{"id":1221,"date":"2023-08-01T14:33:38","date_gmt":"2023-08-01T14:33:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/?post_type=chapter&#038;p=1221"},"modified":"2023-08-10T23:48:35","modified_gmt":"2023-08-10T23:48:35","slug":"solving-a-system-with-gauss-jordan-elimination","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solving-a-system-with-gauss-jordan-elimination\/","title":{"raw":"\u25aa   Solving a System with Gauss-Jordan Elimination","rendered":"\u25aa   Solving a System with Gauss-Jordan Elimination"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use Gauss-Jordan elimination to solve a system of equations represented as an augmented matrix.<\/li>\r\n \t<li>Interpret the solution to a system of equations represented as an augmented matrix.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe have seen how to use Gaussian elimination as\u00a0a tool for solving a system written as an augmented matrix. Now we will see how to use Gauss-Jordan elimination, which is very similar to Gaussian elimination. We will review the same examples we used in the previous section.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a 2 X 2 System by Gauss-Jordan Elimination<\/h3>\r\nSolve the given system by Gauss-Jordan elimination.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}2x+3y=6\\hfill \\\\ \\text{ }x-y=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"264825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"264825\"]\r\n\r\nFirst, we write this as an augmented matrix.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 2&amp; \\hfill 3&amp; \\hfill 6\\\\ \\hfill 1&amp; \\hfill -1&amp; \\hfill \\frac{1}{2}\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n\r\nWe want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.\r\n<div style=\"text-align: center;\">[latex]{R}_{1}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill \\frac{1}{2}\\\\ \\hfill 2&amp; \\hfill 3&amp; \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\r\nWe now have a 1 as the first entry in row 1, column 1. Now let\u2019s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by [latex]-2[\/latex] and then adding the result to row 2.\r\n<div style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill \\frac{1}{2}\\\\ \\hfill 0&amp; \\hfill 5&amp; \\hfill 5\\\\ \\end{array}\\right][\/latex]<\/div>\r\nNext, we need a 1 in row 2, column 2. This can be accomplished by multiplying row 2 by [latex]\\frac{1}{5}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{5}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill \\frac{1}{2}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\\\ \\end{array}\\right][\/latex]<\/p>\r\nLastly, we need a 0 in row 1, column 2, and we can accomplish it by simply adding row 2 to row 1.\r\n<p style=\"text-align: center;\">[latex]{R}_{1}+{R}_{2}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill \\frac{3}{2}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\\\ \\end{array}\\right][\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using \"rref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill \\frac{3}{2}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 1 \\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>The first row of the matrix represents [latex]x=\\frac{3}{2}[\/latex] and the second row of the matrix represents [latex]y=1[\/latex]. No back-substitute is needed in Gauss-Jordan elimination.<\/div>\r\nThe solution is the point [latex]\\left(\\frac{3}{2},1\\right)[\/latex] as we found in the previous section.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Dependent System<\/h3>\r\nSolve the system of equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x+4y=12\\\\ 6x+8y=24\\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"844840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"844840\"]\r\n\r\nFirst, we write this as an augmented matrix.\r\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 3&amp; \\hfill 4&amp; \\hfill 12\\\\ \\hfill 6&amp; \\hfill 8&amp; \\hfill 24\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<div style=\"text-align: center;\"><\/div>\r\n<div class=\"textbox shaded\">\r\n\r\nPerform <strong>row operations<\/strong> on the augmented matrix to try and achieve <strong>reduced\u00a0row-echelon form<\/strong>.\r\n<div style=\"text-align: center;\">[latex]-\\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\hfill 6&amp; \\hfill 8&amp; \\hfill 24\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<div style=\"text-align: center;\"><\/div>\r\n<div style=\"text-align: center;\">[latex]{R}_{1}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 6&amp; \\hfill 8&amp; \\hfill 24\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\end{array}\\right][\/latex]<\/div>\r\n<div><\/div>\r\nNow multiply [latex]\\frac{1}{6}[\/latex] to row 1 to obtsin 1 in row 1, column 1.\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{6}{R}_{1}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 4\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\\\ \\end{array}\\right][\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using \"rref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 4\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\nThe matrix ends up with all zeros in the last row: [latex]0y=0[\/latex]. