{"id":5231,"date":"2018-05-03T21:28:11","date_gmt":"2018-05-03T21:28:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coreq-collegealgebra\/?post_type=chapter&p=5231"},"modified":"2018-05-21T18:35:30","modified_gmt":"2018-05-21T18:35:30","slug":"solve-systems-with-inverses","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/solve-systems-with-inverses\/","title":{"raw":"Solve Systems with Inverses of Matrices*","rendered":"Solve Systems with Inverses of Matrices*"},"content":{"raw":"Nancy plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem?\r\n\r\nThere are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.\r\n

Find the Inverse of a Matrix<\/h2>\r\nWe know that the multiplicative inverse of a real number [latex]a[\/latex] is [latex]{a}^{-1}[\/latex], and [latex]a{a}^{-1}={a}^{-1}a=\\left(\\frac{1}{a}\\right)a=1[\/latex]. For example, [latex]{2}^{-1}=\\frac{1}{2}[\/latex] and [latex]\\left(\\frac{1}{2}\\right)2=1[\/latex]. The multiplicative inverse of a matrix<\/strong> is similar in concept, except that the product of matrix [latex]A[\/latex] and its inverse [latex]{A}^{-1}[\/latex] equals the identity matrix<\/strong>. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by [latex]{I}_{n}[\/latex] where [latex]n[\/latex] represents the dimension of the matrix. The equations below\u00a0are the identity matrices for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a [latex]3\\text{}\\times \\text{}3[\/latex] matrix, respectively.\r\n

[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n

[latex]{I}_{3}=\\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 0& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nThe identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A[\/latex].\r\n\r\nA matrix that has a multiplicative inverse has the properties\r\n

[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/p>\r\nA matrix that has a multiplicative inverse is called an invertible matrix<\/strong>. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[\/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[\/latex] is invertible, then [latex]{A}^{-1}[\/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\\text{}\\times \\text{}2[\/latex] matrix and a third method that can be used on both [latex]2\\text{}\\times \\text{}2[\/latex] and [latex]3\\text{}\\times \\text{}3[\/latex] matrices.\r\n

\r\n

A General Note: The Identity Matrix and Multiplicative Inverse<\/h3>\r\nThe identity matrix<\/strong>, [latex]{I}_{n}[\/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.\r\n

[latex]\\begin{array}{l}\\hfill\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {I}_{2}=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right]\\begin{array}{cccc}& & & \\end{array}{I}_{3}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ \\text{ }2\\times 2\\text{ 3}\\times 3\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/p>\r\nIf [latex]A[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix and [latex]B[\/latex] is an [latex]n\\times n[\/latex]\u00a0matrix such that [latex]AB=BA={I}_{n}[\/latex], then [latex]B={A}^{-1}[\/latex], the multiplicative inverse of a matrix<\/strong> [latex]A[\/latex].\r\n\r\n<\/div>\r\n

\r\n

Example: Showing That the Identity Matrix Acts as a 1<\/h3>\r\nGiven matrix A<\/em>, show that [latex]AI=IA=A[\/latex].\r\n[latex]A=\\left[\\begin{array}{cc}3& 4\\\\ -2& 5\\end{array}\\right][\/latex]\r\n\r\n[reveal-answer q=\"38185\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"38185\"]\r\n\r\nUse matrix multiplication to show that the product of [latex]A[\/latex] and the identity is equal to the product of the identity and A.<\/em>\r\n

[latex]AI=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 3\\cdot 1+4\\cdot 0& \\hfill & \\hfill & \\hfill 3\\cdot 0+4\\cdot 1\\\\ \\hfill -2\\cdot 1+5\\cdot 0& \\hfill & \\hfill & \\hfill -2\\cdot 0+5\\cdot 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n

[latex]AI=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right]=\\left[\\begin{array}{rrrr}\\hfill 1\\cdot 3+0\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 1\\cdot 4+0\\cdot 5\\\\ \\hfill 0\\cdot 3+1\\cdot \\left(-2\\right)& \\hfill & \\hfill & \\hfill 0\\cdot 4+1\\cdot 5\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill 4\\\\ \\hfill -2& \\hfill & \\hfill 5\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

\r\n

How To: Given two matrices, show that one is the multiplicative inverse of the other.\r\n<\/strong><\/h3>\r\n
    \r\n \t
  1. Given matrix [latex]A[\/latex] of order [latex]n\\times n[\/latex] and matrix [latex]B[\/latex] of order [latex]n\\times n[\/latex] multiply [latex]AB[\/latex].<\/li>\r\n \t
  2. If [latex]AB=I[\/latex], then find the product [latex]BA[\/latex]. If [latex]BA=I[\/latex], then [latex]B={A}^{-1}[\/latex] and [latex]A={B}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    Example: Showing That Matrix A<\/em> Is the Multiplicative Inverse of Matrix B<\/em><\/h3>\r\nShow that the given matrices are multiplicative inverses of each other.\r\n

