [latex]x-[\/latex]terms go in the first column, [latex]y-[\/latex]terms in the second column, and [latex]z-[\/latex]terms in the third column. It is very important that each equation is written in standard form [latex]ax+by+cz=d[\/latex] so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, write an augmented matrix<\/h3>\r\n<ol>\r\n \t<li>Write the coefficients of the [latex]x-[\/latex]terms as the numbers down the first column.<\/li>\r\n \t<li>Write the coefficients of the [latex]y-[\/latex]terms as the numbers down the second column.<\/li>\r\n \t<li>If there are [latex]z-[\/latex]terms, write the coefficients as the numbers down the third column.<\/li>\r\n \t<li>Draw a vertical line and write the constants to the right of the line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Augmented Matrix for a System of Equations<\/h3>\r\nWrite the augmented matrix for the given system of equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+2y-z=3\\hfill \\\\ \\text{ }2x-y+2z=6\\hfill \\\\ \\text{ }x - 3y+3z=4\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"863191\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"863191\"]\r\n\r\nThe augmented matrix displays the coefficients of the variables and an additional column for the constants.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 2&amp; \\hfill -1&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill -1&amp; \\hfill 2&amp; \\hfill 6\\\\ \\hfill 1&amp; \\hfill -3&amp; \\hfill 3&amp; \\hfill 4\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the augmented matrix of the given system of equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4x - 3y=11\\\\ 3x+2y=4\\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[reveal-answer q=\"769522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"769522\"][latex]\\left[\\begin{array}{cc|c}\\hfill 4&amp; \\hfill -3&amp; \\hfill 11\\\\ \\hfill 3&amp; \\hfill 2&amp; \\hfill 4\\\\ \\end{array}\\right][\/latex][\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div>[ohm_question height=\"300\"]125538[\/ohm_question]<\/div>\r\n<\/div>\r\n<h2>Writing a System of Equations from an Augmented Matrix<\/h2>\r\nWe can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the <strong>system of equations<\/strong> in standard form.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing a System of Equations from an Augmented Matrix Form<\/h3>\r\nFind the system of equations from the augmented matrix.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill -5&amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill -5&amp; \\hfill -4&amp; \\hfill 5\\\\ \\hfill -3&amp; \\hfill 5&amp; \\hfill 4&amp; \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\r\n[reveal-answer q=\"811290\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"811290\"]\r\nWhen the columns represent the variables [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex],\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill -5&amp; \\hfill -2\\\\ \\hfill 2&amp; \\hfill -5&amp; \\hfill -4&amp; \\hfill 5\\\\ \\hfill -3&amp; \\hfill 5&amp; \\hfill 4&amp; \\hfill 6\\\\ \\end{array}\\right]\\to \\begin{array}{l}x - 3y - 5z=-2\\hfill \\\\ 2x - 5y - 4z=5\\hfill \\\\ -3x+5y+4z=6\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the system of equations from the augmented matrix.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 5\\\\ \\hfill 2&amp; \\hfill -1&amp; \\hfill 3&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill -9\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<div>[reveal-answer q=\"696438\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"696438\"][latex]\\begin{array}{c}x-y+z=5\\\\ 2x-y+3z=1\\\\ y+z=-9\\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div>[ohm_question height=\"325\"]84689[\/ohm_question]<\/div>\r\n<\/div>\r\n<div>\r\n<h2>Row Operations<\/h2>\r\nNow that we can write systems of equations in augmented matrix form, we will examine the various <strong>row operations<\/strong> that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows.\r\n\r\nPerforming row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to <strong>row-echelon form<\/strong>, in which there are ones down the <strong>main diagonal<\/strong> from the upper left corner to the lower right corner and zeros in every position below the main diagonal as shown, or to <strong>reduced row-echelon form<\/strong>, in which there are ones down the main diagonal and zeros in everywhere else.