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for [latex]y[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{4}{3}y=4\\hfill \\\\ \\frac{4}{3}y=4 - x\\hfill \\\\ y=3-\\frac{3}{4}x\\hfill \\end{array}[\/latex]<\/div>\r\nSo the solution to this system is [latex]\\left(x,3-\\frac{3}{4}x\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System of Linear Equations using Matrices<\/h3>\r\nSolve the following system of linear equations using Gaussian Elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x-y+z=8\\\\ \\hfill 2x+3y-z=-2\\\\ \\hfill 3x-2y-9z=9\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"779369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"779369\"]\r\n\r\nFirst, we write the augmented matrix.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 2&amp; \\hfill 3&amp; \\hfill -1&amp; \\hfill -2\\\\ \\hfill 3&amp; \\hfill -2&amp; \\hfill -9&amp; \\hfill 9\\end{array}\\right][\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<p style=\"text-align: left;\">Next, we perform row operations to obtain reduced row-echelon form.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}\\hfill -2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 0&amp; \\hfill 5&amp; \\hfill -3&amp; \\hfill -18\\\\ \\hfill 3&amp; \\hfill -2&amp; \\hfill -9&amp; \\hfill 9\\end{array}\\right]&amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill -3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 0&amp; \\hfill 5&amp; \\hfill -3&amp; \\hfill -18\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -12&amp; \\hfill -15\\end{array}\\right]\\end{array}[\/latex]<\/p>\r\nThe easiest way to obtain a 1 in row 2 of column 1 is to interchange [latex]{R}_{2}[\/latex] and [latex]{R}_{3}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\text{Interchange }{R}_{2}\\text{ and }{R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -12&amp; \\hfill -15\\\\ \\hfill 0&amp; \\hfill 5&amp; \\hfill -3&amp; \\hfill -18\\end{array}\\right][\/latex]<\/p>\r\nThen\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill -5{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -12&amp; \\hfill -15\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 57&amp; \\hfill 57\\end{array}\\right]&amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill -\\frac{1}{57}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -12&amp; \\hfill -15\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/p>\r\nFinally,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill 12{R}_{3}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -1&amp; \\hfill 1&amp; \\hfill 8\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill -3\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\end{array}\\right]&amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill {R}_{2}-{R}_{3}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 4\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill -3\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using \"rref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 4\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill -3\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1\\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\nThe last matrix represents the equivalent system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x=4\\hfill \\\\ \\text{ }y=-3\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/p>\r\nSo, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a 3 x 3 Dependent System<\/h3>\r\nSolve the following system of linear equations using Gaussian Elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill -x - 2y+z=-1\\\\ \\hfill 2x+3y=2\\\\ \\hfill y - 2z=0\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"664612\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664612\"]\r\n\r\nWrite the augmented matrix.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill -1&amp; \\hfill -2&amp; \\hfill 1&amp; \\hfill -1\\\\ \\hfill 2&amp; \\hfill 3&amp; \\hfill 0&amp; \\hfill 2\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n\r\nFirst, multiply row 1 by [latex]-1[\/latex] to get a 1 in row 1, column 1. Then, perform row operations to obtain reduced row-echelon form.\r\n<p style=\"text-align: center;\">[latex]-{R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 2&amp; \\hfill -1&amp; \\hfill 1\\\\ \\hfill 2&amp; \\hfill 3&amp; \\hfill 0&amp; \\hfill 2\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{R}_{2}\\leftrightarrow {R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 2&amp; \\hfill -1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 2&amp; \\hfill 3&amp; \\hfill 0&amp; \\hfill 2\\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|c}\\hfill 1&amp; \\hfill 2&amp; \\hfill -1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill -1&amp; \\hfill 2&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{R}_{2}+{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 2&amp; \\hfill -1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-2{R}_{2}+{R}_{1}={R}_{1}\\to\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 3&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using \"rref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 3&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill -2&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 0\\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\nThe last matrix represents the following system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3z=1\\hfill \\\\ \\text{ }y - 2z=0\\hfill \\\\ \\text{ }0=0\\hfill \\end{array}[\/latex]<\/p>\r\nWe see by the identity [latex]0=0[\/latex] that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the first equation for [latex]x[\/latex] and the second equation for [latex]y[\/latex] in terms of [latex]z[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x=1-3z\\hfill \\\\ \\text{ }y=2z\\hfill \\end{array}[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">So, the generic solution is [latex]\\left(1-3z, 2z, z\\right)[\/latex].