{"id":1376,"date":"2025-07-24T19:14:37","date_gmt":"2025-07-24T19:14:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1376"},"modified":"2026-03-24T17:53:12","modified_gmt":"2026-03-24T17:53:12","slug":"solving-systems-with-inverses-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-inverses-learn-it-4\/","title":{"raw":"Solving Systems with Inverses: Learn It 4","rendered":"Solving Systems with Inverses: Learn It 4"},"content":{"raw":"

Solving a System of Linear Equations Using the Inverse of a Matrix<\/h2>\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 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\n
\r\n

system of equation in matrix form<\/h3>\r\nTo solve a system of linear equations using an inverse matrix<\/strong>, let [latex]A[\/latex] be the 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].\r\n\r\n \r\n\r\nFor example, look at the following system of equations.\r\n

[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<\/strong> is\r\n

[latex]A=\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right][\/latex]<\/p>\r\nThe variable matrix<\/strong> is\r\n

[latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/p>\r\nAnd the constant matrix<\/strong> is\r\n

[latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\r\n \r\n\r\nThen [latex]AX=B[\/latex] looks like\r\n

[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>\r\n\r\n<\/section>

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].\r\n[latex]\\\\[\/latex]\r\nTo 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

[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 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\n<\/section>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.\r\n

[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

Solve the given system of equations using the inverse of a matrix.\r\n

[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

[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>\r\nThen\r\n

[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>\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

[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>\r\nSo,\r\n

[latex]{A}^{-1}=\\left[\\begin{array}{cc}11& -8\\\\ -4& \\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

[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>\r\nThe solution is [latex]\\left(-1,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Can we solve for [latex]X[\/latex] by finding the product [latex]B{A}^{-1}?[\/latex]\r\n<\/strong>\r\n\r\n
\r\n\r\nNo, 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.\r\n

[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\nNotice 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.\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]321760[\/ohm_question]<\/section>
Solve the following system using the inverse of a matrix.\r\n

[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

[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>\r\nFirst, we will find the inverse of [latex]A[\/latex] by augmenting with the identity.\r\n

[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>\r\nMultiply row 1 by [latex]\\frac{1}{5}[\/latex].\r\n

[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>\r\nMultiply row 1 by 4 and add to row 2.\r\n

[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>\r\nAdd row 1 to row 3.\r\n

[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>\r\nMultiply row 2 by \u22123 and add to row 1.\r\n

[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>\r\nMultiply row 3 by 5.\r\n

[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>\r\nMultiply row 3 by [latex]\\frac{1}{5}[\/latex] and add to row 1.\r\n

[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>\r\nMultiply row 3 by [latex]-\\frac{19}{5}[\/latex] and add to row 2.\r\n

[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>\r\nSo,\r\n

[latex]{A}^{-1}=\\left[\\begin{array}{ccc}-2& -3& 1\\\\ -3& 1& -19\\\\ 1& 0& 5\\end{array}\\right][\/latex]<\/p>\r\nMultiply both sides of the equation by [latex]{A}^{-1}[\/latex]. We want\r\n

[latex]{A}^{-1}AX={A}^{-1}B:[\/latex]<\/p>\r\n

[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>\r\nThus,\r\n

[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<\/section>

How To: Given a system of equations, solve with matrix inverses using a calculator\r\n<\/strong>\r\n
    \r\n \t
  1. 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
  2. Enter the multiplication, [latex][A]^-1\\times [B][\/latex], into the calculator, calling up each matrix variable as needed.<\/li>\r\n \t
  3. 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<\/section>
    Solve the system of equations with matrix inverses using a calculator\r\n

    [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 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

    [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>\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

    [latex]{\\left[A\\right]}^{-1}\\times \\left[B\\right][\/latex]<\/p>\r\nEvaluate the expression.\r\n

    [latex]\\left[\\begin{array}{c}-59\\\\ -34\\\\ 252\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm_question hide_question_numbers=1]321761[\/ohm_question]<\/section>","rendered":"

    Solving a System of Linear Equations Using the Inverse of a Matrix<\/h2>\n

    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 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

    \n

    system of equation in matrix form<\/h3>\n

    To solve a system of linear equations using an inverse matrix<\/strong>, let [latex]A[\/latex] be the 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].<\/p>\n

     <\/p>\n

    For example, look at the following system of equations.<\/p>\n

    [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

    From this system, the coefficient matrix<\/strong> is<\/p>\n

    [latex]A=\\left[\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right][\/latex]<\/p>\n

    The variable matrix<\/strong> is<\/p>\n

    [latex]X=\\left[\\begin{array}{c}x\\\\ y\\end{array}\\right][\/latex]<\/p>\n

    And the constant matrix<\/strong> is<\/p>\n

    [latex]B=\\left[\\begin{array}{c}{c}_{1}\\\\ {c}_{2}\\end{array}\\right][\/latex]<\/p>\n

     <\/p>\n

    Then [latex]AX=B[\/latex] looks like<\/p>\n

    [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<\/section>\n

    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].
    \n[latex]\\\\[\/latex]
    \nTo 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

    [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

    The only difference between solving a linear equation and a 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<\/section>\n

    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

    [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

    Solve the given system of equations using the inverse of a matrix.<\/p>\n

    [latex]\\begin{array}{r}\\hfill 3x+8y=5\\\\ \\hfill 4x+11y=7\\end{array}[\/latex]<\/p>\n