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[latex]\\begin{array}{c}\\text{Row-echelon form}\\\\ \\left[\\begin{array}{ccc}1&amp; a&amp; b\\\\ 0&amp; 1&amp; d\\\\ 0&amp; 0&amp; 1\\end{array}\\right]\\end{array}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\begin{array}{c}\\text{Reduced row-echelon form}\\\\ \\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right]\\end{array}[\/latex]<\/span>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong><span style=\"color: #339966;\">[Optional]<\/span><\/strong>\r\n\r\nWe use row operations corresponding to equation operations to obtain a new matrix that is <strong>row-equivalent<\/strong> in a simpler form. Here are the guidelines to obtaining row-echelon form.\r\n<ol>\r\n \t<li>In any nonzero row, the first nonzero number is a 1. It is called a <em>leading<\/em> 1.<\/li>\r\n \t<li>Any all-zero rows are placed at the bottom of the matrix.<\/li>\r\n \t<li>Any leading 1 is below and to the right of a previous leading 1.<\/li>\r\n \t<li>Any column containing a leading 1 has zeros in all other positions in the column.<\/li>\r\n<\/ol>\r\nTo solve a system of equations we can perform the following row operations to convert the <strong>coefficient matrix<\/strong> to row-echelon form and do back-substitution to find the solution.\r\n<ol>\r\n \t<li>Interchange rows. (Notation: [latex]{R}_{i}\\leftrightarrow {R}_{j}[\/latex] )<\/li>\r\n \t<li>Multiply a row by a constant. (Notation: [latex]c{R}_{i}[\/latex] )<\/li>\r\n \t<li>Add the product of a row multiplied by a constant to another row. (Notation: [latex]{R}_{i}+c{R}_{j}[\/latex])<\/li>\r\n<\/ol>\r\nEach of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nAs with other challenging concepts you've encountered in this course, time and patient practice with the process of Gaussian elimination will aid familiarity.\r\n\r\nIf the instructions seem intimidating, work out the example below on paper once or twice, then try to apply the General Note and How To information to the example.\r\n\r\nWork back and forth between them until you gain a firmer understanding. Then try the practice problems.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Gaussian Elimination<\/h3>\r\nThe <strong>Gaussian elimination<\/strong> method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix [latex]A[\/latex] with the number 1 as the entry down the main diagonal and have all zeros below.\r\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill {a}_{11}&amp; \\hfill {a}_{12}&amp; \\hfill {a}_{13}\\\\ \\hfill {a}_{21}&amp; \\hfill {a}_{22}&amp; \\hfill {a}_{23}\\\\ \\hfill {a}_{31}&amp; \\hfill {a}_{32}&amp; \\hfill {a}_{33}\\end{array}\\right]\\stackrel{\\text{After Gaussian elimination}}{\\to }A=\\left[\\begin{array}{rrr}\\hfill 1&amp; \\hfill {b}_{12}&amp; \\hfill {b}_{13}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill {b}_{23}\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/div>\r\n<div><\/div>\r\n<div>The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below.<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an augmented matrix, perform row operations to achieve row-echelon form<\/h3>\r\n<ol>\r\n \t<li>The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.<\/li>\r\n \t<li>Use row operations to obtain zeros down the first column below the first entry of 1.<\/li>\r\n \t<li>Use row operations to obtain a 1 in row 2, column 2.<\/li>\r\n \t<li>Use row operations to obtain zeros down column 2, below the entry of 1.<\/li>\r\n \t<li>Use row operations to obtain a 1 in row 3, column 3.<\/li>\r\n \t<li>Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below.<\/li>\r\n \t<li>If any rows contain all zeros, place them at the bottom.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<strong><span style=\"color: #339966;\">* We will use \"ref\" for row-echelon form and \"rref\" for reduced row-echelon form in a graphing calculator for this process. Refer to \"<a href=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/how-to-use-ti-84-plus-ce-or-any-ti-84-family#Using Matrices in TI\" target=\"_blank\" rel=\"noopener\">Using Matrices to Solve Linear System on TI-84 Plus CE<\/a>\" in \"How to Use TI-84 Plus CE (or any TI-84 family)\" section for calculator direction. Please note that \"ref\" from the graphing calculator might be different from the ones here. However, its \"rref\" forms are always the same regardless of the methods.\u00a0<\/span><\/strong>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Performing Row Operations on a 3\u00d73 Augmented Matrix to Obtain Row-Echelon Form<\/h3>\r\nPerform row operations on the given matrix to obtain row-echelon form and reduced row-echelon form.