{"id":4106,"date":"2022-04-14T18:15:55","date_gmt":"2022-04-14T18:15:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-nonhomogeneous-linear-equations\/"},"modified":"2022-11-09T16:49:29","modified_gmt":"2022-11-09T16:49:29","slug":"skills-review-for-nonhomogeneous-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-nonhomogeneous-linear-equations\/","title":{"raw":"Skills Review for Nonhomogeneous Linear Equations","rendered":"Skills Review for Nonhomogeneous Linear Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve systems of equations<\/li>\r\n \t<li>Evaluate the determinant of a 2 x 2 matrix<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Nonhomogeneous Linear Equations section, we will learn how to solve yet another type of second-order differential equation. Here we will review how to solve systems of equations in two variables and evaluate the determinant of a 2 x 2 matrix.\r\n<h2>Solve a System of Equations in Two Variables<\/h2>\r\n<strong><em>(also in Module 4, Skills Review for Maxima\/Minima Problems and Lagrange Multipliers)<\/em><\/strong>\r\nA <strong>system of linear equations<\/strong> consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.\r\n\r\nWe will first look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}2x+y=15\\\\ 3x-y=5\\end{gathered}[\/latex]<\/div>\r\nThe <em>solution<\/em> to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}2\\left(4\\right)+\\left(7\\right)&amp;=15&amp;&amp;\\text{True} \\\\ 3\\left(4\\right)-\\left(7\\right)&amp;=5&amp;&amp;\\text{True} \\end{align}[\/latex]<\/div>\r\nTwo of the most common ways to solve a <strong>system of linear equations<\/strong> are the\u00a0<strong>substitution method<\/strong> and the <strong>a<\/strong><strong>ddition (elimination) method<\/strong>.\r\n<h3>The Substitution Method<\/h3>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\r\n<ol>\r\n \t<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\r\n \t<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\r\n \t<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\r\n \t<li>Check the solution in both equations.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE: SOLVING A SYSTEM OF EQUATIONS USING THE SUBSTITUTION METHOD<\/h3>\r\nSolve the following system of equations by substitution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ 2x-5y&amp;=1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"786744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"786744\"]\r\n\r\nFirst, we will solve the first equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ y&amp;=x - 5 \\end{align}[\/latex]<\/p>\r\nNow we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 5y&amp;=1 \\\\ 2x - 5\\left(x - 5\\right)&amp;=1 \\\\ 2x - 5x+25&amp;=1 \\\\ -3x&amp;=-24 \\\\ x&amp;=8 \\end{align}[\/latex]<\/p>\r\nNow, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-\\left(8\\right)+y&amp;=-5 \\\\ y&amp;=3 \\end{align}[\/latex]<\/p>\r\nOur solution is [latex]\\left(8,3\\right)[\/latex].\r\n\r\nCheck the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ -\\left(8\\right)+\\left(3\\right)&amp;=-5 &amp;&amp; \\text{True} \\\\[3mm] 2x - 5y&amp;=1 \\\\ 2\\left(8\\right)-5\\left(3\\right)&amp;=1 &amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]22573[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>The Addition (Elimination) Method<\/h3>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\r\n<ol>\r\n \t<li>Write both equations with <em>x<\/em>- and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\r\n \t<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\r\n \t<li>Solve the resulting equation for the remaining variable.<\/li>\r\n \t<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\r\n \t<li>Check the solution by substituting the values into the other equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: SOLVING A SYSTEM OF EQUATIONS USING THE ADDITION METHOD<\/h3>\r\nSolve the given system of equations by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"924657\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924657\"]\r\n\r\nBoth equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\\\ \\hline 3y&amp;=2\\end{align}[\/latex]<\/p>\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3y&amp;=2 \\\\ y&amp;=\\dfrac{2}{3} \\end{align}[\/latex]<\/p>\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=3 \\\\ -x+\\frac{2}{3}&amp;=3 \\\\ -x&amp;=3-\\frac{2}{3} \\\\ -x&amp;=\\frac{7}{3} \\\\ x&amp;=-\\frac{7}{3} \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\r\nCheck the solution in the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ \\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)&amp;= \\\\ -\\frac{7}{3}+\\frac{4}{3}&amp;= \\\\ -\\frac{3}{3}&amp;= \\\\ -1&amp;=-1&amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/>\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[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: SOLVING A SYSTEM OF EQUATIONS USING THE ADDITION METHOD<\/h3>\r\nSolve the given system of equations by the <strong>addition method<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11 \\\\ x - 2y&amp;=11 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"883001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"883001\"]\r\n\r\nAdding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11 \\\\ -3\\left(x - 2y\\right)&amp;=-3\\left(11\\right) &amp;&amp; \\text{Multiply both sides by }-3 \\\\ -3x+6y&amp;=-33 &amp;&amp; \\text{Use the distributive property}. \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now, let\u2019s add them.