<\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>The General Solution to a Dependent 3 X 3 System in Gauss-Jordan Elimination<\/h3>\r\nRecall that when you solve a dependent system of linear equations in two variables using elimination or substitution, you can write the solution [latex](x,y)[\/latex] in terms of y, because\u00a0there are infinitely many (x,y) pairs that will satisfy a dependent system of equations, and they all fall on the line\u00a0[latex](my+b, y)[\/latex]. Now that you are working in three dimensions, the solution will represent a plane, so you would write it in the general form [latex](m_{1}z+b_{1}, m_{2}z+b_{2}, z)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question height=\"340\"]6369[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can any system of linear equations be solved by Gauss-Jordan elimination?<\/strong>\r\n\r\n<em>Yes, a system of linear equations of any size can be solved by Gauss-Jordan elimination.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve with matrices using a calculator<\/h3>\r\n<ol>\r\n \t<li>Save the augmented matrix as a matrix variable [latex]\\left[A\\right],\\left[B\\right],\\left[C\\right]\\text{,} \\dots [\/latex].<\/li>\r\n \t<li>Use the <strong>rref(<\/strong> function in the calculator, calling up each matrix variable as needed.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Systems of Equations Using a Calculator<\/h3>\r\nSolve the system of equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 5x+3y+9z=-1\\\\ \\hfill -2x+3y-z=-2\\\\ \\hfill -x - 4y+5z=1\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"641879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"641879\"]\r\n\r\nWrite the augmented matrix for the system of equations.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 5&amp; \\hfill 3&amp; \\hfill 9&amp; \\hfill -1\\\\ \\hfill -2&amp; \\hfill 3&amp; \\hfill -1&amp; \\hfill -2\\\\ \\hfill -1&amp; \\hfill -4&amp; \\hfill 5&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nOn the matrix page of the calculator, enter the augmented matrix above as the matrix variable [latex]\\left[A\\right][\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left[A\\right]=\\left[\\begin{array}{rrrr}\\hfill 5&amp; \\hfill 3&amp; \\hfill 9&amp; \\hfill -1\\\\ \\hfill -2&amp; \\hfill 3&amp; \\hfill -1&amp; \\hfill -2\\\\ \\hfill -1&amp; \\hfill -4&amp; \\hfill 5&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nUse the <strong>rref(<\/strong> function in the calculator, calling up the matrix variable [latex]\\left[A\\right][\/latex].\r\n<p style=\"text-align: center;\">[latex]\\text{rref}\\left(A\\right)[\/latex]<\/p>\r\nEvaluate.\r\n<p style=\"text-align: center;\">[latex]\\text{rref}\\left(\\left[A\\right]\\right)=\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill \\frac{61}{187}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill -\\frac{92}{187}\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill -\\frac{24}{187}\\end{array}\\right][\/latex]<\/p>\r\nSo, the solution is [latex]\\left(\\frac{61}{187},-\\frac{92}{187},-\\frac{24}{187}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Applications of Systems of Equations<\/h2>\r\nNow we will turn to the applications for which systems of equations are used. Again, we will use same examples but use Gauss-Jordan elimination instead of Gaussian elimination.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying 2 \u00d7 2 Matrices to Finance<\/h3>\r\nCarolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?\r\n\r\n[reveal-answer q=\"591342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"591342\"]\r\n\r\nWe have a system of two equations in two variables. Let [latex]x=[\/latex] the amount invested at 10.5% interest, and [latex]y=[\/latex] the amount invested at 12% interest.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y=12,000\\hfill \\\\ 0.105x+0.12y=1,335\\hfill \\end{array}[\/latex]<\/p>\r\nAs a matrix, we have\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 12,000\\\\ \\hfill 0.105&amp; \\hfill 0.12&amp; \\hfill 1,335\\\\ \\end{array}\\right][\/latex]<\/p>\r\nand its reduced row-echelon form is\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 7,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 5,000\\\\ \\end{array}\\right][\/latex]<\/p>\r\nThen,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x=7,000\\\\ \\hfill y=5,000 \\end{array}[\/latex]<\/p>\r\nSo, $7,000 was invested at 10.5% interest and $5,000 was invested at 12% interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question height=\"215\"]2520[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying 3 \u00d7 3 Matrices to Finance<\/h3>\r\nAva invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?