{"id":334,"date":"2023-06-21T13:22:58","date_gmt":"2023-06-21T13:22:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solve-a-system-with-gaussian-elimination\/"},"modified":"2023-08-02T08:24:35","modified_gmt":"2023-08-02T08:24:35","slug":"solve-a-system-with-gaussian-elimination","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solve-a-system-with-gaussian-elimination\/","title":{"raw":"\u25aa   Solving a System with Gaussian Elimination","rendered":"\u25aa   Solving a System with Gaussian Elimination"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use Gaussian 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 write a <strong>system of equations<\/strong> with an <strong>augmented matrix\u00a0<\/strong>and then how to use row operations and back-substitution to obtain <strong>row-echelon form<\/strong>. Now we will use Gaussian Elimination as a tool for solving a system written as an augmented matrix.\u00a0In our first example, we will show you the process for using Gaussian Elimination on a system of two equations in two variables.\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nGain practice by working out the given examples on paper, perhaps more than once or twice, before trying practice problems.\r\n\r\nAs with any challenging concept in mathematics, patient practice and effort aid understanding.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a 2 X 2 System by Gaussian Elimination<\/h3>\r\nSolve the given system by Gaussian 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\nWe only have one more step, to multiply 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\n<strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using \"ref\" 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{3}{2}&amp; \\hfill 3\\\\ \\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>Use back-substitution. The second row of the matrix represents [latex]y=1[\/latex]. Back-substitute [latex]y=1[\/latex] into the first equation.<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x-\\left(1\\right)=\\frac{1}{2}\\hfill \\\\ \\text{ }x=\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solution is the point [latex]\\left(\\frac{3}{2},1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the given system by Gaussian elimination.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4x+3y=11\\hfill \\\\ \\text{ }\\text{}\\text{}x - 3y=-1\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[reveal-answer q=\"761791\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"761791\"][latex]\\left(2,1\\right)[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>In our next example, we will solve a system of two equations in two variables that is dependent. Recall that a dependent system has an infinite number of solutions and the result of row operations on its augmented matrix will be an equation such as [latex]0=0[\/latex]. We also review writing the general solution to a dependent system.<\/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<p style=\"text-align: center;\">[latex]A=\\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]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n\r\nPerform <strong>row operations<\/strong> on the augmented matrix to try and achieve <strong>row-echelon form<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}-\\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]\\hfill \\\\ {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]\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using \"ref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[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}3x+4y=12\\hfill \\\\ \\text{ }4y=12 - 3x\\hfill \\\\ \\text{ }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\nNow, we will take row-echelon form a step further to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.\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 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\n<strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using \"ref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill -\\frac{2}{3}&amp; \\hfill -3&amp; \\hfill 3\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{15}{13}&amp; \\hfill -\\frac{24}{13}\\\\ \\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-y+z=8\\hfill \\\\ \\text{ }y - 12z=-15\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/p>\r\nUsing back-substitution, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nRecall that there are three possible outcomes for solutions to linear systems. \u00a0In the previous example, the solution\u00a0[latex]\\left(4,-3,1\\right)[\/latex] represents a point in three dimensional space. This point represents the intersection of three planes. \u00a0In the next example, we solve a system using row operations and find that it represents a dependent system. \u00a0A dependent system in 3 dimensions can be represented by two planes that are identical, much like in 2 dimensions where a dependent system represents two lines that are identical.