{"id":2292,"date":"2016-11-03T20:03:35","date_gmt":"2016-11-03T20:03:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2292"},"modified":"2025-10-15T22:34:53","modified_gmt":"2025-10-15T22:34:53","slug":"solve-a-system-using-an-inverse","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/chapter\/solve-a-system-using-an-inverse\/","title":{"raw":"Solving a System Using an Inverse","rendered":"Solving a System Using an Inverse"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve a 2x2 system using an inverse.<\/li>\r\n \t<li>Solve a 3x3 systems using an inverse.<\/li>\r\n \t<li>Solve a system with a calculator.<\/li>\r\n<\/ul>\r\n<\/div>\r\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]\r\n\r\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.\r\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>\r\nFrom this system, the coefficient matrix is\r\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>\r\nThe variable matrix is\r\n<p style=\"text-align: center\">[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/p>\r\nAnd the constant matrix is\r\n<p style=\"text-align: center\">[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\r\nThen [latex]AX=B[\/latex] looks like\r\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>\r\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,\r\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>\r\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.\r\n\r\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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Solving a System of Equations Using the Inverse of a Matrix<\/h3>\r\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].\u00a0Multiply both sides by the inverse of [latex]A[\/latex] to obtain the solution.\r\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>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>If the coefficient matrix does not have an inverse, does that mean the system has no solution?<\/strong>\r\n\r\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>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a 2 \u00d7 2 System Using the Inverse of a Matrix<\/h3>\r\nSolve the given system of equations using the inverse of a matrix.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"167361\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"167361\"]\r\n\r\nWrite the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.\r\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>\r\nThen\r\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>\r\nFirst, we need to calculate [latex]{A}^{-1}[\/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:\r\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>\r\nSo,\r\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>\r\nNow we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[\/latex].\r\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>\r\nThe solution is [latex]\\left(-1,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we solve for [latex]X[\/latex] by finding the product [latex]B{A}^{-1}?[\/latex]<\/strong>\r\n\r\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>\r\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>\r\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>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a 3 \u00d7 3 System Using the Inverse of a Matrix<\/h3>\r\nSolve the following system using the inverse of a matrix.\r\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>\r\n[reveal-answer q=\"837562\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"837562\"]\r\n\r\nWrite the equation [latex]AX=B[\/latex].\r\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>\r\nFirst, we will find the inverse of [latex]A[\/latex] by augmenting with the identity.\r\n<p style=\"text-align: center\">[latex]\\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}|\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 1 by [latex]\\frac{1}{5}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ -4&amp; -11&amp; -41\\\\ -1&amp; -3&amp; -11\\end{array}|\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 1 by 4 and add to row 2.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ -1&amp; -3&amp; -11\\end{array}|\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nAdd row 1 to row 3.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 3&amp; \\frac{56}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; \\frac{1}{5}\\end{array}|\\begin{array}{ccc}\\frac{1}{5}&amp; 0&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ \\frac{1}{5}&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 2 by \u22123 and add to row 1.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; -\\frac{1}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; \\frac{1}{5}\\end{array}|\\begin{array}{ccc}-\\frac{11}{5}&amp; -3&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ \\frac{1}{5}&amp; 0&amp; 1\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 3 by 5.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; -\\frac{1}{5}\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; 1\\end{array}|\\begin{array}{ccc}-\\frac{11}{5}&amp; -3&amp; 0\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; \\frac{19}{5}\\\\ 0&amp; 0&amp; 1\\end{array}|\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ \\frac{4}{5}&amp; 1&amp; 0\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{ccc}1&amp; 0&amp; 0\\\\ 0&amp; 1&amp; 0\\\\ 0&amp; 0&amp; 1\\end{array}|\\begin{array}{ccc}-2&amp; -3&amp; 1\\\\ -3&amp; 1&amp; -19\\\\ 1&amp; 0&amp; 5\\end{array}\\right][\/latex]<\/p>\r\nSo,\r\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>\r\nMultiply both sides of the equation by [latex]{A}^{-1}[\/latex]. We want\r\n<p style=\"text-align: center\">[latex]{A}^{-1}AX={A}^{-1}B:[\/latex]<\/p>\r\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>\r\nThus,\r\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>\r\nThe solution is [latex]\\left(1,2,0\\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 system using the inverse of the coefficient matrix.\r\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>\r\n[reveal-answer q=\"514137\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"514137\"]\r\n\r\n[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve with matrix inverses using a calculator<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165135503570\">\r\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>\r\n \t<li>Enter the multiplication into the calculator, calling up each matrix variable as needed.<\/li>\r\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>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a Calculator to Solve a System of Equations with Matrix Inverses<\/h3>\r\nSolve the system of equations with matrix inverses using a calculator\r\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>\r\n[reveal-answer q=\"714265\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"714265\"]\r\n\r\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].\r\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>\r\nOn the home screen of the calculator, type in the multiplication to solve for [latex]X[\/latex], calling up each matrix variable as needed.\r\n<p style=\"text-align: center\">[latex]{\\left[A\\right]}^{-1}\\times \\left[B\\right][\/latex]<\/p>\r\nEvaluate the expression.\r\n<p style=\"text-align: center\">[latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\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<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].\u00a0Multiply 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}{rrr}\\hfill 5& \\hfill 15& \\hfill 56\\\\ \\hfill -4& \\hfill -11& \\hfill -41\\\\ \\hfill -1& \\hfill -3& \\hfill -11\\end{array}|\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 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}1& 3& \\frac{56}{5}\\\\ -4& -11& -41\\\\ -1& -3& -11\\end{array}|\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 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}1& 3& \\frac{56}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ -1& -3& -11\\end{array}|\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ \\frac{4}{5}& 1& 0\\\\ 0& 0& 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}1& 3& \\frac{56}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& \\frac{1}{5}\\end{array}|\\begin{array}{ccc}\\frac{1}{5}& 0& 0\\\\ \\frac{4}{5}& 1& 0\\\\ \\frac{1}{5}& 0& 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}1& 0& -\\frac{1}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& \\frac{1}{5}\\end{array}|\\begin{array}{ccc}-\\frac{11}{5}& -3& 0\\\\ \\frac{4}{5}& 1& 0\\\\ \\frac{1}{5}& 0& 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}1& 0& -\\frac{1}{5}\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& 1\\end{array}|\\begin{array}{ccc}-\\frac{11}{5}& -3& 0\\\\ \\frac{4}{5}& 1& 0\\\\ 1& 0& 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}1& 0& 0\\\\ 0& 1& \\frac{19}{5}\\\\ 0& 0& 1\\end{array}|\\begin{array}{ccc}-2& -3& 1\\\\ \\frac{4}{5}& 1& 0\\\\ 1& 0& 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}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}|\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 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-2292\">\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":21,"menu_order":12,"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 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