{"id":1806,"date":"2015-11-12T18:30:44","date_gmt":"2015-11-12T18:30:44","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1806"},"modified":"2015-11-12T18:30:44","modified_gmt":"2015-11-12T18:30:44","slug":"finding-the-inverse-of-a-matrix","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/finding-the-inverse-of-a-matrix\/","title":{"raw":"Finding the Inverse of a Matrix","rendered":"Finding the Inverse of a Matrix"},"content":{"raw":"

We 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.\n<\/p>

[latex]{I}_{2}=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill 0\\\\ \\hfill 0& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\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]<\/div>\nThe identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A[\/latex].\n\nA matrix that has a multiplicative inverse has the properties\n
[latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/div>\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.\n
\n

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

Example 1: Showing That the Identity Matrix Acts as a 1<\/h3>\nGiven matrix A<\/em>, show that [latex]AI=IA=A[\/latex].\n
[latex]A=\\left[\\begin{array}{cc}3& 4\\\\ -2& 5\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
\n

Solution<\/h3>\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>\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]<\/div>\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]<\/div>\n<\/div>\n
\n

How To: Given two matrices, show that one is the multiplicative inverse of the other.\n<\/strong><\/h3>\n
  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>\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>\n<\/ol><\/div>\n
    \n

    Example 2: Showing That Matrix A<\/em> Is the Multiplicative Inverse of Matrix B<\/em><\/h3>\nShow that the given matrices are multiplicative inverses of each other.\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]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\nMultiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.\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]<\/div>\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]<\/div>\n[latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.\n\n<\/div>\n
    \n

    Try It 1<\/h3>\nShow that the following two matrices are inverses of each other.\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]<\/div>\n
    Solution<\/a><\/div>\n<\/div>\n

    Finding the Multiplicative Inverse Using Matrix Multiplication<\/h2>\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>.\n
    \n

    Example 3: Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\nUse matrix multiplication to find the inverse of the given matrix.\n
    [latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill & \\hfill -2\\\\ \\hfill 2& \\hfill & \\hfill -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\nFor this method, we multiply [latex]A[\/latex] by a matrix containing unknown constants and set it equal to the identity.\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]<\/div>\nFind the product of the two matrices on the left side of the equal sign.\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]<\/div>\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.\n
    [latex]\\begin{array}{c}1a - 2c=1\\text{ }{R}_{1}\\\\ 2a - 3c=0\\text{ }{R}_{2}\\end{array}[\/latex]<\/div>\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].\n
    [latex]\\begin{array}{r}\\hfill 1a - 2c=1\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/div>\nBack-substitute to solve for [latex]a[\/latex].\n
    [latex]\\begin{array}{r}\\hfill a - 2\\left(-2\\right)=1\\\\ \\hfill a+4=1\\\\ \\hfill a=-3\\end{array}[\/latex]<\/div>\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.\n
    [latex]\\begin{array}{rr}\\hfill 1b - 2d=0& \\hfill {R}_{1}\\\\ \\hfill 2b - 3d=1& \\hfill {R}_{2}\\end{array}[\/latex]<\/div>\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].\n
    [latex]\\begin{array}{r}\\hfill 1b - 2d=0\\\\ \\hfill \\frac{0+1d=1}{d=1}\\\\ \\hfill \\end{array}[\/latex]<\/div>\nOnce more, back-substitute and solve for [latex]b[\/latex].\n
    [latex]\\begin{array}{r}\\hfill b - 2\\left(1\\right)=0\\\\ \\hfill b - 2=0\\\\ \\hfill b=2\\end{array}[\/latex]<\/div>\n
    [latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill 2\\\\ \\hfill -2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n

    Finding the Multiplicative Inverse by Augmenting with the Identity<\/h2>\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].\n\nFor example, given\n
    [latex]A=\\left[\\begin{array}{rrr}\\hfill 2& \\hfill & \\hfill 1\\\\ \\hfill 5& \\hfill & \\hfill 3\\end{array}\\right][\/latex]<\/div>\naugment [latex]A[\/latex] with the identity\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]<\/div>\nPerform row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.\n
    1. Switch row 1 and row 2.\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]<\/div><\/li>\n\t
    2. Multiply row 2 by [latex]-2[\/latex] and add to row 1.\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]<\/div><\/li>\n\t
    3. Multiply row 1 by [latex]-2[\/latex] and add to row 2.\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]<\/div><\/li>\n\t
    4. Add row 2 to row 1.\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]<\/div><\/li>\n\t
    5. Multiply row 2 by [latex]-1[\/latex].\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]<\/div><\/li>\n<\/ol>\nThe matrix we have found is [latex]{A}^{-1}[\/latex].\n
      [latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill & \\hfill -1\\\\ \\hfill -5& \\hfill & \\hfill 2\\end{array}\\right][\/latex]<\/div>\n

      Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h2>\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.\n\nIf [latex]A[\/latex] is a [latex]2\\times 2[\/latex] matrix, such as\n
      [latex]A=\\left[\\begin{array}{rrr}\\hfill a& \\hfill & \\hfill b\\\\ \\hfill c& \\hfill & \\hfill d\\end{array}\\right][\/latex]<\/div>\nthe multiplicative inverse of [latex]A[\/latex] is given by the formula\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]<\/div>\nwhere [latex]ad-bc\\ne 0[\/latex]. If [latex]ad-bc=0[\/latex], then [latex]A[\/latex] has no inverse.\n
      \n

      Example 4: Using the Formula to Find the Multiplicative Inverse of Matrix A<\/em><\/h3>\nUse the formula to find the multiplicative inverse of\n
      [latex]A=\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\nUsing the formula, we have\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]<\/div>\n<\/div>\n
      \n

      Analysis of the Solution<\/h3>\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.\n
      [latex]\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}|\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\nPerform row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.\n
      1. Multiply row 1 by [latex]-2[\/latex] and add to row 2.\n
        [latex]\\left[\\begin{array}{cc}1& -2\\\\ 0& 1\\end{array}|\\begin{array}{cc}1& 0\\\\ -2& 1\\end{array}\\right][\/latex]<\/div><\/li>\n\t
      2. Multiply row 1 by 2 and add to row 1.\n
        [latex]\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}|\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div><\/li>\n<\/ol>\nSo, we have verified our original solution.\n
        [latex]{A}^{-1}=\\left[\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
        \n

        Try It 2<\/h3>\nUse the formula to find the inverse of matrix [latex]A[\/latex]. Verify your answer by augmenting with the identity matrix.\n
        [latex]A=\\left[\\begin{array}{cc}1& -1\\\\ 2& 3\\end{array}\\right][\/latex]<\/div>\n
        Solution<\/a><\/div>\n<\/div>\n
        \n

        Example 5: Finding the Inverse of the Matrix, If It Exists<\/h3>\nFind the inverse, if it exists, of the given matrix.\n
        [latex]A=\\left[\\begin{array}{cc}3& 6\\\\ 1& 2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
        \n

        Solution<\/h3>\nWe will use the method of augmenting with the identity.\n
        [latex]\\left[\\begin{array}{cc}3& 6\\\\ 1& 3\\end{array}|\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\n
        1. Switch row 1 and row 2.\n
          [latex]\\left[\\begin{array}{cc}1& 3\\\\ 3& 6\\text{ }\\end{array}\\text{ }\\text{ }\\text{ }|\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right][\/latex]<\/div><\/li>\n\t
        2. Multiply row 1 by \u22123 and add it to row 2.\n
          [latex]\\left[\\begin{array}{cc}1& 2\\\\ 0& 0\\end{array}|\\begin{array}{cc}1& 0\\\\ -3& 1\\end{array}\\right][\/latex]<\/div><\/li>\n\t
        3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/li>\n<\/ol><\/div>\n

          Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h2>\nUnfortunately, we do not have a formula similar to the one for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix to find the inverse of a [latex]3\\text{}\\times \\text{}3[\/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use row operations<\/strong> to obtain the inverse.\n\nGiven a [latex]3\\text{}\\times \\text{}3[\/latex]\u00a0matrix\n
          [latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\naugment [latex]A[\/latex] with the identity matrix\n
          [latex]A|I=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\text{ }|\\text{ }\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\nTo begin, we write the augmented matrix<\/strong> with the identity on the right and [latex]A[\/latex] on the left. Performing elementary row operations<\/strong> so that the identity matrix<\/strong> appears on the left, we will obtain the inverse matrix<\/strong> on the right. We will find the inverse of this matrix in the next example.\n
          \n

          How To: Given a [latex]3\\times 3[\/latex] matrix, find the inverse<\/h3>\n
          1. Write the original matrix augmented with the identity matrix on the right.<\/li>\n\t
          2. Use elementary row operations so that the identity appears on the left.<\/li>\n\t
          3. What is obtained on the right is the inverse of the original matrix.<\/li>\n\t
          4. Use matrix multiplication to show that [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].<\/li>\n<\/ol><\/div>\n
            \n

