{"id":410,"date":"2019-07-15T22:44:49","date_gmt":"2019-07-15T22:44:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/solve-a-system-using-an-inverse\/"},"modified":"2019-07-15T22:44:49","modified_gmt":"2019-07-15T22:44:49","slug":"solve-a-system-using-an-inverse","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/solve-a-system-using-an-inverse\/","title":{"raw":"Solving a System Using an Inverse","rendered":"Solving a System Using an Inverse"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Solve a 2x2 system using an inverse.<\/li>\n \t<li>Solve a 3x3 systems using an inverse.<\/li>\n \t<li>Solve a system with a calculator.<\/li>\n<\/ul>\n<\/div>\nSolving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: [latex]X[\/latex] is the matrix representing the variables of the system, and [latex]B[\/latex] is the matrix representing the constants. Using <strong>matrix multiplication<\/strong>, we may define a system of equations with the same number of equations as variables as [latex]AX=B[\/latex]\n\nTo solve a system of linear equations using an <strong>inverse matrix<\/strong>, let [latex]A[\/latex] be the <strong>coefficient matrix<\/strong>, let [latex]X[\/latex] be the variable matrix, and let [latex]B[\/latex] be the constant matrix. Thus, we want to solve a system [latex]AX=B[\/latex]. For example, look at the following system of equations.\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/p>\nFrom this system, the coefficient matrix is\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right][\/latex]<\/p>\nThe variable matrix is\n<p style=\"text-align: center;\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/p>\nAnd the constant matrix is\n<p style=\"text-align: center;\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\nThen [latex]AX=B[\/latex] looks like\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\nRecall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex]\\left({2}^{-1}\\right)2=\\left(\\frac{1}{2}\\right)2=1[\/latex]. To solve a single linear equation [latex]ax=b[\/latex] for [latex]x[\/latex], we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of [latex]a[\/latex]. Thus,\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{ }ax=b\\\\ \\text{ }\\left(\\frac{1}{a}\\right)ax=\\left(\\frac{1}{a}\\right)b\\\\ \\left({a}^{-1}\\text{ }\\right)ax=\\left({a}^{-1}\\right)b\\\\ \\left[\\left({a}^{-1}\\right)a\\right]x=\\left({a}^{-1}\\right)b\\\\ \\text{ }1x=\\left({a}^{-1}\\right)b\\\\ \\text{ }x=\\left({a}^{-1}\\right)b\\end{array}[\/latex]<\/p>\nThe only difference between solving a linear equation and a <strong>system of equations<\/strong> written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same\u2014to isolate the variable.\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nAs you work through the following examples, keep in mind that matrix operations aren't bound to the same property characteristics as operations on real numbers. Since matrix multiplication is not commutative, order matters!\n\nOtherwise, the analogy of isolating the variable in the equation is a good one. Comparing a new concept or process to similar familiar material is a good way to develop intuition.\n\n<\/div>\nWe will investigate this idea in detail, but it is helpful to begin with a [latex]2\\times 2[\/latex] system and then move on to a [latex]3\\times 3[\/latex] system.\n<div class=\"textbox\">\n<h3>A General Note: Solving a System of Equations Using the Inverse of a Matrix<\/h3>\nGiven a system of equations, write the coefficient matrix [latex]A[\/latex], the variable matrix [latex]X[\/latex], and the constant matrix [latex]B[\/latex]. Then [latex]AX=B[\/latex].&nbsp;Multiply both sides by the inverse of [latex]A[\/latex] to obtain the solution.\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<strong>If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/strong>\n\n<em>No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution or be dependent and have infinitely many solutions.<\/em>\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\nSolve the given system of equations using the inverse of a matrix.