    [latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"812081\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"812081\"]\r\n\r\nMultiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.\r\n

    [latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill 1\\left(-9\\right)+5\\left(2\\right)& \\hfill & \\hfill 1\\left(-5\\right)+5\\left(1\\right)\\\\ \\hfill -2\\left(-9\\right)-9\\left(2\\right)& \\hfill & \\hfill -2\\left(-5\\right)-9\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n

    [latex]\\begin{array}{l}BA=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill & \\hfill -5\\\\ \\hfill 2& \\hfill & \\hfill 1\\end{array}\\right]\\cdot \\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 5\\\\ \\hfill -2& \\hfill & \\hfill -9\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -9\\left(1\\right)-5\\left(-2\\right)& \\hfill & \\hfill -9\\left(5\\right)-5\\left(-9\\right)\\\\ \\hfill 2\\left(1\\right)+1\\left(-2\\right)& \\hfill & \\hfill 2\\left(-5\\right)+1\\left(-9\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& & 0\\\\ 0& & 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    Try It<\/h3>\r\nShow that the following two matrices are inverses of each other.\r\n

    [latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right],B=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"159815\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"159815\"]\r\n

    [latex]\\begin{array}{l}AB=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1\\left(-3\\right)+4\\left(1\\right)& \\hfill & \\hfill 1\\left(-4\\right)+4\\left(1\\right)\\\\ \\hfill -1\\left(-3\\right)+-3\\left(1\\right)& \\hfill & \\hfill -1\\left(-4\\right)+-3\\left(1\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\\\ BA=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill -4\\\\ \\hfill 1& \\hfill & \\hfill 1\\end{array}\\right]\\begin{array}{r}\\hfill \\end{array}\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 4\\\\ \\hfill -1& \\hfill & \\hfill -3\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill -3\\left(1\\right)+-4\\left(-1\\right)& \\hfill & \\hfill -3\\left(4\\right)+-4\\left(-3\\right)\\\\ \\hfill 1\\left(1\\right)+1\\left(-1\\right)& \\hfill & \\hfill 1\\left(4\\right)+1\\left(-3\\right)\\end{array}\\right]=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\r\nWe can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication<\/strong>.\r\n
    \r\n

    Example: Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\r\nUse matrix multiplication to find the inverse of the given matrix.\r\n

    [latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill -2\\\\ \\hfill 2& \\hfill & \\hfill -3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"31546\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"31546\"]\r\n\r\nFor this method, we multiply [latex]A[\/latex] by a matrix containing unknown constants and set it equal to the identity.\r\n

    [latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill a& \\hfill b\\\\ \\hfill c& \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nFind the product of the two matrices on the left side of the equal sign.\r\n

    [latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill a& \\hfill b\\\\ \\hfill c& \\hfill d\\end{array}\\right]=\\left[\\begin{array}{rr}\\hfill 1a - 2c& \\hfill 1b - 2d\\\\ \\hfill 2a - 3c& \\hfill 2b - 3d\\end{array}\\right][\/latex]<\/p>\r\nNext, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.\r\n

    [latex]\\begin{array}{c}1a - 2c=1\\text{ }{R}_{1}\\\\ 2a - 3c=0\\text{ }{R}_{2}\\end{array}[\/latex]<\/p>\r\nUsing row operations, multiply and add as follows: [latex]\\left(-2\\right){R}_{1}+{R}_{2}\\to {R}_{2}[\/latex]. Add the equations, and solve for [latex]c[\/latex].\r\n

    [latex]\\begin{array}{r}\\hfill 1a - 2c=1\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/p>\r\nBack-substitute to solve for [latex]a[\/latex].\r\n

    [latex]\\begin{array}{r}\\hfill a - 2\\left(-2\\right)=1\\\\ \\hfill a+4=1\\\\ \\hfill a=-3\\end{array}[\/latex]<\/p>\r\nWrite another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.\r\n

    [latex]\\begin{array}{rr}\\hfill 1b - 2d=0& \\hfill {R}_{1}\\\\ \\hfill 2b - 3d=1& \\hfill {R}_{2}\\end{array}[\/latex]<\/p>\r\nUsing row operations, multiply and add as follows: [latex]\\left(-2\\right){R}_{1}+{R}_{2}={R}_{2}[\/latex]. Add the two equations and solve for [latex]d[\/latex].\r\n

    [latex]\\begin{array}{r}\\hfill 1b - 2d=0\\\\ \\hfill \\frac{0+1d=1}{d=1}\\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nOnce more, back-substitute and solve for [latex]b[\/latex].\r\n

    [latex]\\begin{array}{r}\\hfill b - 2\\left(1\\right)=0\\\\ \\hfill b - 2=0\\\\ \\hfill b=2\\end{array}[\/latex]<\/p>\r\n