            Example 6: Finding the Inverse of a 3 \u00d7 3 Matrix<\/h3>\nGiven the [latex]3\\times 3[\/latex] matrix [latex]A[\/latex], find the inverse.\n
            [latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
            \n

            Solution<\/h3>\nAugment [latex]A[\/latex] with the identity matrix, and then begin row operations until the identity matrix replaces [latex]A[\/latex]. The matrix on the right will be the inverse of [latex]A[\/latex].\n
            [latex]\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\stackrel{\\text{Interchange }{R}_{2}\\text{and }{R}_{1}}{\\to }\\left[\\begin{array}{ccc}3& 3& 1\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{ccc}0& 1& 0\\\\ 1& 0& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n
            [latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n
            [latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 0& 1& 0\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n
            [latex]{R}_{3}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 2& 3& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
            [latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 3& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 3& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
            [latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right][\/latex]<\/div>\nThus,\n
            [latex]{A}^{-1}=B=\\left[\\begin{array}{ccc}-1& 1& 0\\\\ -1& 0& 1\\\\ 6& -2& -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
            \n

            Analysis of the Solution<\/h3>\nTo prove that [latex]B={A}^{-1}[\/latex], let\u2019s multiply the two matrices together to see if the product equals the identity, if [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].\n
            [latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{ccc}2\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& 2\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& 2\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 3\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& 3\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& 3\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 2\\left(-1\\right)+4\\left(-1\\right)+1\\left(6\\right)& 2\\left(1\\right)+4\\left(0\\right)+1\\left(-2\\right)& 2\\left(0\\right)+4\\left(1\\right)+1\\left(-3\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n
            [latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -1\\left(2\\right)+1\\left(3\\right)+0\\left(2\\right)& \\hfill -1\\left(3\\right)+1\\left(3\\right)+0\\left(4\\right)& \\hfill -1\\left(1\\right)+1\\left(1\\right)+0\\left(1\\right)\\\\ \\hfill -1\\left(2\\right)+0\\left(3\\right)+1\\left(2\\right)& \\hfill -1\\left(3\\right)+0\\left(3\\right)+1\\left(4\\right)& \\hfill -1\\left(1\\right)+0\\left(1\\right)+1\\left(1\\right)\\\\ \\hfill 6\\left(2\\right)+-2\\left(3\\right)+-3\\left(2\\right)& \\hfill 6\\left(3\\right)+-2\\left(3\\right)+-3\\left(4\\right)& \\hfill 6\\left(1\\right)+-2\\left(1\\right)+-3\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\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}[\/latex]<\/div>\n<\/div>\n
            \n

            Try It 3<\/h3>\nFind the inverse of the [latex]3\\times 3[\/latex] matrix.\n
            [latex]A=\\left[\\begin{array}{ccc}2& -17& 11\\\\ -1& 11& -7\\\\ 0& 3& -2\\end{array}\\right][\/latex]<\/div>\n
            Solution<\/a><\/div>\n<\/div>","rendered":"

            We 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.\n<\/p>\n

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

            The identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A[\/latex].<\/p>\n

            A matrix that has a multiplicative inverse has the properties<\/p>\n

            [latex]\\begin{array}{l}A{A}^{-1}=I\\\\ {A}^{-1}A=I\\end{array}[\/latex]<\/div>\n

            A 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.<\/p>\n

            \n

            A General Note: The Identity Matrix and Multiplicative Inverse<\/h3>\n

            The identity matrix<\/strong>, [latex]{I}_{n}[\/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.<\/p>\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]<\/div>\n

            If [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].<\/p>\n<\/div>\n

            \n

            Example 1: Showing That the Identity Matrix Acts as a 1<\/h3>\n

            Given matrix A<\/em>, show that [latex]AI=IA=A[\/latex].<\/p>\n

            [latex]A=\\left[\\begin{array}{cc}3& 4\\\\ -2& 5\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
            \n

            Solution<\/h3>\n

            Use 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><\/p>\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]<\/div>\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]<\/div>\n<\/div>\n
            \n

            How To: Given two matrices, show that one is the multiplicative inverse of the other.
            \n<\/strong><\/h3>\n
              \n
            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>\n
            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>\n<\/ol>\n<\/div>\n
              \n