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=\u221211 \\\\ \u22123x+6y&amp;=\u221233 \\\\ \\hline 11y&amp;=\u221244 \\\\ y&amp;=\u22124 \\end{align}[\/latex]<\/p>\r\nFor the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11\\\\ 3x+5\\left(-4\\right)&amp;=-11\\\\ 3x - 20&amp;=-11\\\\ 3x&amp;=9\\\\ x&amp;=3\\end{align}[\/latex]<\/p>\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11 \\\\ \\left(3\\right)-2\\left(-4\\right)&amp;=3+8 \\\\ &amp;=11 &amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Evaluate the Determinant of a 2\u00d72 Matrix<\/h2>\r\n<strong><em>(also in Module 2, Skills Review for the Dot Product, Cross Product, and Equations of Lines and Planes in Space)<\/em><\/strong>\r\nA determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a <strong>square matrix<\/strong> to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an <strong>invertible matrix<\/strong> and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section.\r\n<div class=\"textbox shaded\">\r\n<h3>A General Note: Find the Determinant of a 2 \u00d7 2 Matrix<\/h3>\r\nThe <strong>determinant<\/strong> of a [latex]2\\text{ }\\times \\text{ }2[\/latex] matrix, given\r\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}a&amp; b\\\\ c&amp; d\\end{array}\\right][\/latex]<\/div>\r\nis defined as\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181431\/CNX_Precalc_Figure_09_08_0012.jpg\" alt=\"\" width=\"487\" height=\"59\" \/>\r\n\r\nNotice the change in notation. There are several ways to indicate the determinant, including [latex]\\mathrm{det}\\left(A\\right)[\/latex] and replacing the brackets in a matrix with straight lines, [latex]|A|[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Determinant of a 2 \u00d7 2 Matrix<\/h3>\r\nFind the determinant of the given matrix.\r\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}5&amp; 2\\\\ -6&amp; 3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"149250\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"149250\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{det}\\left(A\\right)&amp;=\\left\\rvert\\begin{array}{cc}5&amp; 2\\\\ -6&amp; 3\\end{array}\\right\\rvert\\hfill \\\\ &amp;=5\\left(3\\right)-\\left(-6\\right)\\left(2\\right)\\hfill \\\\ &amp;=27\\hfill \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]6397[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve systems of equations<\/li>\n<li>Evaluate the determinant of a 2 x 2 matrix<\/li>\n<\/ul>\n<\/div>\n<p>In the Nonhomogeneous Linear Equations section, we will learn how to solve yet another type of second-order differential equation. Here we will review how to solve systems of equations in two variables and evaluate the determinant of a 2 x 2 matrix.<\/p>\n<h2>Solve a System of Equations in Two Variables<\/h2>\n<p><strong><em>(also in Module 4, Skills Review for Maxima\/Minima Problems and Lagrange Multipliers)<\/em><\/strong><br \/>\nA <strong>system of linear equations<\/strong> consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.<\/p>\n<p>We will first look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}2x+y=15\\\\ 3x-y=5\\end{gathered}[\/latex]<\/div>\n<p>The <em>solution<\/em> to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}2\\left(4\\right)+\\left(7\\right)&=15&&\\text{True} \\\\ 3\\left(4\\right)-\\left(7\\right)&=5&&\\text{True} \\end{align}[\/latex]<\/div>\n<p>Two of the most common ways to solve a <strong>system of linear equations<\/strong> are the\u00a0<strong>substitution method<\/strong> and the <strong>a<\/strong><strong>ddition (elimination) method<\/strong>.<\/p>\n<h3>The Substitution Method<\/h3>\n<div class=\"textbox\">\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\n<ol>\n<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\n<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\n<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\n<li>Check the solution in both equations.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE: SOLVING A SYSTEM OF EQUATIONS USING THE SUBSTITUTION METHOD<\/h3>\n<p>Solve the following system of equations by substitution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ 2x-5y&=1 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786744\">Show Solution<\/span><\/p>\n<div id=\"q786744\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will solve the first equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ y&=x - 5 \\end{align}[\/latex]<\/p>\n<p>Now we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 5y&=1 \\\\ 2x - 5\\left(x - 5\\right)&=1 \\\\ 2x - 5x+25&=1 \\\\ -3x&=-24 \\\\ x&=8 \\end{align}[\/latex]<\/p>\n<p>Now, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-\\left(8\\right)+y&=-5 \\\\ y&=3 \\end{align}[\/latex]<\/p>\n<p>Our solution is [latex]\\left(8,3\\right)[\/latex].<\/p>\n<p>Check the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ -\\left(8\\right)+\\left(3\\right)&=-5 && \\text{True} \\\\[3mm] 2x - 5y&=1 \\\\ 2\\left(8\\right)-5\\left(3\\right)&=1 && \\text{True} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm22573\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=22573&theme=oea&iframe_resize_id=ohm22573\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>The Addition (Elimination) Method<\/h3>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\n<ol>\n<li>Write both equations with <em>x<\/em>&#8211; and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\n<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\n<li>Solve the resulting equation for the remaining variable.