\r\n\r\n[reveal-answer q=\"357628\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"357628\"]\r\n\r\nWe have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at 5% interest, let [latex]y[\/latex] be the amount invested at 8% interest, and let [latex]z[\/latex] be the amount invested at 9% interest. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y+z=10,000\\hfill \\\\ 0.05x+0.08y+0.09z=770\\hfill \\\\ \\text{ }2x-z=0\\hfill \\end{array}[\/latex]<\/p>\r\nAs a matrix, we have\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0.05&amp; \\hfill 0.08&amp; \\hfill 0.09&amp; \\hfill 770\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n\r\nNow, we either perform row operation or use \"rref\" to achieve reduced row-echelon form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ -0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 0.03&amp; \\hfill 0.04&amp; \\hfill 270\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -1&amp; \\hfill 0\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ -2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 0.03&amp; \\hfill 0.04&amp; \\hfill 270\\\\ \\hfill 0&amp; -2&amp; \\hfill -3&amp; \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ \\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 9,000\\\\ \\hfill 0&amp; -2&amp; \\hfill -3&amp; \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ 2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 9,000\\\\ \\hfill 0&amp; 0&amp; \\hfill -\\frac{1}{3}&amp; \\hfill -2,000 \\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-3{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 9,000\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 6,000\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-\\frac{3}{4}{R}_{3}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 1,000\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 6,000\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-{R}_{2}-{R}_{3}+{R}_1={R}_{1}\\to\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 3,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 1,000\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 6,000\\end{array}\\right][\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using \"rref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 3,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 1,000\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 6,000\\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\nThe last matrix represents the following system:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x=3,000\\\\ \\hfill y=1,000\\\\ \\hfill z=6,000\\end{array}[\/latex]<\/p>\r\nSo, the answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use Gauss-Jordan elimination to solve a system of equations represented as an augmented matrix.<\/li>\n<li>Interpret the solution to a system of equations represented as an augmented matrix.<\/li>\n<\/ul>\n<\/div>\n<p>We have seen how to use Gaussian elimination as\u00a0a tool for solving a system written as an augmented matrix. Now we will see how to use Gauss-Jordan elimination, which is very similar to Gaussian elimination. We will review the same examples we used in the previous section.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 2 X 2 System by Gauss-Jordan Elimination<\/h3>\n<p>Solve the given system by Gauss-Jordan elimination.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}2x+3y=6\\hfill \\\\ \\text{ }x-y=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q264825\">Show Solution<\/span><\/p>\n<div id=\"q264825\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we write this as an augmented matrix.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 2& \\hfill 3& \\hfill 6\\\\ \\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\end{array}\\right][\/latex]<\/div>\n<div>\n<div class=\"textbox shaded\">\n<p>We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.<\/p>\n<div style=\"text-align: center;\">[latex]{R}_{1}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\hfill 2& \\hfill 3& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\n<p>We now have a 1 as the first entry in row 1, column 1. Now let\u2019s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by [latex]-2[\/latex] and then adding the result to row 2.<\/p>\n<div style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\hfill 0& \\hfill 5& \\hfill 5\\\\ \\end{array}\\right][\/latex]<\/div>\n<p>Next, we need a 1 in row 2, column 2. This can be accomplished by multiplying row 2 by [latex]\\frac{1}{5}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{5}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 1\\\\ \\end{array}\\right][\/latex]<\/p>\n<p>Lastly, we need a 0 in row 1, column 2, and we can accomplish it by simply adding row 2 to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]{R}_{1}+{R}_{2}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 0& \\hfill \\frac{3}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 1\\\\ \\end{array}\\right][\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using &#8220;rref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 0& \\hfill \\frac{3}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 1 \\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<div>The first row of the matrix represents [latex]x=\\frac{3}{2}[\/latex] and the second row of the matrix represents [latex]y=1[\/latex]. No back-substitute is needed in Gauss-Jordan elimination.