\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/p>\n[reveal-answer q=\"167361\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"167361\"]\n\nWrite the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right],X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right],B=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/p>\nThen\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/p>\nFirst, we need to calculate [latex]{A}^{-1}[\/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{A}^{-1} &amp; =\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d&amp; -b\\\\ -c&amp; a\\end{array}\\right]\\hfill \\\\ &amp; =\\frac{1}{3\\left(11\\right)-8\\left(4\\right)}\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; 3\\end{array}\\right]\\hfill \\\\ &amp; =\\frac{1}{1}\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; 3\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\nSo,\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}11&amp; -8\\\\ -4&amp; \\text{ }\\text{ }3\\end{array}\\right][\/latex]<\/p>\nNow we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\hfill \\\\ \\left[\\begin{array}{rr}\\hfill 11&amp; \\hfill -8\\\\ \\hfill -4&amp; \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{cc}3&amp; 8\\\\ 4&amp; 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 11&amp; \\hfill -8\\\\ \\hfill -4&amp; \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right]\\hfill \\\\ \\left[\\begin{array}{cc}1&amp; 0\\\\ 0&amp; 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 11\\left(5\\right)+\\left(-8\\right)7\\\\ \\hfill -4\\left(5\\right)+3\\left(7\\right)\\end{array}\\right]\\hfill \\\\ \\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill -1\\\\ \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\nThe solution is [latex]\\left(-1,1\\right)[\/latex].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<strong>Can we solve for [latex]X[\/latex] by finding the product [latex]B{A}^{-1}?[\/latex]<\/strong>\n\n<em>No, recall that matrix multiplication is not commutative, so [latex]{A}^{-1}B\\ne B{A}^{-1}[\/latex]. Consider our steps for solving the matrix equation.<\/em>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/p>\n<em>Notice in the first step we multiplied both sides of the equation by [latex]{A}^{-1}[\/latex], but the [latex]{A}^{-1}[\/latex] was to the left of [latex]A[\/latex] on the left side and to the left of [latex]B[\/latex] on the right side. Because matrix multiplication is not commutative, order matters.<\/em>\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\nSolve the following system using the inverse of a matrix.\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 5x+15y+56z=35\\\\ \\hfill -4x - 11y - 41z=-26\\\\ \\hfill -x - 3y - 11z=-7\\end{array}[\/latex]<\/p>\n[reveal-answer q=\"837562\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"837562\"]\n\nWrite the equation [latex]AX=B[\/latex].\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc}5&amp; 15&amp; 56\\\\ -4&amp; -11&amp; -41\\\\ -1&amp; -3&amp; -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\nFirst, we will find the inverse of [latex]A[\/latex] by augmenting with the identity.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 5&amp; \\hfill 15&amp; \\hfill 56&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill0\\\\ \\hfill -4&amp; \\hfill -11&amp; \\hfill -41&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill-1&amp; \\hfill -3&amp; \\hfill -11&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\nMultiply row 1 by [latex]\\frac{1}{5}[\/latex].\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 3&amp; \\hfill \\frac{56}{5}&amp; \\hfill \\frac{1}{5}&amp; \\hfill 0&amp; \\hfill0\\\\ \\hfill -4&amp; \\hfill -11&amp; \\hfill -41&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill-1&amp; \\hfill -3&amp; \\hfill -11&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\nMultiply row 1 by 4 and add to row 2.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 3&amp; \\hfill \\frac{56}{5}&amp; \\hfill \\frac{1}{5}&amp; \\hfill 0&amp; \\hfill0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{19}{5}&amp; \\hfill \\frac{4}{5}&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill-1&amp; \\hfill -3&amp; \\hfill -11&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\nAdd row 1 to row 3.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 3&amp; \\hfill \\frac{56}{5}&amp; \\hfill \\frac{1}{5}&amp; \\hfill 0&amp; \\hfill0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{19}{5}&amp; \\hfill \\frac{4}{5}&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill0&amp; \\hfill 0&amp; \\hfill \\frac{1}{5}&amp; \\hfill \\frac{1}{5}&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\nMultiply row 2 by \u22123 and add to row 1.