    [latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill 2\\\\ \\hfill -2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Finding the Multiplicative Inverse by Augmenting with the Identity<\/h3>\r\nAnother way to find the multiplicative inverse<\/strong> is by augmenting with the identity. When matrix [latex]A[\/latex] is transformed into [latex]I[\/latex], the augmented matrix [latex]I[\/latex] transforms into [latex]{A}^{-1}[\/latex].\r\n\r\nFor example, given\r\n

    [latex]A=\\left[\\begin{array}{rrr}\\hfill 2& \\hfill & \\hfill 1\\\\ \\hfill 5& \\hfill & \\hfill 3\\end{array}\\right][\/latex]<\/p>\r\naugment [latex]A[\/latex] with the identity\r\n

    [latex]\\left[\\begin{array}{rr}\\hfill 2& \\hfill 1\\\\ \\hfill 5& \\hfill 3\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/p>\r\nPerform row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.\r\n

      \r\n \t
    1. Switch row 1 and row 2.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 5& \\hfill 3\\\\ \\hfill 2& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0\\end{array}\\right][\/latex]<\/li>\r\n \t
    2. Multiply row 2 by [latex]-2[\/latex] and add to row 1.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 2& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2& \\hfill 1\\\\ \\hfill 1& \\hfill 0\\end{array}\\right][\/latex]<\/li>\r\n \t
    3. Multiply row 1 by [latex]-2[\/latex] and add to row 2.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill -2& \\hfill 1\\\\ \\hfill 5& \\hfill -2\\end{array}\\right][\/latex]<\/li>\r\n \t
    4. Add row 2 to row 1.\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3& \\hfill -1\\\\ \\hfill 5& \\hfill -2\\end{array}\\right][\/latex]<\/li>\r\n \t
    5. Multiply row 2 by [latex]-1[\/latex].\r\n[latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 0\\\\ \\hfill 0& \\hfill 1\\end{array}\\text{ }|\\text{ }\\begin{array}{rr}\\hfill 3& \\hfill -1\\\\ \\hfill -5& \\hfill 2\\end{array}\\right][\/latex]<\/li>\r\n<\/ol>\r\nThe matrix we have found is [latex]{A}^{-1}[\/latex].\r\n

      [latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill -1\\\\ \\hfill -5& \\hfill & \\hfill 2\\end{array}\\right][\/latex]<\/p>\r\n\r\n

      Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h3>\r\nWhen we need to find the multiplicative inverse<\/strong> of a [latex]2\\times 2[\/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.\r\n\r\nIf [latex]A[\/latex] is a [latex]2\\times 2[\/latex] matrix, such as\r\n

      [latex]A=\\left[\\begin{array}{rrr}\\hfill a& \\hfill & \\hfill b\\\\ \\hfill c& \\hfill & \\hfill d\\end{array}\\right][\/latex]<\/p>\r\nthe multiplicative inverse of [latex]A[\/latex] is given by the formula\r\n

      [latex]{A}^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{rrr}\\hfill d& \\hfill & \\hfill -b\\\\ \\hfill -c& \\hfill & \\hfill a\\end{array}\\right][\/latex]<\/p>\r\nwhere [latex]ad-bc\\ne 0[\/latex]. If [latex]ad-bc=0[\/latex], then [latex]A[\/latex] has no inverse.\r\n

      \r\n

      Example: Using the Formula to Find the Multiplicative Inverse of Matrix A<\/em><\/h3>\r\nUse the formula to find the multiplicative inverse of\r\n

      [latex]A=\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"11636\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"11636\"]\r\n\r\nUsing the formula, we have\r\n

      [latex]\\begin{array}{l}{A}^{-1}=\\frac{1}{\\left(1\\right)\\left(-3\\right)-\\left(-2\\right)\\left(2\\right)}\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\\\ =\\frac{1}{-3+4}\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n

      Analysis of the Solution<\/h4>\r\nWe can check that our formula works by using one of the other methods to calculate the inverse. Let\u2019s augment [latex]A[\/latex] with the identity.\r\n

      [latex]\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}|\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/p>\r\nPerform row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.\r\n

        \r\n \t
      1. Multiply row 1 by [latex]-2[\/latex] and add to row 2.\r\n[latex]\\left[\\begin{array}{cc}1& -2\\\\ 0& 1\\end{array}|\\begin{array}{cc}1& 0\\\\ -2& 1\\end{array}\\right][\/latex]<\/li>\r\n \t
      2. Multiply row 1 by 2 and add to row 1.\r\n[latex]\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}|\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/li>\r\n<\/ol>\r\nSo, we have verified our original solution.\r\n

        [latex]{A}^{-1}=\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

        \r\n

        Try It<\/h3>\r\nUse the formula to find the inverse of matrix [latex]A[\/latex]. Verify your answer by augmenting with the identity matrix.\r\n

        [latex]A=\\left[\\begin{array}{cc}1& -1\\\\ 2& 3\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"161972\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"161972\"][latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}& \\frac{1}{5}\\\\ -\\frac{2}{5}& \\frac{1}{5}\\end{array}\\right][\/latex][\/hidden-answer]\r\n