<\/div>\n<p>The solution is the point [latex]\\left(\\frac{3}{2},1\\right)[\/latex] as we found in the previous section.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Dependent System<\/h3>\n<p>Solve the system of equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x+4y=12\\\\ 6x+8y=24\\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q844840\">Show Solution<\/span><\/p>\n<div id=\"q844840\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we write this as an augmented matrix.<\/p>\n<div style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 3& \\hfill 4& \\hfill 12\\\\ \\hfill 6& \\hfill 8& \\hfill 24\\\\ \\end{array}\\right][\/latex]<\/div>\n<div style=\"text-align: center;\"><\/div>\n<div class=\"textbox shaded\">\n<p>Perform <strong>row operations<\/strong> on the augmented matrix to try and achieve <strong>reduced\u00a0row-echelon form<\/strong>.<\/p>\n<div style=\"text-align: center;\">[latex]-\\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 6& \\hfill 8& \\hfill 24\\\\ \\end{array}\\right][\/latex]<\/div>\n<div style=\"text-align: center;\"><\/div>\n<div style=\"text-align: center;\">[latex]{R}_{1}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 6& \\hfill 8& \\hfill 24\\\\ \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\end{array}\\right][\/latex]<\/div>\n<div><\/div>\n<p>Now multiply [latex]\\frac{1}{6}[\/latex] to row 1 to obtsin 1 in row 1, column 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{6}{R}_{1}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill \\frac{4}{3}& \\hfill 4\\\\ \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\end{array}\\right][\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using &#8220;rref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill \\frac{4}{3}& \\hfill 4\\\\ \\hfill 0& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<p>The matrix ends up with all zeros in the last row: [latex]0y=0[\/latex]. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for [latex]y[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{4}{3}y=4\\hfill \\\\ \\frac{4}{3}y=4 - x\\hfill \\\\ y=3-\\frac{3}{4}x\\hfill \\end{array}[\/latex]<\/div>\n<p>So the solution to this system is [latex]\\left(x,3-\\frac{3}{4}x\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a System of Linear Equations using Matrices<\/h3>\n<p>Solve the following system of linear equations using Gaussian Elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x-y+z=8\\\\ \\hfill 2x+3y-z=-2\\\\ \\hfill 3x-2y-9z=9\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q779369\">Show Solution<\/span><\/p>\n<div id=\"q779369\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we write the augmented matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 2& \\hfill 3& \\hfill -1& \\hfill -2\\\\ \\hfill 3& \\hfill -2& \\hfill -9& \\hfill 9\\end{array}\\right][\/latex]<\/p>\n<div class=\"textbox shaded\">\n<p style=\"text-align: left;\">Next, we perform row operations to obtain reduced row-echelon form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}\\hfill -2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 5& \\hfill -3& \\hfill -18\\\\ \\hfill 3& \\hfill -2& \\hfill -9& \\hfill 9\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 5& \\hfill -3& \\hfill -18\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\end{array}\\right]\\end{array}[\/latex]<\/p>\n<p>The easiest way to obtain a 1 in row 2 of column 1 is to interchange [latex]{R}_{2}[\/latex] and [latex]{R}_{3}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Interchange }{R}_{2}\\text{ and }{R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\\\ \\hfill 0& \\hfill 5& \\hfill -3& \\hfill -18\\end{array}\\right][\/latex]<\/p>\n<p>Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill -5{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\\\ \\hfill 0& \\hfill 0& \\hfill 57& \\hfill 57\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -\\frac{1}{57}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/p>\n<p>Finally,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill 12{R}_{3}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill -3\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 1\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill {R}_{2}-{R}_{3}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 4\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill -3\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using &#8220;rref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 4\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill -3\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 1\\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<p>The last matrix represents the equivalent system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x=4\\hfill \\\\ \\text{ }y=-3\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/p>\n<p>So, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 3 x 3 Dependent System<\/h3>\n<p>Solve the following system of linear equations using Gaussian Elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill -x - 2y+z=-1\\\\ \\hfill 2x+3y=2\\\\ \\hfill y - 2z=0\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q664612\">Show Solution<\/span><\/p>\n<div id=\"q664612\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the augmented matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill -1& \\hfill -2& \\hfill 1& \\hfill -1\\\\ \\hfill 2& \\hfill 3& \\hfill 0& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<div class=\"textbox shaded\">\n<p>First, multiply row 1 by [latex]-1[\/latex] to get a 1 in row 1, column 1. Then, perform row operations to obtain reduced row-echelon form.