              Example 2: Showing That Matrix A<\/em> Is the Multiplicative Inverse of Matrix B<\/em><\/h3>\n

              Show that the given matrices are multiplicative inverses of each other.<\/p>\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]<\/div>\n<\/div>\n
              \n

              Solution<\/h3>\n

              Multiply [latex]AB[\/latex] and [latex]BA[\/latex]. If both products equal the identity, then the two matrices are inverses of each other.<\/p>\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]<\/div>\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]<\/div>\n

              [latex]A[\/latex] and [latex]B[\/latex] are inverses of each other.<\/p>\n<\/div>\n

              \n

              Try It 1<\/h3>\n

              Show that the following two matrices are inverses of each other.<\/p>\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]<\/div>\n
              Solution<\/a><\/div>\n<\/div>\n

              Finding the Multiplicative Inverse Using Matrix Multiplication<\/h2>\n

              We 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>.<\/p>\n

              \n

              Example 3: Finding the Multiplicative Inverse Using Matrix Multiplication<\/h3>\n

              Use matrix multiplication to find the inverse of the given matrix.<\/p>\n

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

              Solution<\/h3>\n

              For this method, we multiply [latex]A[\/latex] by a matrix containing unknown constants and set it equal to the identity.<\/p>\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]<\/div>\n

              Find the product of the two matrices on the left side of the equal sign.<\/p>\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]<\/div>\n

              Next, 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.<\/p>\n

              [latex]\\begin{array}{c}1a - 2c=1\\text{ }{R}_{1}\\\\ 2a - 3c=0\\text{ }{R}_{2}\\end{array}[\/latex]<\/div>\n

              Using 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].<\/p>\n

              [latex]\\begin{array}{r}\\hfill 1a - 2c=1\\\\ \\hfill 0+1c=-2\\\\ \\hfill c=-2\\end{array}[\/latex]<\/div>\n

              Back-substitute to solve for [latex]a[\/latex].<\/p>\n

              [latex]\\begin{array}{r}\\hfill a - 2\\left(-2\\right)=1\\\\ \\hfill a+4=1\\\\ \\hfill a=-3\\end{array}[\/latex]<\/div>\n

              Write 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.<\/p>\n

              [latex]\\begin{array}{rr}\\hfill 1b - 2d=0& \\hfill {R}_{1}\\\\ \\hfill 2b - 3d=1& \\hfill {R}_{2}\\end{array}[\/latex]<\/div>\n

              Using 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].<\/p>\n

              [latex]\\begin{array}{r}\\hfill 1b - 2d=0\\\\ \\hfill \\frac{0+1d=1}{d=1}\\\\ \\hfill \\end{array}[\/latex]<\/div>\n

              Once more, back-substitute and solve for [latex]b[\/latex].<\/p>\n

              [latex]\\begin{array}{r}\\hfill b - 2\\left(1\\right)=0\\\\ \\hfill b - 2=0\\\\ \\hfill b=2\\end{array}[\/latex]<\/div>\n
              [latex]{A}^{-1}=\\left[\\begin{array}{rrr}\\hfill -3& \\hfill & \\hfill 2\\\\ \\hfill -2& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n

              Finding the Multiplicative Inverse by Augmenting with the Identity<\/h2>\n

              Another 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].<\/p>\n

              For example, given<\/p>\n

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

              augment [latex]A[\/latex] with the identity<\/p>\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]<\/div>\n

              Perform row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.<\/p>\n

                \n
              1. Switch row 1 and row 2.\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]<\/div>\n<\/li>\n
              2. Multiply row 2 by [latex]-2[\/latex] and add to row 1.\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]<\/div>\n<\/li>\n
              3. Multiply row 1 by [latex]-2[\/latex] and add to row 2.\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]<\/div>\n<\/li>\n
              4. Add row 2 to row 1.\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]<\/div>\n<\/li>\n
              5. Multiply row 2 by [latex]-1[\/latex].\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]<\/div>\n<\/li>\n<\/ol>\n

                The matrix we have found is [latex]{A}^{-1}[\/latex].<\/p>\n

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

                Finding the Multiplicative Inverse of 2\u00d72 Matrices Using a Formula<\/h2>\n

                When 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.<\/p>\n

                If [latex]A[\/latex] is a [latex]2\\times 2[\/latex] matrix, such as<\/p>\n

                [latex]A=\\left[\\begin{array}{rrr}\\hfill a& \\hfill & \\hfill b\\\\ \\hfill c& \\hfill & \\hfill d\\end{array}\\right][\/latex]<\/div>\n

                the multiplicative inverse of [latex]A[\/latex] is given by the formula<\/p>\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]<\/div>\n

                where [latex]ad-bc\\ne 0[\/latex]. If [latex]ad-bc=0[\/latex], then [latex]A[\/latex] has no inverse.<\/p>\n