<\/li>\n<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\n<li>Check the solution by substituting the values into the other equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: SOLVING A SYSTEM OF EQUATIONS USING THE ADDITION METHOD<\/h3>\n<p>Solve the given system of equations by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ -x+y&=3 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924657\">Show Solution<\/span><\/p>\n<div id=\"q924657\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} x+2y&=-1 \\\\ -x+y&=3 \\\\ \\hline 3y&=2\\end{align}[\/latex]<\/p>\n<p>Now that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3y&=2 \\\\ y&=\\dfrac{2}{3} \\end{align}[\/latex]<\/p>\n<p>Then, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=3 \\\\ -x+\\frac{2}{3}&=3 \\\\ -x&=3-\\frac{2}{3} \\\\ -x&=\\frac{7}{3} \\\\ x&=-\\frac{7}{3} \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ \\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)&= \\\\ -\\frac{7}{3}+\\frac{4}{3}&= \\\\ -\\frac{3}{3}&= \\\\ -1&=-1&& \\text{True} \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm115130\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&#38;theme=oea&#38;iframe_resize_id=ohm115130&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: SOLVING A SYSTEM OF EQUATIONS USING THE ADDITION METHOD<\/h3>\n<p>Solve the given system of equations by the <strong>addition method<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11 \\\\ x - 2y&=11 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883001\">Show Solution<\/span><\/p>\n<div id=\"q883001\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11 \\\\ -3\\left(x - 2y\\right)&=-3\\left(11\\right) && \\text{Multiply both sides by }-3 \\\\ -3x+6y&=-33 && \\text{Use the distributive property}. \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now, let\u2019s add them.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=\u221211 \\\\ \u22123x+6y&=\u221233 \\\\ \\hline 11y&=\u221244 \\\\ y&=\u22124 \\end{align}[\/latex]<\/p>\n<p>For the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11\\\\ 3x+5\\left(-4\\right)&=-11\\\\ 3x - 20&=-11\\\\ 3x&=9\\\\ x&=3\\end{align}[\/latex]<\/p>\n<p>Our solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11 \\\\ \\left(3\\right)-2\\left(-4\\right)&=3+8 \\\\ &=11 && \\text{True} \\end{align}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Evaluate the Determinant of a 2\u00d72 Matrix<\/h2>\n<p><strong><em>(also in Module 2, Skills Review for the Dot Product, Cross Product, and Equations of Lines and Planes in Space)<\/em><\/strong><br \/>\nA determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a <strong>square matrix<\/strong> to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an <strong>invertible matrix<\/strong> and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section.<\/p>\n<div class=\"textbox shaded\">\n<h3>A General Note: Find the Determinant of a 2 \u00d7 2 Matrix<\/h3>\n<p>The <strong>determinant<\/strong> of a [latex]2\\text{ }\\times \\text{ }2[\/latex] matrix, given<\/p>\n<div style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}a& b\\\\ c& d\\end{array}\\right][\/latex]<\/div>\n<p>is defined as<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181431\/CNX_Precalc_Figure_09_08_0012.jpg\" alt=\"\" width=\"487\" height=\"59\" \/><\/p>\n<p>Notice the change in notation. There are several ways to indicate the determinant, including [latex]\\mathrm{det}\\left(A\\right)[\/latex] and replacing the brackets in a matrix with straight lines, [latex]|A|[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Determinant of a 2 \u00d7 2 Matrix<\/h3>\n<p>Find the determinant of the given matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}5& 2\\\\ -6& 3\\end{array}\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q149250\">Show Solution<\/span><\/p>\n<div id=\"q149250\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{det}\\left(A\\right)&=\\left\\rvert\\begin{array}{cc}5& 2\\\\ -6& 3\\end{array}\\right\\rvert\\hfill \\\\ &=5\\left(3\\right)-\\left(-6\\right)\\left(2\\right)\\hfill \\\\ &=27\\hfill \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6397\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6397&theme=oea&iframe_resize_id=ohm6397\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-4106\">\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>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/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":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4106","chapter","type-chapter","status-publish","hentry"],"part":4150,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4106","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4106\/revisions"}],"predecessor-version":[{"id":6495,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4106\/revisions\/6495"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4150"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4106\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4106"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4106"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4106"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4106"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}