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 0&amp; \\hfill -\\frac{1}{5}&amp; \\hfill -\\frac{11}{5}&amp; \\hfill -3&amp; \\hfill0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{19}{5}&amp; \\hfill \\frac{4}{5}&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill0&amp; \\hfill 0&amp; \\hfill \\frac{1}{5}&amp; \\hfill \\frac{1}{5}&amp; \\hfill 0&amp; \\hfill 1\\end{array}\\right][\/latex]<\/p>\nMultiply row 3 by 5.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 0&amp; \\hfill -\\frac{1}{5}&amp; \\hfill -\\frac{11}{5}&amp; \\hfill -3&amp; \\hfill0\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{19}{5}&amp; \\hfill \\frac{4}{5}&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\nMultiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill -2&amp; \\hfill -3&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{19}{5}&amp; \\hfill \\frac{4}{5}&amp; \\hfill 1&amp; \\hfill 0\\\\ \\hfill0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\nMultiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1&amp; \\hfill 0&amp; \\hfill 0&amp; \\hfill -2&amp; \\hfill -3&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill -3&amp; \\hfill 1&amp; \\hfill -19\\\\ \\hfill0&amp; \\hfill 0&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right][\/latex]<\/p>\nSo,\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ -3&amp; 1&amp; -19\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\nMultiply both sides of the equation by [latex]{A}^{-1}[\/latex]. We want\n<p style=\"text-align: center;\">[latex]{A}^{-1}AX={A}^{-1}B:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rrr}\\hfill -2&amp; \\hfill -3&amp; \\hfill 1\\\\ \\hfill -3&amp; \\hfill 1&amp; \\hfill -19\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill 5&amp; \\hfill 15&amp; \\hfill 56\\\\ \\hfill -4&amp; \\hfill -11&amp; \\hfill -41\\\\ \\hfill -1&amp; \\hfill -3&amp; \\hfill -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -2&amp; \\hfill -3&amp; \\hfill 1\\\\ \\hfill -3&amp; \\hfill 1&amp; \\hfill -19\\\\ \\hfill 1&amp; \\hfill 0&amp; \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\nThus,\n<p style=\"text-align: center;\">[latex]{A}^{-1}B=\\left[\\begin{array}{r}\\hfill -70+78 - 7\\\\ \\hfill -105 - 26+133\\\\ \\hfill 35+0 - 35\\end{array}\\right]=\\left[\\begin{array}{c}1\\\\ 2\\\\ 0\\end{array}\\right][\/latex]<\/p>\nThe solution is [latex]\\left(1,2,0\\right)[\/latex].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nSolve the system using the inverse of the coefficient matrix.\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }2x - 17y+11z=0\\hfill \\\\ \\text{ }-x+11y - 7z=8\\hfill \\\\ \\text{ }3y - 2z=-2\\hfill \\end{array}[\/latex]<\/p>\n[reveal-answer q=\"514137\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"514137\"]\n\n[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex][\/hidden-answer]\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve with matrix inverses using a calculator<strong>\n<\/strong><\/h3>\n<ol id=\"fs-id1165135503570\">\n \t<li>Save the coefficient matrix and the constant matrix as matrix variables [latex]\\left[A\\right][\/latex] and [latex]\\left[B\\right][\/latex].<\/li>\n \t<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\n \t<li>If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\nSolve the system of equations with matrix inverses using a calculator\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+3y+z=32\\hfill \\\\ 3x+3y+z=-27\\hfill \\\\ 2x+4y+z=-2\\hfill \\end{array}[\/latex]<\/p>\n[reveal-answer q=\"714265\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"714265\"]\n\nOn the matrix page of the calculator, enter the <strong>coefficient matrix<\/strong> as the matrix variable [latex]\\left[A\\right][\/latex], and enter the constant matrix as the matrix variable [latex]\\left[B\\right][\/latex].\n<p style=\"text-align: center;\">[latex]\\left[A\\right]=\\left[\\begin{array}{ccc}2&amp; 3&amp; 1\\\\ 3&amp; 3&amp; 1\\\\ 2&amp; 4&amp; 1\\end{array}\\right],\\text{ }\\left[B\\right]=\\left[\\begin{array}{c}32\\\\ -27\\\\ -2\\end{array}\\right][\/latex]<\/p>\nOn the home screen of the calculator, type in the multiplication to solve for [latex]X[\/latex], calling up each matrix variable as needed.\n<p style=\"text-align: center;\">[latex]{\\left[A\\right]}^{-1}\\times \\left[B\\right][\/latex]<\/p>\nEvaluate the expression.\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve a 2&#215;2 system using an inverse.