<\/p>\n<p style=\"text-align: center;\">[latex]-{R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 2& \\hfill 3& \\hfill 0& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{R}_{2}\\leftrightarrow {R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 2& \\hfill 3& \\hfill 0& \\hfill 2\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill -1& \\hfill 2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{R}_{2}+{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-2{R}_{2}+{R}_{1}={R}_{1}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 3& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using &#8220;rref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 3& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<p>The last matrix represents the following system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3z=1\\hfill \\\\ \\text{ }y - 2z=0\\hfill \\\\ \\text{ }0=0\\hfill \\end{array}[\/latex]<\/p>\n<p>We see by the identity [latex]0=0[\/latex] that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the first equation for [latex]x[\/latex] and the second equation for [latex]y[\/latex] in terms of [latex]z[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x=1-3z\\hfill \\\\ \\text{ }y=2z\\hfill \\end{array}[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">So, the generic solution is [latex]\\left(1-3z, 2z, z\\right)[\/latex].<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>The General Solution to a Dependent 3 X 3 System in Gauss-Jordan Elimination<\/h3>\n<p>Recall that when you solve a dependent system of linear equations in two variables using elimination or substitution, you can write the solution [latex](x,y)[\/latex] in terms of y, because\u00a0there are infinitely many (x,y) pairs that will satisfy a dependent system of equations, and they all fall on the line\u00a0[latex](my+b, y)[\/latex]. Now that you are working in three dimensions, the solution will represent a plane, so you would write it in the general form [latex](m_{1}z+b_{1}, m_{2}z+b_{2}, z)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6369\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6369&theme=oea&iframe_resize_id=ohm6369&show_question_numbers\" width=\"100%\" height=\"340\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can any system of linear equations be solved by Gauss-Jordan elimination?<\/strong><\/p>\n<p><em>Yes, a system of linear equations of any size can be solved by Gauss-Jordan elimination.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve with matrices using a calculator<\/h3>\n<ol>\n<li>Save the augmented matrix as a matrix variable [latex]\\left[A\\right],\\left[B\\right],\\left[C\\right]\\text{,} \\dots[\/latex].<\/li>\n<li>Use the <strong>rref(<\/strong> function in the calculator, calling up each matrix variable as needed.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Systems of Equations Using a Calculator<\/h3>\n<p>Solve the system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 5x+3y+9z=-1\\\\ \\hfill -2x+3y-z=-2\\\\ \\hfill -x - 4y+5z=1\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q641879\">Show Solution<\/span><\/p>\n<div id=\"q641879\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the augmented matrix for the system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 5& \\hfill 3& \\hfill 9& \\hfill -1\\\\ \\hfill -2& \\hfill 3& \\hfill -1& \\hfill -2\\\\ \\hfill -1& \\hfill -4& \\hfill 5& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [latex]\\left[A\\right][\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left[A\\right]=\\left[\\begin{array}{rrrr}\\hfill 5& \\hfill 3& \\hfill 9& \\hfill -1\\\\ \\hfill -2& \\hfill 3& \\hfill -1& \\hfill -2\\\\ \\hfill -1& \\hfill -4& \\hfill 5& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Use the <strong>rref(<\/strong> function in the calculator, calling up the matrix variable [latex]\\left[A\\right][\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rref}\\left(A\\right)[\/latex]<\/p>\n<p>Evaluate.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rref}\\left(\\left[A\\right]\\right)=\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill \\frac{61}{187}\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill -\\frac{92}{187}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill -\\frac{24}{187}\\end{array}\\right][\/latex]<\/p>\n<p>So, the solution is [latex]\\left(\\frac{61}{187},-\\frac{92}{187},-\\frac{24}{187}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Applications of Systems of Equations<\/h2>\n<p>Now we will turn to the applications for which systems of equations are used. Again, we will use same examples but use Gauss-Jordan elimination instead of Gaussian elimination.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Applying 2 \u00d7 2 Matrices to Finance<\/h3>\n<p>Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q591342\">Show Solution<\/span><\/p>\n<div id=\"q591342\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a system of two equations in two variables. Let [latex]x=[\/latex] the amount invested at 10.5% interest, and [latex]y=[\/latex] the amount invested at 12% interest.