{"id":1374,"date":"2025-07-24T19:14:09","date_gmt":"2025-07-24T19:14:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1374"},"modified":"2026-03-24T16:28:51","modified_gmt":"2026-03-24T16:28:51","slug":"solving-systems-with-gaussian-elimination-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-gaussian-elimination-learn-it-3\/","title":{"raw":"Solving Systems with Gaussian Elimination: Learn It 2","rendered":"Solving Systems with Gaussian Elimination: Learn It 2"},"content":{"raw":"

Performing Row Operations on a Matrix<\/h2>\r\nPerforming row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form<\/strong>, in which there are ones down the main diagonal<\/strong> from the upper left corner to the lower right corner and zeros in every position below the main diagonal as shown.\r\n
[latex]\\begin{array}{c}\\text{Row-echelon form}\\\\ \\left[\\begin{array}{ccc}1& a& b\\\\ 0& 1& d\\\\ 0& 0& 1\\end{array}\\right]\\end{array}[\/latex]<\/div>\r\n
\r\n

Row-echelon Form<\/h3>\r\nRow echelon form<\/strong> is a specific arrangement of a matrix used to solve systems of linear equations. In this form, each row has a leading entry of [latex]1[\/latex] (called a leading [latex]1[\/latex]) that appears to the right of the leading [latex]1[\/latex] in the row above it. All entries below each leading [latex]1[\/latex] are zeros.\r\n\r\n \r\n

[latex]\\begin{array}{c}\\text{Row-echelon form}\\\\ \\left[\\begin{array}{ccc}1& a& b\\\\ 0& 1& d\\\\ 0& 0& 1\\end{array}\\right]\\end{array}[\/latex]<\/p>\r\n\r\n<\/section>We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent<\/strong> in a simpler form. Here are the guidelines to obtaining row-echelon form.\r\n

    \r\n \t
  1. In any nonzero row, the first nonzero number is a 1. It is called a leading<\/em> 1.<\/li>\r\n \t
  2. Any all-zero rows are placed at the bottom of the matrix.<\/li>\r\n \t
  3. Any leading 1 is below and to the right of a previous leading 1.<\/li>\r\n \t
  4. Any column containing a leading 1 has zeros in all other positions in the column.<\/li>\r\n<\/ol>\r\nEach of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.\r\n\r\n
    \r\n