<\/li>\n<li>Solve a 3&#215;3 systems using an inverse.<\/li>\n<li>Solve a system with a calculator.<\/li>\n<\/ul>\n<\/div>\n<p>Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: [latex]X[\/latex] is the matrix representing the variables of the system, and [latex]B[\/latex] is the matrix representing the constants. Using <strong>matrix multiplication<\/strong>, we may define a system of equations with the same number of equations as variables as [latex]AX=B[\/latex]<\/p>\n<p>To solve a system of linear equations using an <strong>inverse matrix<\/strong>, let [latex]A[\/latex] be the <strong>coefficient matrix<\/strong>, let [latex]X[\/latex] be the variable matrix, and let [latex]B[\/latex] be the constant matrix. Thus, we want to solve a system [latex]AX=B[\/latex]. For example, look at the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/p>\n<p>From this system, the coefficient matrix is<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right][\/latex]<\/p>\n<p>The variable matrix is<\/p>\n<p style=\"text-align: center;\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/p>\n<p>And the constant matrix is<\/p>\n<p style=\"text-align: center;\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\n<p>Then [latex]AX=B[\/latex] looks like<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\n<p>Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex]\\left({2}^{-1}\\right)2=\\left(\\frac{1}{2}\\right)2=1[\/latex]. To solve a single linear equation [latex]ax=b[\/latex] for [latex]x[\/latex], we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of [latex]a[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{ }ax=b\\\\ \\text{ }\\left(\\frac{1}{a}\\right)ax=\\left(\\frac{1}{a}\\right)b\\\\ \\left({a}^{-1}\\text{ }\\right)ax=\\left({a}^{-1}\\right)b\\\\ \\left[\\left({a}^{-1}\\right)a\\right]x=\\left({a}^{-1}\\right)b\\\\ \\text{ }1x=\\left({a}^{-1}\\right)b\\\\ \\text{ }x=\\left({a}^{-1}\\right)b\\end{array}[\/latex]<\/p>\n<p>The only difference between solving a linear equation and a <strong>system of equations<\/strong> written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same\u2014to isolate the variable.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>As you work through the following examples, keep in mind that matrix operations aren&#8217;t bound to the same property characteristics as operations on real numbers. Since matrix multiplication is not commutative, order matters!<\/p>\n<p>Otherwise, the analogy of isolating the variable in the equation is a good one. Comparing a new concept or process to similar familiar material is a good way to develop intuition.<\/p>\n<\/div>\n<p>We will investigate this idea in detail, but it is helpful to begin with a [latex]2\\times 2[\/latex] system and then move on to a [latex]3\\times 3[\/latex] system.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Solving a System of Equations Using the Inverse of a Matrix<\/h3>\n<p>Given a system of equations, write the coefficient matrix [latex]A[\/latex], the variable matrix [latex]X[\/latex], and the constant matrix [latex]B[\/latex]. Then [latex]AX=B[\/latex].&nbsp;Multiply both sides by the inverse of [latex]A[\/latex] to obtain the solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/strong><\/p>\n<p><em>No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution or be dependent and have infinitely many solutions.<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\n<p>Solve the given system of equations using the inverse of a matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q167361\">Show Solution<\/span><\/p>\n<div id=\"q167361\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.<\/p>\n<p style=\"text-align: center;\">[latex]A=\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right],X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right],B=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/p>\n<p>Then<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right][\/latex]<\/p>\n<p>First, we need to calculate [latex]{A}^{-1}[\/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{A}^{-1} & =\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d& -b\\\\ -c& a\\end{array}\\right]\\hfill \\\\ & =\\frac{1}{3\\left(11\\right)-8\\left(4\\right)}\\left[\\begin{array}{cc}11& -8\\\\ -4& 3\\end{array}\\right]\\hfill \\\\ & =\\frac{1}{1}\\left[\\begin{array}{cc}11& -8\\\\ -4& 3\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{cc}11& -8\\\\ -4& \\text{ }\\text{ }3\\end{array}\\right][\/latex]<\/p>\n<p>Now we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\hfill \\\\ \\left[\\begin{array}{rr}\\hfill 11& \\hfill -8\\\\ \\hfill -4& \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{cc}3& 8\\\\ 4& 11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 11& \\hfill -8\\\\ \\hfill -4& \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}5\\\\ 7\\end{array}\\right]\\hfill \\\\ \\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 11\\left(5\\right)+\\left(-8\\right)7\\\\ \\hfill -4\\left(5\\right)+3\\left(7\\right)\\end{array}\\right]\\hfill \\\\ \\left[\\begin{array}{c}x\\\\ y\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill -1\\\\ \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we solve for [latex]X[\/latex] by finding the product [latex]B{A}^{-1}?[\/latex]<\/strong><\/p>\n<p><em>No, recall that matrix multiplication is not commutative, so [latex]{A}^{-1}B\\ne B{A}^{-1}[\/latex]. Consider our steps for solving the matrix equation.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\left({A}^{-1}\\right)AX=\\left({A}^{-1}\\right)B\\\\ \\hfill \\left[\\left({A}^{-1}\\right)A\\right]X=\\left({A}^{-1}\\right)B\\\\ \\hfill IX=\\left({A}^{-1}\\right)B\\\\ \\hfill X=\\left({A}^{-1}\\right)B\\end{array}[\/latex]<\/p>\n<p><em>Notice in the first step we multiplied both sides of the equation by [latex]{A}^{-1}[\/latex], but the [latex]{A}^{-1}[\/latex] was to the left of [latex]A[\/latex] on the left side and to the left of [latex]B[\/latex] on the right side. Because matrix multiplication is not commutative, order matters.<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\n<p>Solve the following system using the inverse of a matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill 5x+15y+56z=35\\\\ \\hfill -4x - 11y - 41z=-26\\\\ \\hfill -x - 3y - 11z=-7\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q837562\">Show Solution<\/span><\/p>\n<div id=\"q837562\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the equation [latex]AX=B[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc}5& 15& 56\\\\ -4& -11& -41\\\\ -1& -3& -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\n<p>First, we will find the inverse of [latex]A[\/latex] by augmenting with the identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 5& \\hfill 15& \\hfill 56& \\hfill 1& \\hfill 0& \\hfill0\\\\ \\hfill -4& \\hfill -11& \\hfill -41& \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill-1& \\hfill -3& \\hfill -11& \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 1 by [latex]\\frac{1}{5}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 3& \\hfill \\frac{56}{5}& \\hfill \\frac{1}{5}& \\hfill 0& \\hfill0\\\\ \\hfill -4& \\hfill -11& \\hfill -41& \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill-1& \\hfill -3& \\hfill -11& \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 1 by 4 and add to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 3& \\hfill \\frac{56}{5}& \\hfill \\frac{1}{5}& \\hfill 0& \\hfill0\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{19}{5}& \\hfill \\frac{4}{5}& \\hfill 1& \\hfill 0\\\\ \\hfill-1& \\hfill -3& \\hfill -11& \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Add row 1 to row 3.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 3& \\hfill \\frac{56}{5}& \\hfill \\frac{1}{5}& \\hfill 0& \\hfill0\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{19}{5}& \\hfill \\frac{4}{5}& \\hfill 1& \\hfill 0\\\\ \\hfill0& \\hfill 0& \\hfill \\frac{1}{5}& \\hfill \\frac{1}{5}& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 2 by \u22123 and add to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 0& \\hfill -\\frac{1}{5}& \\hfill -\\frac{11}{5}& \\hfill -3& \\hfill0\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{19}{5}& \\hfill \\frac{4}{5}& \\hfill 1& \\hfill 0\\\\ \\hfill0& \\hfill 0& \\hfill \\frac{1}{5}& \\hfill \\frac{1}{5}& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 3 by 5.