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill -5&amp; \\hfill 6&amp; \\hfill 6\\\\ \\hfill -3&amp; \\hfill 3&amp; \\hfill 4&amp; \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\r\n[reveal-answer q=\"637835\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"637835\"]\r\n<div class=\"textbox shaded\">\r\n\r\nThe first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by [latex]-2[\/latex] and add it to row 2. Then replace row 2 with the result.\r\n<div style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill -3&amp; \\hfill 3&amp; \\hfill 4&amp; \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\r\nNext, obtain a zero in row 3, column 1.\r\n<div style=\"text-align: center;\">[latex]3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill -6&amp; \\hfill 16&amp; \\hfill 15\\\\ \\end{array}\\right][\/latex]<\/div>\r\nNext, obtain a zero in row 3, column 2.\r\n<div style=\"text-align: center;\">[latex]6{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 4&amp; \\hfill 15\\\\ \\end{array}\\right][\/latex]<\/div>\r\nThe last step is to obtain a 1 in row 3, column 3.\r\n<div style=\"text-align: center;\">[latex]\\frac{1}{4}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<\/div>\r\nSo, the row-echelon form is\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n\r\nNow, obtain a zero in row 2, column 3.\r\n<p style=\"text-align: center;\">[latex]2{R}_{3}+{R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -3&amp; \\hfill 4 &amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill \\frac{15}{2}\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]3{R}_{2}-4{R}_{3}+{R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0 &amp; \\hfill \\frac{21}{2}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill \\frac{15}{2}\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\nThus, the reduced row-echelon form is\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0 &amp; \\hfill \\frac{21}{2}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill \\frac{15}{2}\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\r\n\r\n<div style=\"text-align: center;\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the system of equations in row-echelon form using \"ref\" in a graphing calculator. Then write it in reduced row-echelon form using \"rref\" in a graphing calculator.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x - 2y+3z=9\\hfill \\\\ \\text{ }-x+3y=-4\\hfill \\\\ 2x - 5y+5z=17\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"971348\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"971348\"]<\/p>\r\n<p style=\"text-align: left;\">1) Row-echelon form:<\/p>\r\n<p style=\"padding-left: 30px; text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -\\frac{5}{2}&amp; \\hfill \\frac{5}{2}&amp; \\hfill \\frac{17}{2}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 5&amp; \\hfill 9\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 2\\\\ \\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: left;\">2) Reduced row-echelon form:<\/p>\r\n<p style=\"padding-left: 30px; text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill -1\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 2\\\\ \\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n<div>[ohm_question height=\"315\"]125521[\/ohm_question]<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the augmented matrix for a system of equations.<\/li>\n<li>Perform row operations on an augmented matrix.<\/li>\n<\/ul>\n<\/div>\n<p>A <strong>matrix<\/strong> can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an <strong>augmented matrix<\/strong>.<\/p>\n<p>For example, consider the following [latex]2\\times 2[\/latex] system of equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x+4y=7\\\\ 4x - 2y=5\\end{array}[\/latex]<\/div>\n<p>We can write this system as an augmented matrix:<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 3& \\hfill 4& \\hfill 7\\\\ \\hfill 4& \\hfill -2& \\hfill 5\\\\ \\end{array}\\right][\/latex]<\/div>\n<p>We can also write a matrix containing just the coefficients. This is called the <strong>coefficient matrix<\/strong>.