\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 <strong>row operations<\/strong> to obtain 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<strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using \"ref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill \\frac{3}{2}&amp; \\hfill 0&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}x+2y-z=1\\hfill \\\\ y - 2z=0\\hfill \\\\ 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 second equation for [latex]y[\/latex] and substituting it into the first equation we can solve for [latex]z[\/latex] in terms of [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y-z=1\\hfill \\\\ y=2z\\hfill \\\\ \\hfill \\\\ x+2\\left(2z\\right)-z=1\\hfill \\\\ x+3z=1\\hfill \\\\ z=\\frac{1-x}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nNow we substitute the expression for [latex]z[\/latex] into the second equation to solve for [latex]y[\/latex] in terms of [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 2z=0\\hfill \\\\ z=\\frac{1-x}{3}\\hfill \\\\ \\hfill \\\\ y - 2\\left(\\frac{1-x}{3}\\right)=0\\hfill \\\\ y=\\frac{2 - 2x}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe generic solution is [latex]\\left(x,\\frac{2 - 2x}{3},\\frac{1-x}{3}\\right)[\/latex].\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<\/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 x, 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](x, mx+b)[\/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](x, m_{1}x+b_{1}, m_{2}x+b_{2})[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nDo you recall how you wrote the general solution to dependent linear systems in the previous module? Writing the general solution for these systems in terms of [latex]x[\/latex] is no different.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system using Gaussian Elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x+4y-z=4\\\\ 2x+5y+8z=15\\\\ x+3y - 3z=1\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"589600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589600\"]\r\n\r\n[latex]\\left(1,1,1\\right)[\/latex][\/hidden-answer]\r\n\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 Gaussian elimination?<\/strong>\r\n\r\n<em>Yes, a system of linear equations of any size can be solved by Gaussian 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>ref(<\/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}{rrrrrrr}\\hfill 5&amp; \\hfill &amp; \\hfill 3&amp; \\hfill &amp; \\hfill 9&amp; \\hfill &amp; \\hfill -1\\\\ \\hfill -2&amp; \\hfill &amp; \\hfill 3&amp; \\hfill &amp; \\hfill -1&amp; \\hfill &amp; \\hfill -2\\\\ \\hfill -1&amp; \\hfill &amp; \\hfill -4&amp; \\hfill &amp; \\hfill 5&amp; \\hfill &amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nUse the <strong>ref(<\/strong> function in the calculator, calling up the matrix variable [latex]\\left[A\\right][\/latex].\r\n<p style=\"text-align: center;\">[latex]\\text{ref}\\left(\\left[A\\right]\\right)[\/latex]<\/p>\r\nEvaluate.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ \\left[\\begin{array}{rrrr}\\hfill 1&amp; \\hfill \\frac{3}{5}&amp; \\hfill \\frac{9}{5}&amp; \\hfill -\\frac{1}{5}\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{13}{21}&amp; \\hfill -\\frac{4}{7}\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill -\\frac{24}{187}\\end{array}\\right]\\to \\begin{array}{l} \\text{ }x+\\frac{3}{5}y+\\frac{9}{5}z=-\\frac{1}{5}\\hfill \\\\ \\text{ }y+\\frac{13}{21}z=-\\frac{4}{7}\\hfill \\\\ \\text{ }z=-\\frac{24}{187}\\hfill \\end{array}\\hfill \\end{array}<\/p>\r\n[\/latex]\r\n\r\nUsing back-substitution, 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.\u00a0In the next example we determine how much money was invested at two different rates given the sum of the interest earned by both accounts.\r\n<div class=\"textbox examples\">\r\n<h3>Tip for success<\/h3>\r\nSetting up a system of equations in the following examples uses the same ideas you have used before to write a system of linear equations to model a situation. The only difference is that now you'll use matrices and Gaussian elimination to solve the system.\r\n\r\n<\/div>\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}{r}\\hfill x+y=12,000\\\\ \\hfill 0.105x+0.12y=1,335\\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\n\r\n<div class=\"textbox shaded\">\r\n\r\nMultiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 12,000\\\\ \\hfill 0&amp; \\hfill 0.015&amp; \\hfill 75\\\\ \\end{array}\\right][\/latex]<\/p>\r\n<strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using \"ref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 12,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 5,000\\\\ \\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\nThen,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0.015y=75 \\hfill \\\\ y=5,000 \\hfill \\end{array}[\/latex]<\/p>\r\nSo,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y=12,000 \\hfill \\\\ x=12,000-y \\hfill \\\\ x=12,000 - 5,000 \\hfill \\\\ x=7,000 \\hfill \\end{array}[\/latex]<\/p>\r\nThus, $7,000 was invested at 10.5% interest and $5,000 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]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 perform Gaussian elimination to achieve 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<strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using \"ref\" on a graphing calculator:<\/span><\/strong>\r\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill -\\frac{1}{2}&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{3}{2}&amp; \\hfill 10000\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 6000\\\\ \\end{array}\\right][\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\nThe third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].