                \n

                Example 4: Using the Formula to Find the Multiplicative Inverse of Matrix A<\/em><\/h3>\n

                Use the formula to find the multiplicative inverse of<\/p>\n

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

                Solution<\/h3>\n

                Using the formula, we have<\/p>\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]<\/div>\n<\/div>\n
                \n

                Analysis of the Solution<\/h3>\n

                We 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.<\/p>\n

                [latex]\\left[\\begin{array}{cc}1& -2\\\\ 2& -3\\end{array}|\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\n

                Perform row operations<\/strong> with the goal of turning [latex]A[\/latex] into the identity.<\/p>\n

                  \n
                1. Multiply row 1 by [latex]-2[\/latex] and add to row 2.\n
                  [latex]\\left[\\begin{array}{cc}1& -2\\\\ 0& 1\\end{array}|\\begin{array}{cc}1& 0\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n
                2. Multiply row 1 by 2 and add to row 1.\n
                  [latex]\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}|\\begin{array}{cc}-3& 2\\\\ -2& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n<\/ol>\n

                  So, we have verified our original solution.<\/p>\n

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

                  Try It 2<\/h3>\n

                  Use the formula to find the inverse of matrix [latex]A[\/latex]. Verify your answer by augmenting with the identity matrix.<\/p>\n

                  [latex]A=\\left[\\begin{array}{cc}1& -1\\\\ 2& 3\\end{array}\\right][\/latex]<\/div>\n
                  Solution<\/a><\/div>\n<\/div>\n
                  \n

                  Example 5: Finding the Inverse of the Matrix, If It Exists<\/h3>\n

                  Find the inverse, if it exists, of the given matrix.<\/p>\n

                  [latex]A=\\left[\\begin{array}{cc}3& 6\\\\ 1& 2\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
                  \n

                  Solution<\/h3>\n

                  We will use the method of augmenting with the identity.<\/p>\n

                  [latex]\\left[\\begin{array}{cc}3& 6\\\\ 1& 3\\end{array}|\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right][\/latex]<\/div>\n
                    \n
                  1. Switch row 1 and row 2.\n
                    [latex]\\left[\\begin{array}{cc}1& 3\\\\ 3& 6\\text{ }\\end{array}\\text{ }\\text{ }\\text{ }|\\begin{array}{cc}0& 1\\\\ 1& 0\\end{array}\\right][\/latex]<\/div>\n<\/li>\n
                  2. Multiply row 1 by \u22123 and add it to row 2.\n
                    [latex]\\left[\\begin{array}{cc}1& 2\\\\ 0& 0\\end{array}|\\begin{array}{cc}1& 0\\\\ -3& 1\\end{array}\\right][\/latex]<\/div>\n<\/li>\n
                  3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.<\/li>\n<\/ol>\n<\/div>\n

                    Finding the Multiplicative Inverse of 3\u00d73 Matrices<\/h2>\n

                    Unfortunately, we do not have a formula similar to the one for a [latex]2\\text{}\\times \\text{}2[\/latex] matrix to find the inverse of a [latex]3\\text{}\\times \\text{}3[\/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use row operations<\/strong> to obtain the inverse.<\/p>\n

                    Given a [latex]3\\text{}\\times \\text{}3[\/latex]\u00a0matrix<\/p>\n

                    [latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\n

                    augment [latex]A[\/latex] with the identity matrix<\/p>\n

                    [latex]A|I=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\text{ }|\\text{ }\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n

                    To begin, we write the augmented matrix<\/strong> with the identity on the right and [latex]A[\/latex] on the left. Performing elementary row operations<\/strong> so that the identity matrix<\/strong> appears on the left, we will obtain the inverse matrix<\/strong> on the right. We will find the inverse of this matrix in the next example.<\/p>\n

                    \n

                    How To: Given a [latex]3\\times 3[\/latex] matrix, find the inverse<\/h3>\n
                      \n
                    1. Write the original matrix augmented with the identity matrix on the right.<\/li>\n
                    2. Use elementary row operations so that the identity appears on the left.<\/li>\n
                    3. What is obtained on the right is the inverse of the original matrix.<\/li>\n
                    4. Use matrix multiplication to show that [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].<\/li>\n<\/ol>\n<\/div>\n
                      \n