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3& 4\\\\ 4& -2\\end{array}\\right][\/latex]<\/div>\n<p>A three-by-three <strong>system of equations<\/strong> such as<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x-y-z=0\\hfill \\\\ \\text{ }x+y=5\\hfill \\\\ \\text{ }2x - 3z=2\\hfill \\end{array}[\/latex]<\/div>\n<p>has a coefficient matrix<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{rrr}\\hfill 3& \\hfill -1& \\hfill -1\\\\ \\hfill 1& \\hfill 1& \\hfill 0\\\\ \\hfill 2& \\hfill 0& \\hfill -3\\end{array}\\right][\/latex]<\/div>\n<p>and is represented by the augmented matrix<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 3& \\hfill -1& \\hfill -1& \\hfill 0\\\\ \\hfill 1& \\hfill 1& \\hfill 0& \\hfill 5\\\\ \\hfill 2& \\hfill 0& \\hfill -3& \\hfill 2\\\\ \\end{array}\\right][\/latex]<\/p>\n<p>Notice that the matrix is written so that the variables line up in their own columns: [latex]x-[\/latex]terms go in the first column, [latex]y-[\/latex]terms in the second column, and [latex]z-[\/latex]terms in the third column. It is very important that each equation is written in standard form [latex]ax+by+cz=d[\/latex] so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, write an augmented matrix<\/h3>\n<ol>\n<li>Write the coefficients of the [latex]x-[\/latex]terms as the numbers down the first column.<\/li>\n<li>Write the coefficients of the [latex]y-[\/latex]terms as the numbers down the second column.<\/li>\n<li>If there are [latex]z-[\/latex]terms, write the coefficients as the numbers down the third column.<\/li>\n<li>Draw a vertical line and write the constants to the right of the line.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Augmented Matrix for a System of Equations<\/h3>\n<p>Write the augmented matrix for the given system of equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+2y-z=3\\hfill \\\\ \\text{ }2x-y+2z=6\\hfill \\\\ \\text{ }x - 3y+3z=4\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q863191\">Show Solution<\/span><\/p>\n<div id=\"q863191\" class=\"hidden-answer\" style=\"display: none\">\n<p>The augmented matrix displays the coefficients of the variables and an additional column for the constants.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 3\\\\ \\hfill 2& \\hfill -1& \\hfill 2& \\hfill 6\\\\ \\hfill 1& \\hfill -3& \\hfill 3& \\hfill 4\\\\ \\end{array}\\right][\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the augmented matrix of the given system of equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4x - 3y=11\\\\ 3x+2y=4\\end{array}[\/latex]<\/div>\n<div><\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q769522\">Show Solution<\/span><\/p>\n<div id=\"q769522\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left[\\begin{array}{cc|c}\\hfill 4& \\hfill -3& \\hfill 11\\\\ \\hfill 3& \\hfill 2& \\hfill 4\\\\ \\end{array}\\right][\/latex]<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div><iframe loading=\"lazy\" id=\"ohm125538\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=125538&theme=oea&iframe_resize_id=ohm125538&show_question_numbers\" width=\"100%\" height=\"300\"><\/iframe><\/div>\n<\/div>\n<h2>Writing a System of Equations from an Augmented Matrix<\/h2>\n<p>We can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the <strong>system of equations<\/strong> in standard form.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing a System of Equations from an Augmented Matrix Form<\/h3>\n<p>Find the system of equations from the augmented matrix.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill -5& \\hfill -2\\\\ \\hfill 2& \\hfill -5& \\hfill -4& \\hfill 5\\\\ \\hfill -3& \\hfill 5& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q811290\">Show Solution<\/span><\/p>\n<div id=\"q811290\" class=\"hidden-answer\" style=\"display: none\">\nWhen the columns represent the variables [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex],<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill -5& \\hfill -2\\\\ \\hfill 2& \\hfill -5& \\hfill -4& \\hfill 5\\\\ \\hfill -3& \\hfill 5& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right]\\to \\begin{array}{l}x - 3y - 5z=-2\\hfill \\\\ 2x - 5y - 4z=5\\hfill \\\\ -3x+5y+4z=6\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the system of equations from the augmented matrix.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 5\\\\ \\hfill 2& \\hfill -1& \\hfill 3& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill 1& \\hfill -9\\\\ \\end{array}\\right][\/latex]<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q696438\">Show Solution<\/span><\/p>\n<div id=\"q696438\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}x-y+z=5\\\\ 2x-y+3z=1\\\\ y+z=-9\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div><iframe loading=\"lazy\" id=\"ohm84689\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=84689&theme=oea&iframe_resize_id=ohm84689&show_question_numbers\" width=\"100%\" height=\"325\"><\/iframe><\/div>\n<\/div>\n<div>\n<h2>Row Operations<\/h2>\n<p>Now that we can write systems of equations in augmented matrix form, we will examine the various <strong>row operations<\/strong> that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows.<\/p>\n<p>Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to <strong>row-echelon form<\/strong>, in which there are ones down the <strong>main diagonal<\/strong> from the upper left corner to the lower right corner and zeros in every position below the main diagonal as shown, or to <strong>reduced row-echelon form<\/strong>, in which there are ones down the main diagonal and zeros in everywhere else.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\begin{array}{c}\\text{Row-echelon form}\\\\ \\left[\\begin{array}{ccc}1& a& b\\\\ 0& 1& d\\\\ 0& 0& 1\\end{array}\\right]\\end{array}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\begin{array}{c}\\text{Reduced row-echelon form}\\\\ \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\end{array}[\/latex]<\/span><\/p>\n<div class=\"textbox shaded\">\n<p><strong><span style=\"color: #339966;\">[Optional]<\/span><\/strong><\/p>\n<p>We use row operations corresponding to equation operations to obtain a new matrix that is <strong>row-equivalent<\/strong> in a simpler form. Here are the guidelines to obtaining row-echelon form.<\/p>\n<ol>\n<li>In any nonzero row, the first nonzero number is a 1. It is called a <em>leading<\/em> 1.<\/li>\n<li>Any all-zero rows are placed at the bottom of the matrix.<\/li>\n<li>Any leading 1 is below and to the right of a previous leading 1.<\/li>\n<li>Any column containing a leading 1 has zeros in all other positions in the column.<\/li>\n<\/ol>\n<p>To solve a system of equations we can perform the following row operations to convert the <strong>coefficient matrix<\/strong> to row-echelon form and do back-substitution to find the solution.<\/p>\n<ol>\n<li>Interchange rows. (Notation: [latex]{R}_{i}\\leftrightarrow {R}_{j}[\/latex] )<\/li>\n<li>Multiply a row by a constant. (Notation: [latex]c{R}_{i}[\/latex] )<\/li>\n<li>Add the product of a row multiplied by a constant to another row. (Notation: [latex]{R}_{i}+c{R}_{j}[\/latex])<\/li>\n<\/ol>\n<p>Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>As with other challenging concepts you&#8217;ve encountered in this course, time and patient practice with the process of Gaussian elimination will aid familiarity.<\/p>\n<p>If the instructions seem intimidating, work out the example below on paper once or twice, then try to apply the General Note and How To information to the example.<\/p>\n<p>Work back and forth between them until you gain a firmer understanding. Then try the practice problems.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Gaussian Elimination<\/h3>\n<p>The <strong>Gaussian elimination<\/strong> method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix [latex]A[\/latex] with the number 1 as the entry down the main diagonal and have all zeros below.<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{rrr}\\hfill {a}_{11}& \\hfill {a}_{12}& \\hfill {a}_{13}\\\\ \\hfill {a}_{21}& \\hfill {a}_{22}& \\hfill {a}_{23}\\\\ \\hfill {a}_{31}& \\hfill {a}_{32}& \\hfill {a}_{33}\\end{array}\\right]\\stackrel{\\text{After Gaussian elimination}}{\\to }A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill {b}_{12}& \\hfill {b}_{13}\\\\ \\hfill 0& \\hfill 1& \\hfill {b}_{23}\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<div><\/div>\n<div>The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below.<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an augmented matrix, perform row operations to achieve row-echelon form<\/h3>\n<ol>\n<li>The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.<\/li>\n<li>Use row operations to obtain zeros down the first column below the first entry of 1.<\/li>\n<li>Use row operations to obtain a 1 in row 2, column 2.<\/li>\n<li>Use row operations to obtain zeros down column 2, below the entry of 1.<\/li>\n<li>Use row operations to obtain a 1 in row 3, column 3.<\/li>\n<li>Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below.<\/li>\n<li>If any rows contain all zeros, place them at the bottom.<\/li>\n<\/ol>\n<\/div>\n<p><strong><span style=\"color: #339966;\">* We will use &#8220;ref&#8221; for row-echelon form and &#8220;rref&#8221; for reduced row-echelon form in a graphing calculator for this process. Refer to &#8220;<a href=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/how-to-use-ti-84-plus-ce-or-any-ti-84-family#Using Matrices in TI\" target=\"_blank\" rel=\"noopener\">Using Matrices to Solve Linear System on TI-84 Plus CE<\/a>&#8221; in &#8220;How to Use TI-84 Plus CE (or any TI-84 family)&#8221; section for calculator direction. Please note that &#8220;ref&#8221; from the graphing calculator might be different from the ones here. However, its &#8220;rref&#8221; forms are always the same regardless of the methods.