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 0& \\hfill -\\frac{1}{5}& \\hfill -\\frac{11}{5}& \\hfill -3& \\hfill0\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{19}{5}& \\hfill \\frac{4}{5}& \\hfill 1& \\hfill 0\\\\ \\hfill0& \\hfill 0& \\hfill 1& \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill -2& \\hfill -3& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{19}{5}& \\hfill \\frac{4}{5}& \\hfill 1& \\hfill 0\\\\ \\hfill0& \\hfill 0& \\hfill 1& \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|ccc}\\hfill 1& \\hfill 0& \\hfill 0& \\hfill -2& \\hfill -3& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill 0& \\hfill -3& \\hfill 1& \\hfill -19\\\\ \\hfill0& \\hfill 0& \\hfill 1& \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right][\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}=\\left[\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\n<p>Multiply both sides of the equation by [latex]{A}^{-1}[\/latex]. We want<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}AX={A}^{-1}B:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{rrr}\\hfill -2& \\hfill -3& \\hfill 1\\\\ \\hfill -3& \\hfill 1& \\hfill -19\\\\ \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill 5& \\hfill 15& \\hfill 56\\\\ \\hfill -4& \\hfill -11& \\hfill -41\\\\ \\hfill -1& \\hfill -3& \\hfill -11\\end{array}\\right]\\text{ }\\left[\\begin{array}{c}x\\\\ y\\\\ z\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -2& \\hfill -3& \\hfill 1\\\\ \\hfill -3& \\hfill 1& \\hfill -19\\\\ \\hfill 1& \\hfill 0& \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{r}\\hfill 35\\\\ \\hfill -26\\\\ \\hfill -7\\end{array}\\right][\/latex]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]{A}^{-1}B=\\left[\\begin{array}{r}\\hfill -70+78 - 7\\\\ \\hfill -105 - 26+133\\\\ \\hfill 35+0 - 35\\end{array}\\right]=\\left[\\begin{array}{c}1\\\\ 2\\\\ 0\\end{array}\\right][\/latex]<\/p>\n<p>The solution is [latex]\\left(1,2,0\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system using the inverse of the coefficient matrix.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }2x - 17y+11z=0\\hfill \\\\ \\text{ }-x+11y - 7z=8\\hfill \\\\ \\text{ }3y - 2z=-2\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q514137\">Show Solution<\/span><\/p>\n<div id=\"q514137\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve with matrix inverses using a calculator<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165135503570\">\n<li>Save the coefficient matrix and the constant matrix as matrix variables [latex]\\left[A\\right][\/latex] and [latex]\\left[B\\right][\/latex].<\/li>\n<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\n<li>If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\n<p>Solve the system of equations with matrix inverses using a calculator<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+3y+z=32\\hfill \\\\ 3x+3y+z=-27\\hfill \\\\ 2x+4y+z=-2\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q714265\">Show Solution<\/span><\/p>\n<div id=\"q714265\" class=\"hidden-answer\" style=\"display: none\">\n<p>On the matrix page of the calculator, enter the <strong>coefficient matrix<\/strong> as the matrix variable [latex]\\left[A\\right][\/latex], and enter the constant matrix as the matrix variable [latex]\\left[B\\right][\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left[A\\right]=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right],\\text{ }\\left[B\\right]=\\left[\\begin{array}{c}32\\\\ -27\\\\ -2\\end{array}\\right][\/latex]<\/p>\n<p>On the home screen of the calculator, type in the multiplication to solve for [latex]X[\/latex], calling up each matrix variable as needed.<\/p>\n<p style=\"text-align: center;\">[latex]{\\left[A\\right]}^{-1}\\times \\left[B\\right][\/latex]<\/p>\n<p>Evaluate the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/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-410\">\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><\/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":17533,"menu_order":15,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"}]","CANDELA_OUTCOMES_GUID":"f39333f2-de91-462f-8a07-dcbf13b6b6f0","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-410","chapter","type-chapter","status-publish","hentry"],"part":395,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/410","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/410\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/parts\/395"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/410\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/media?parent=410"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=410"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/contributor?post=410"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/license?post=410"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}