    Gaussian Elimination<\/h3>\r\nThe Gaussian elimination<\/strong> method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix [latex]A[\/latex] with the number 1 as the entry down the main diagonal and have all zeros below.\r\n
    [latex]A=\\left[\\begin{array}{rrr}\\hfill {a}_{11}& \\hfill {a}_{12}& \\hfill {a}_{13}\\\\ \\hfill {a}_{21}& \\hfill {a}_{22}& \\hfill {a}_{23}\\\\ \\hfill {a}_{31}& \\hfill {a}_{32}& \\hfill {a}_{33}\\end{array}\\right]\\stackrel{\\text{After Gaussian elimination}}{\\to }A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill {b}_{12}& \\hfill {b}_{13}\\\\ \\hfill 0& \\hfill 1& \\hfill {b}_{23}\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\r\n
    <\/div>\r\n
    The first step of the Gaussian strategy includes obtaining a [latex]1[\/latex] as the first entry, so that row [latex]1[\/latex] may be used to alter the rows below.<\/div>\r\n<\/section>
    How To: Given an augmented matrix, perform row operations to achieve row-echelon form<\/strong>\r\n
      \r\n \t
    1. The first equation should have a leading coefficient of [latex]1[\/latex]. Interchange rows or multiply by a constant, if necessary.<\/li>\r\n \t
    2. Use row operations to obtain zeros down the first column below the first entry of [latex]1[\/latex].<\/li>\r\n \t
    3. Use row operations to obtain a [latex]1[\/latex] in row 2, column 2.<\/li>\r\n \t
    4. Use row operations to obtain zeros down column 2, below the entry of [latex]1[\/latex].<\/li>\r\n \t
    5. Use row operations to obtain a [latex]1[\/latex] in row 3, column 3.<\/li>\r\n \t
    6. Continue this process for all rows until there is a [latex]1[\/latex] in every entry down the main diagonal and there are only zeros below.<\/li>\r\n \t
    7. If any rows contain all zeros, place them at the bottom.<\/li>\r\n<\/ol>\r\n<\/section>
      Perform row operations on the given matrix to obtain row-echelon form.\r\n
      [latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 2& \\hfill -5& \\hfill 6& \\hfill 6\\\\ \\hfill -3& \\hfill 3& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\r\n[reveal-answer q=\"637835\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"637835\"]\r\n\r\nThe first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by [latex]-2[\/latex] and add it to row 2. Then replace row 2 with the result.\r\n
      [latex]-2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill -3& \\hfill 3& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\r\nNext, obtain a zero in row 3, column 1.\r\n
      [latex]3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill -6& \\hfill 16& \\hfill 15\\\\ \\end{array}\\right][\/latex]<\/div>\r\nNext, obtain a zero in row 3, column 2.\r\n
      [latex]6{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill 0& \\hfill 4& \\hfill 15\\\\ \\end{array}\\right][\/latex]<\/div>\r\nThe last step is to obtain a 1 in row 3, column 3.\r\n
      [latex]\\frac{1}{4}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill -6\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill \\frac{15}{4}\\\\ \\end{array}\\right][\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>We have seen how to write a system of equations<\/strong> with an augmented matrix\u00a0<\/strong>and then how to use row operations and back-substitution to obtain row-echelon form<\/strong>. Now we will use Gaussian Elimination as a tool for solving a system written as an augmented matrix.\u00a0In our first example, we will show you the process for using Gaussian Elimination on a system of two equations in two variables.\r\n\r\n
      Solve the given system by Gaussian elimination.\r\n
      [latex]\\begin{array}{l}2x+3y=6\\hfill \\\\ \\text{ }x-y=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n
      \r\n\r\n[reveal-answer q=\"656723\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"656723\"]\r\n\r\nFirst, we write this as an augmented matrix.\r\n\r\n
      [latex]\\left[\\begin{array}{cc|c}\\hfill 2& \\hfill 3& \\hfill 6\\\\ \\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\end{array}\\right][\/latex]<\/center>\r\nWe want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.\r\n\r\n
      [latex]{R}_{1}\\leftrightarrow {R}_{2}\\implies \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\hfill 2& \\hfill 3& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/center>\r\nWe now have a 1 as the first entry in row 1, column 1. Now let\u2019s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by [latex]-2[\/latex] and then adding the result to row 2.\r\n\r\n
      [latex]-2{R}_{1}+{R}_{2} \\rightarrow {R}_{2}\\implies \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\hfill 0& \\hfill 5& \\hfill 5\\\\ \\end{array}\\right][\/latex]<\/center>\r\nWe only have one more step, to multiply row 2 by [latex]\\frac{1}{5}[\/latex].\r\n\r\n
      [latex]\\frac{1}{5}{R}_{2} \\rightarrow {R}_{2}\\implies \\left[\\begin{array}{cc|c}\\hfill 1& \\hfill -1& \\hfill \\frac{1}{2}\\\\ \\hfill 0& \\hfill 1& \\hfill 1\\\\ \\end{array}\\right][\/latex]<\/center>\r\n
      Now we can either use back-substitution or get a 0 in row 1, column 2.<\/div>\r\n
        \r\n \t
      • Use back-substitution. The second row of the matrix represents [latex]y=1[\/latex]. Back-substitute [latex]y=1[\/latex] into the first equation.\r\n
        [latex]\\begin{array}{l}x-\\left(1\\right)=\\frac{1}{2}\\hfill \\\\ \\text{ }x=\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/center><\/li>\r\n \t
      • \u00a0Get a 0 in row 1, column 2.<\/li>\r\n<\/ul>\r\n
        [latex]R_1 + R_2 \\rightarrow R_1 \\implies \\left[\\begin{array}{cc|c} 1 & -1 & \\frac{1}{2} \\\\ 0 & 1 & 1 \\\\ \\end{array}\\right] + \\left[\\begin{array}{cc|c} 0 & 1 & 1 \\\\ 0 & 0 & 0 \\\\ \\end{array}\\right] = \\left[\\begin{array}{cc|c} 1 & 0 & \\frac{3}{2} \\\\ 0 & 1 & 1 \\\\ \\end{array}\\right][\/latex]<\/center>\r\nThe solution is the point [latex]\\left(\\frac{3}{2},1\\right)[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
        [ohm_question hide_question_numbers=1]321749[\/ohm_question]<\/section>In our next example, we will solve a system of two equations in two variables that is dependent. Recall that a dependent system has an infinite number of solutions and the result of row operations on its augmented matrix will be an equation such as [latex]0=0[\/latex]. We also review writing the general solution to a dependent system.\r\n\r\n
        Solve the system of equations.\r\n
        [latex]\\begin{array}{l}3x+4y=12\\\\ 6x+8y=24\\end{array}[\/latex]<\/center>\r\n[reveal-answer q=\"844840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"844840\"]\r\nPerform row operations<\/strong> on the augmented matrix to try and achieve row-echelon form<\/strong>.\r\n
        [latex]A=\\left[\\begin{array}{cc|c}\\hfill 3& \\hfill 4& \\hfill 12\\\\ \\hfill 6& \\hfill 8& \\hfill 24\\\\ \\end{array}\\right][\/latex]<\/center>\r\n
        [latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}-\\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\\to \\left[\\begin{array}{cc|c}\\hfill 0& \\hfill 0& \\hfill 0\\\\ \\hfill 6& \\hfill 8& \\hfill 24\\\\ \\end{array}\\right]\\hfill \\\\ {R}_{1}\\leftrightarrow {R}_{2}\\to \\left[\\begin{array}{cc|c}\\hfill 6& \\hfill 8& \\hfill 24\\\\ \\hfill 0& \\hfill 0& \\hfill 0\\\\ \\end{array}\\right]\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/center>\r\nThe matrix ends up with all zeros in the last row: [latex]0y=0[\/latex]. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for [latex]y[\/latex].\r\n
        [latex]\\begin{array}{l}3x+4y=12\\hfill \\\\ \\text{ }4y=12 - 3x\\hfill \\\\ \\text{ }y=3-\\frac{3}{4}x\\hfill \\end{array}[\/latex]<\/center>\r\nSo the solution to this system is [latex]\\left(x,3-\\frac{3}{4}x\\right)[\/latex].[\/hidden-answer]<\/section>","rendered":"