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y=12,000\\hfill \\\\ 0.105x+0.12y=1,335\\hfill \\end{array}[\/latex]<\/p>\n<p>As a matrix, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 1& \\hfill 12,000\\\\ \\hfill 0.105& \\hfill 0.12& \\hfill 1,335\\\\ \\end{array}\\right][\/latex]<\/p>\n<p>and its reduced row-echelon form is<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 0& \\hfill 7,000\\\\ \\hfill 0& \\hfill 1& \\hfill 5,000\\\\ \\end{array}\\right][\/latex]<\/p>\n<p>Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x=7,000\\\\ \\hfill y=5,000 \\end{array}[\/latex]<\/p>\n<p>So, $7,000 was invested at 10.5% interest and $5,000 was invested at 12% interest.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm2520\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2520&theme=oea&iframe_resize_id=ohm2520&show_question_numbers\" width=\"100%\" height=\"215\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Applying 3 \u00d7 3 Matrices to Finance<\/h3>\n<p>Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q357628\">Show Solution<\/span><\/p>\n<div id=\"q357628\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at 5% interest, let [latex]y[\/latex] be the amount invested at 8% interest, and let [latex]z[\/latex] be the amount invested at 9% interest. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y+z=10,000\\hfill \\\\ 0.05x+0.08y+0.09z=770\\hfill \\\\ \\text{ }2x-z=0\\hfill \\end{array}[\/latex]<\/p>\n<p>As a matrix, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0.05& \\hfill 0.08& \\hfill 0.09& \\hfill 770\\\\ \\hfill 2& \\hfill 0& \\hfill -1& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<div class=\"textbox shaded\">\n<p>Now, we either perform row operation or use &#8220;rref&#8221; to achieve reduced row-echelon form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ -0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 0.03& \\hfill 0.04& \\hfill 270\\\\ \\hfill 2& \\hfill 0& \\hfill -1& \\hfill 0\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ -2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 0.03& \\hfill 0.04& \\hfill 270\\\\ \\hfill 0& -2& \\hfill -3& \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ \\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{4}{3}& \\hfill 9,000\\\\ \\hfill 0& -2& \\hfill -3& \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ 2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{4}{3}& \\hfill 9,000\\\\ \\hfill 0& 0& \\hfill -\\frac{1}{3}& \\hfill -2,000 \\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-3{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{4}{3}& \\hfill 9,000\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 6,000\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-\\frac{3}{4}{R}_{3}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 1,000\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 6,000\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-{R}_{2}-{R}_{3}+{R}_1={R}_{1}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 3,000\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 1,000\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 6,000\\end{array}\\right][\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the reduced row-echelon form can be found by using &#8220;rref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">\u00a0[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill 3,000\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill 1,000\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 6,000\\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<p>The last matrix represents the following system:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x=3,000\\\\ \\hfill y=1,000\\\\ \\hfill z=6,000\\end{array}[\/latex]<\/p>\n<p>So, the answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.<\/p>\n<\/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-1221\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li> Solving a System with Gauss-Jordan Elimination. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":705214,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\" Solving a System with Gauss-Jordan Elimination\",\"author\":\"Michelle Eunhee Chung\",\"organization\":\"Georgia State University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["mchung12"],"pb_section_license":""},"chapter-type":[],"contributor":[62],"license":[],"class_list":["post-1221","chapter","type-chapter","status-publish","hentry","contributor-mchung12"],"part":323,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1221","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/705214"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1221\/revisions"}],"predecessor-version":[{"id":1241,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1221\/revisions\/1241"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/323"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1221\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=1221"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1221"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1221"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=1221"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}