                      Example 6: Finding the Inverse of a 3 \u00d7 3 Matrix<\/h3>\n

                      Given the [latex]3\\times 3[\/latex] matrix [latex]A[\/latex], find the inverse.<\/p>\n

                      [latex]A=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
                      \n

                      Solution<\/h3>\n

                      Augment [latex]A[\/latex] with the identity matrix, and then begin row operations until the identity matrix replaces [latex]A[\/latex]. The matrix on the right will be the inverse of [latex]A[\/latex].<\/p>\n

                      [latex]\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\stackrel{\\text{Interchange }{R}_{2}\\text{and }{R}_{1}}{\\to }\\left[\\begin{array}{ccc}3& 3& 1\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{ccc}0& 1& 0\\\\ 1& 0& 0\\\\ 0& 0& 1\\end{array}\\right][\/latex]<\/div>\n
                      [latex]-{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 2& 4& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n
                      [latex]-{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 2& 3& 1\\\\ 0& 1& 0\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n
                      [latex]{R}_{3}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 2& 3& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 1& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
                      [latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 3& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 3& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
                      [latex]-3{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}|\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right][\/latex]<\/div>\n

                      Thus,<\/p>\n

                      [latex]{A}^{-1}=B=\\left[\\begin{array}{ccc}-1& 1& 0\\\\ -1& 0& 1\\\\ 6& -2& -3\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
                      \n

                      Analysis of the Solution<\/h3>\n

                      To prove that [latex]B={A}^{-1}[\/latex], let\u2019s multiply the two matrices together to see if the product equals the identity, if [latex]A{A}^{-1}=I[\/latex] and [latex]{A}^{-1}A=I[\/latex].<\/p>\n

                      [latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ A{A}^{-1}=\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{ccc}2\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& 2\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& 2\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 3\\left(-1\\right)+3\\left(-1\\right)+1\\left(6\\right)& 3\\left(1\\right)+3\\left(0\\right)+1\\left(-2\\right)& 3\\left(0\\right)+3\\left(1\\right)+1\\left(-3\\right)\\\\ 2\\left(-1\\right)+4\\left(-1\\right)+1\\left(6\\right)& 2\\left(1\\right)+4\\left(0\\right)+1\\left(-2\\right)& 2\\left(0\\right)+4\\left(1\\right)+1\\left(-3\\right)\\end{array}\\right]\\hfill \\\\ =\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n
                      [latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ {A}^{-1}A=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 1& \\hfill 0\\\\ \\hfill -1& \\hfill 0& \\hfill 1\\\\ \\hfill 6& \\hfill -2& \\hfill -3\\end{array}\\right]\\text{ }\\left[\\begin{array}{ccc}2& 3& 1\\\\ 3& 3& 1\\\\ 2& 4& 1\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -1\\left(2\\right)+1\\left(3\\right)+0\\left(2\\right)& \\hfill -1\\left(3\\right)+1\\left(3\\right)+0\\left(4\\right)& \\hfill -1\\left(1\\right)+1\\left(1\\right)+0\\left(1\\right)\\\\ \\hfill -1\\left(2\\right)+0\\left(3\\right)+1\\left(2\\right)& \\hfill -1\\left(3\\right)+0\\left(3\\right)+1\\left(4\\right)& \\hfill -1\\left(1\\right)+0\\left(1\\right)+1\\left(1\\right)\\\\ \\hfill 6\\left(2\\right)+-2\\left(3\\right)+-3\\left(2\\right)& \\hfill 6\\left(3\\right)+-2\\left(3\\right)+-3\\left(4\\right)& \\hfill 6\\left(1\\right)+-2\\left(1\\right)+-3\\left(1\\right)\\end{array}\\right]\\hfill \\\\ =\\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}[\/latex]<\/div>\n<\/div>\n
                      \n

                      Try It 3<\/h3>\n

                      Find the inverse of the [latex]3\\times 3[\/latex] matrix.<\/p>\n

                      [latex]A=\\left[\\begin{array}{ccc}2& -17& 11\\\\ -1& 11& -7\\\\ 0& 3& -2\\end{array}\\right][\/latex]<\/div>\n
                      Solution<\/a><\/div>\n<\/div>\n\n\t\t\t
                      \n\t\t\t

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