\r\n\r\nThe second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex].\r\n\r\nSubstituting [latex]z=6,000[\/latex], we get\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y+\\frac{4}{3}\\left(6,000\\right)=9,000 \\hfill \\\\ y+8,000=9,000 \\hfill \\\\ y=1,000 \\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The first row tells us [latex]x+y+z=10,000[\/latex]. Substituting [latex]y=1,000[\/latex] and [latex]z=6,000[\/latex], we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+1,000+6,000=10,000\\hfill \\\\ x=3,000\\text{ }\\hfill \\end{array}[\/latex]<\/p>\r\nThe answer is $3,000 invested at 5% interest, $1,000 at 8%, and $6,000 at 9% 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\nA small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.\r\n\r\n[reveal-answer q=\"65592\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65592\"]\r\n\r\nWe have a system of three equations in three variables. Let [latex]x[\/latex] be the amount of loan at 7% interest, let [latex]y[\/latex] be the amount of loan at 8% interest, and let [latex]z[\/latex] be the amount of loan at 10% interest. Then,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y+z=1,500,000\\hfill \\\\ 0.07x+0.08y+0.10z=130,500\\hfill \\\\ z=4x\\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 1,500,000\\\\ \\hfill 0.07&amp; \\hfill 0.08&amp; \\hfill 0.10&amp; \\hfill 130,500\\\\ \\hfill 4&amp; \\hfill 0&amp; \\hfill -1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\nand its row-echelon form is\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 0&amp; \\hfill -\\frac{1}{4}&amp; \\hfill 0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{5}{4}&amp; \\hfill 1,500,000\\\\ \\hfill 0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 600,000\\end{array}\\right][\/latex]<\/p>\r\nUsing back-substitution we can find that [latex]$150,000[\/latex] at [latex]7\\%[\/latex], [latex]$750,000[\/latex] at [latex]8\\%[\/latex], and [latex]$600,000[\/latex] at\u00a0[latex]10\\%[\/latex] were borrowed.\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 Gaussian 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 write a <strong>system of equations<\/strong> with an <strong>augmented matrix\u00a0<\/strong>and then how to use row operations and back-substitution to obtain <strong>row-echelon form<\/strong>. Now we will use Gaussian Elimination as a tool for solving a system written as an augmented matrix.\u00a0In our first example, we will show you the process for using Gaussian Elimination on a system of two equations in two variables.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Gain practice by working out the given examples on paper, perhaps more than once or twice, before trying practice problems.<\/p>\n<p>As with any challenging concept in mathematics, patient practice and effort aid understanding.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 2 X 2 System by Gaussian Elimination<\/h3>\n<p>Solve the given system by Gaussian 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>We only have one more step, to multiply 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><strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using &#8220;ref&#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{3}{2}& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill 1\\\\ \\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<div>Use back-substitution. The second row of the matrix represents [latex]y=1[\/latex]. Back-substitute [latex]y=1[\/latex] into the first equation.<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x-\\left(1\\right)=\\frac{1}{2}\\hfill \\\\ \\text{ }x=\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solution is the point [latex]\\left(\\frac{3}{2},1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the given system by Gaussian elimination.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4x+3y=11\\hfill \\\\ \\text{ }\\text{}\\text{}x - 3y=-1\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q761791\">Show Solution<\/span><\/p>\n<div id=\"q761791\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(2,1\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>In our next example, we will solve a system of two equations in two variables that is dependent. Recall that a dependent system has an infinite number of solutions and the result of row operations on its augmented matrix will be an equation such as [latex]0=0[\/latex]. We also review writing the general solution to a dependent system.