        Performing Row Operations on a Matrix<\/h2>\n

        Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form<\/strong>, in which there are ones down the main diagonal<\/strong> from the upper left corner to the lower right corner and zeros in every position below the main diagonal as shown.<\/p>\n

        [latex]\\begin{array}{c}\\text{Row-echelon form}\\\\ \\left[\\begin{array}{ccc}1& a& b\\\\ 0& 1& d\\\\ 0& 0& 1\\end{array}\\right]\\end{array}[\/latex]<\/div>\n
        \n

        Row-echelon Form<\/h3>\n

        Row echelon form<\/strong> is a specific arrangement of a matrix used to solve systems of linear equations. In this form, each row has a leading entry of [latex]1[\/latex] (called a leading [latex]1[\/latex]) that appears to the right of the leading [latex]1[\/latex] in the row above it. All entries below each leading [latex]1[\/latex] are zeros.<\/p>\n

         <\/p>\n

        [latex]\\begin{array}{c}\\text{Row-echelon form}\\\\ \\left[\\begin{array}{ccc}1& a& b\\\\ 0& 1& d\\\\ 0& 0& 1\\end{array}\\right]\\end{array}[\/latex]<\/p>\n<\/section>\n

        We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent<\/strong> in a simpler form. Here are the guidelines to obtaining row-echelon form.<\/p>\n

          \n
        1. In any nonzero row, the first nonzero number is a 1. It is called a leading<\/em> 1.<\/li>\n
        2. Any all-zero rows are placed at the bottom of the matrix.<\/li>\n
        3. Any leading 1 is below and to the right of a previous leading 1.<\/li>\n
        4. Any column containing a leading 1 has zeros in all other positions in the column.<\/li>\n<\/ol>\n

          Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.<\/p>\n

          \n

          Gaussian Elimination<\/h3>\n

          The Gaussian elimination<\/strong> method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix [latex]A[\/latex] with the number 1 as the entry down the main diagonal and have all zeros below.<\/p>\n

          [latex]A=\\left[\\begin{array}{rrr}\\hfill {a}_{11}& \\hfill {a}_{12}& \\hfill {a}_{13}\\\\ \\hfill {a}_{21}& \\hfill {a}_{22}& \\hfill {a}_{23}\\\\ \\hfill {a}_{31}& \\hfill {a}_{32}& \\hfill {a}_{33}\\end{array}\\right]\\stackrel{\\text{After Gaussian elimination}}{\\to }A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill {b}_{12}& \\hfill {b}_{13}\\\\ \\hfill 0& \\hfill 1& \\hfill {b}_{23}\\\\ \\hfill 0& \\hfill 0& \\hfill 1\\end{array}\\right][\/latex]<\/div>\n
          <\/div>\n
          The first step of the Gaussian strategy includes obtaining a [latex]1[\/latex] as the first entry, so that row [latex]1[\/latex] may be used to alter the rows below.<\/div>\n<\/section>\n
          How To: Given an augmented matrix, perform row operations to achieve row-echelon form<\/strong><\/p>\n
            \n
          1. The first equation should have a leading coefficient of [latex]1[\/latex]. Interchange rows or multiply by a constant, if necessary.<\/li>\n
          2. Use row operations to obtain zeros down the first column below the first entry of [latex]1[\/latex].<\/li>\n
          3. Use row operations to obtain a [latex]1[\/latex] in row 2, column 2.<\/li>\n
          4. Use row operations to obtain zeros down column 2, below the entry of [latex]1[\/latex].<\/li>\n
          5. Use row operations to obtain a [latex]1[\/latex] in row 3, column 3.<\/li>\n
          6. Continue this process for all rows until there is a [latex]1[\/latex] in every entry down the main diagonal and there are only zeros below.<\/li>\n
          7. If any rows contain all zeros, place them at the bottom.<\/li>\n<\/ol>\n<\/section>\n
            Perform row operations on the given matrix to obtain row-echelon form.<\/p>\n
            [latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -3& \\hfill 4& \\hfill 3\\\\ \\hfill 2& \\hfill -5& \\hfill 6& \\hfill 6\\\\ \\hfill -3& \\hfill 3& \\hfill 4& \\hfill 6\\\\ \\end{array}\\right][\/latex]<\/div>\n