\u00a0<\/span><\/strong><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Performing Row Operations on a 3\u00d73 Augmented Matrix to Obtain Row-Echelon Form<\/h3>\n<p>Perform row operations on the given matrix to obtain row-echelon form and reduced row-echelon form.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 2& \\hfill -5& \\hfill 6& \\hfill 6\\\\ \\hfill -3& \\hfill 3& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q637835\">Show Solution<\/span><\/p>\n<div id=\"q637835\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"textbox shaded\">\n<p>The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by [latex]-2[\/latex] and add it to row 2. Then replace row 2 with the result.<\/p>\n<div style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill -3& \\hfill 3& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\n<p>Next, obtain a zero in row 3, column 1.<\/p>\n<div style=\"text-align: center;\">[latex]3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill -6& \\hfill 16& \\hfill 15\\\\ \\end{array}\\right][\/latex]<\/div>\n<p>Next, obtain a zero in row 3, column 2.<\/p>\n<div style=\"text-align: center;\">[latex]6{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 4& \\hfill 15\\\\ \\end{array}\\right][\/latex]<\/div>\n<p>The last step is to obtain a 1 in row 3, column 3.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{1}{4}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/div>\n<\/div>\n<p>So, the row-echelon form is<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\n<div class=\"textbox shaded\">\n<p>Now, obtain a zero in row 2, column 3.<\/p>\n<p style=\"text-align: center;\">[latex]2{R}_{3}+{R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4 & \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill \\frac{15}{2}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]3{R}_{2}-4{R}_{3}+{R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0 & \\hfill \\frac{21}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill \\frac{15}{2}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\n<\/div>\n<p>Thus, the reduced row-echelon form is<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0 & \\hfill \\frac{21}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill \\frac{15}{2}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/p>\n<div style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the system of equations in row-echelon form using &#8220;ref&#8221; in a graphing calculator. Then write it in reduced row-echelon form using &#8220;rref&#8221; in a graphing calculator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x - 2y+3z=9\\hfill \\\\ \\text{ }-x+3y=-4\\hfill \\\\ 2x - 5y+5z=17\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q971348\">Show Solution<\/span><\/p>\n<div id=\"q971348\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">1) Row-echelon form:<\/p>\n<p style=\"padding-left: 30px; text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -\\frac{5}{2}& \\hfill \\frac{5}{2}& \\hfill \\frac{17}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 5& \\hfill 9\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 2\\\\ \\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: left;\">2) Reduced row-echelon form:<\/p>\n<p style=\"padding-left: 30px; text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill -1\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 2\\\\ \\end{array}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div><iframe loading=\"lazy\" id=\"ohm125521\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=125521&theme=oea&iframe_resize_id=ohm125521&show_question_numbers\" width=\"100%\" height=\"315\"><\/iframe><\/div>\n<\/div>\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-333\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Reduced Row-Echelon . <strong>Authored by<\/strong>: Michelle Eunhee Chung. <strong>Provided by<\/strong>: Georgia State University. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 125538. <strong>Authored by<\/strong>: Smart,Jim. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 84689. <strong>Authored by<\/strong>: Shelton,Alison. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: MathAS Community License CC-BY + GPL<\/li><\/ul><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":395986,"menu_order":14,"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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