<\/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 style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc|c}\\hfill 3& \\hfill 4& \\hfill 12\\\\ \\hfill 6& \\hfill 8& \\hfill 24\\\\ \\end{array}\\right][\/latex]<\/p>\n<div class=\"textbox shaded\">\n<p>Perform <strong>row operations<\/strong> on the augmented matrix to try and achieve <strong>row-echelon form<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}-\\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]\\hfill \\\\ {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]\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using &#8220;ref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[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}3x+4y=12\\hfill \\\\ \\text{ }4y=12 - 3x\\hfill \\\\ \\text{ }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<p>Now, we will take row-echelon form a step further to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.<\/p>\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 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><strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using &#8220;ref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -\\frac{2}{3}& \\hfill -3& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{15}{13}& \\hfill -\\frac{24}{13}\\\\ \\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-y+z=8\\hfill \\\\ \\text{ }y - 12z=-15\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/p>\n<p>Using back-substitution, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Recall that there are three possible outcomes for solutions to linear systems. \u00a0In the previous example, the solution\u00a0[latex]\\left(4,-3,1\\right)[\/latex] represents a point in three dimensional space. This point represents the intersection of three planes. \u00a0In the next example, we solve a system using row operations and find that it represents a dependent system. \u00a0A dependent system in 3 dimensions can be represented by two planes that are identical, much like in 2 dimensions where a dependent system represents two lines that are identical.<\/p>\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 <strong>row operations<\/strong> to obtain 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><strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using &#8220;ref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill \\frac{3}{2}& \\hfill 0& \\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}x+2y-z=1\\hfill \\\\ y - 2z=0\\hfill \\\\ 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 second equation for [latex]y[\/latex] and substituting it into the first equation we can solve for [latex]z[\/latex] in terms of [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y-z=1\\hfill \\\\ y=2z\\hfill \\\\ \\hfill \\\\ x+2\\left(2z\\right)-z=1\\hfill \\\\ x+3z=1\\hfill \\\\ z=\\frac{1-x}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Now we substitute the expression for [latex]z[\/latex] into the second equation to solve for [latex]y[\/latex] in terms of [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 2z=0\\hfill \\\\ z=\\frac{1-x}{3}\\hfill \\\\ \\hfill \\\\ y - 2\\left(\\frac{1-x}{3}\\right)=0\\hfill \\\\ y=\\frac{2 - 2x}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The generic solution is [latex]\\left(x,\\frac{2 - 2x}{3},\\frac{1-x}{3}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>The General Solution to a Dependent 3 X 3 System<\/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 x, 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](x, mx+b)[\/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](x, m_{1}x+b_{1}, m_{2}x+b_{2})[\/latex].<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Do you recall how you wrote the general solution to dependent linear systems in the previous module? Writing the general solution for these systems in terms of [latex]x[\/latex] is no different.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system using Gaussian Elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x+4y-z=4\\\\ 2x+5y+8z=15\\\\ x+3y - 3z=1\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q589600\">Show Solution<\/span><\/p>\n<div id=\"q589600\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(1,1,1\\right)[\/latex]<\/p><\/div>\n<\/div>\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 Gaussian elimination?<\/strong><\/p>\n<p><em>Yes, a system of linear equations of any size can be solved by Gaussian 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>ref(<\/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}{rrrrrrr}\\hfill 5& \\hfill & \\hfill 3& \\hfill & \\hfill 9& \\hfill & \\hfill -1\\\\ \\hfill -2& \\hfill & \\hfill 3& \\hfill & \\hfill -1& \\hfill & \\hfill -2\\\\ \\hfill -1& \\hfill & \\hfill -4& \\hfill & \\hfill 5& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Use the <strong>ref(<\/strong> function in the calculator, calling up the matrix variable [latex]\\left[A\\right][\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\text{ref}\\left(\\left[A\\right]\\right)[\/latex]<\/p>\n<p>Evaluate.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ \\left[\\begin{array}{rrrr}\\hfill 1& \\hfill \\frac{3}{5}& \\hfill \\frac{9}{5}& \\hfill -\\frac{1}{5}\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{13}{21}& \\hfill -\\frac{4}{7}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill -\\frac{24}{187}\\end{array}\\right]\\to \\begin{array}{l} \\text{ }x+\\frac{3}{5}y+\\frac{9}{5}z=-\\frac{1}{5}\\hfill \\\\ \\text{ }y+\\frac{13}{21}z=-\\frac{4}{7}\\hfill \\\\ \\text{ }z=-\\frac{24}{187}\\hfill \\end{array}\\hfill \\end{array}<\/p>\n<p>[\/latex]<\/p>\n<p>Using back-substitution, 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.\u00a0In the next example we determine how much money was invested at two different rates given the sum of the interest earned by both accounts.<\/p>\n<div class=\"textbox examples\">\n<h3>Tip for success<\/h3>\n<p>Setting up a system of equations in the following examples uses the same ideas you have used before to write a system of linear equations to model a situation. The only difference is that now you&#8217;ll use matrices and Gaussian elimination to solve the system.<\/p>\n<\/div>\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}{r}\\hfill x+y=12,000\\\\ \\hfill 0.105x+0.12y=1,335\\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<div class=\"textbox shaded\">\n<p>Multiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 1& \\hfill 12,000\\\\ \\hfill 0& \\hfill 0.015& \\hfill 75\\\\ \\end{array}\\right][\/latex]<\/p>\n<p><strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using &#8220;ref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 1& \\hfill 12,000\\\\ \\hfill 0& \\hfill 1& \\hfill 5,000\\\\ \\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<p>Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0.015y=75 \\hfill \\\\ y=5,000 \\hfill \\end{array}[\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y=12,000 \\hfill \\\\ x=12,000-y \\hfill \\\\ x=12,000 - 5,000 \\hfill \\\\ x=7,000 \\hfill \\end{array}[\/latex]<\/p>\n<p>Thus, $7,000 was invested at 10.5% interest and $5,000 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=\"150\"><\/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 perform Gaussian elimination to achieve 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><strong><span style=\"color: #339966;\">Or the row-echelon form can be found by using &#8220;ref&#8221; on a graphing calculator:<\/span><\/strong><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #339966;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill -\\frac{1}{2}& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{3}{2}& \\hfill 10000\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 6000\\\\ \\end{array}\\right][\/latex]<\/span><\/p>\n<\/div>\n<p>The third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].<\/p>\n<p>The second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex].<\/p>\n<p>Substituting [latex]z=6,000[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y+\\frac{4}{3}\\left(6,000\\right)=9,000 \\hfill \\\\ y+8,000=9,000 \\hfill \\\\ y=1,000 \\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">The first row tells us [latex]x+y+z=10,000[\/latex]. Substituting [latex]y=1,000[\/latex] and [latex]z=6,000[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+1,000+6,000=10,000\\hfill \\\\ x=3,000\\text{ }\\hfill \\end{array}[\/latex]<\/p>\n<p>The answer is $3,000 invested at 5% interest, $1,000 at 8%, and $6,000 at 9% interest.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q65592\">Show Solution<\/span><\/p>\n<div id=\"q65592\" 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 of loan at 7% interest, let [latex]y[\/latex] be the amount of loan at 8% interest, and let [latex]z[\/latex] be the amount of loan at 10% interest. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y+z=1,500,000\\hfill \\\\ 0.07x+0.08y+0.10z=130,500\\hfill \\\\ z=4x\\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 1,500,000\\\\ \\hfill 0.07& \\hfill 0.08& \\hfill 0.10& \\hfill 130,500\\\\ \\hfill 4& \\hfill 0& \\hfill -1& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p>and its row-echelon form is<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 0& \\hfill -\\frac{1}{4}& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{5}{4}& \\hfill 1,500,000\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 600,000\\end{array}\\right][\/latex]<\/p>\n<p>Using back-substitution we can find that [latex]$150,000[\/latex] at [latex]7\\%[\/latex], [latex]$750,000[\/latex] at [latex]8\\%[\/latex], and [latex]$600,000[\/latex] at\u00a0[latex]10\\%[\/latex] were borrowed.<\/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-334\">\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><\/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 6369. <strong>Authored by<\/strong>: Lippman,David, mb Sousa,James. <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 2520. <strong>Authored by<\/strong>: Langkamp,Greg. <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><\/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":15,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax 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