{"id":14223,"date":"2018-06-15T19:27:44","date_gmt":"2018-06-15T19:27:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/using-cramers-rule-to-solve-a-system-of-three-equations-in-three-variables\/"},"modified":"2022-03-21T21:42:40","modified_gmt":"2022-03-21T21:42:40","slug":"using-cramers-rule-to-solve-a-system-of-three-equations-in-three-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/using-cramers-rule-to-solve-a-system-of-three-equations-in-three-variables\/","title":{"raw":"Using Cramer\u2019s Rule to Solve a System of Three Equations in Three Variables","rendered":"Using Cramer\u2019s Rule to Solve a System of Three Equations in Three Variables"},"content":{"raw":"

Evaluating the Determinant of a 3 \u00d7 3 Matrix<\/h2>\r\nFinding the determinant of a 2\u00d72 matrix is straightforward, but finding the determinant of a 3\u00d73 matrix is more complicated. One method is to augment the 3\u00d73 matrix with a repetition of the first two columns, giving a 3\u00d75 matrix. Then we calculate the sum of the products of entries down<\/em> each of the three diagonals (upper left to lower right), and subtract the products of entries up<\/em> each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example.\r\n\r\nFind the determinant<\/strong> of the 3\u00d73 matrix.\r\n
[latex]A=\\left[\\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\\\ {a}_{2}& {b}_{2}& {c}_{2}\\\\ {a}_{3}& {b}_{3}& {c}_{3}\\end{array}\\right][\/latex]<\/div>\r\n
    \r\n \t
  1. Augment [latex]A[\/latex] with the first two columns.\r\n
    [latex]\\mathrm{det}\\left(A\\right)=\\left|\\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\\\ {a}_{2}& {b}_{2}& {c}_{2}\\\\ {a}_{3}& {b}_{3}& {c}_{3}\\end{array}\\right|\\left.\\begin{array}{c}{a}_{1}\\\\ {a}_{2}\\\\ {a}_{3}\\end{array}\\begin{array}{c}{b}_{1}\\\\ {b}_{2}\\\\ {b}_{3}\\end{array}\\right|[\/latex]<\/div><\/li>\r\n \t
  2. From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal.<\/li>\r\n \t
  3. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal.<\/li>\r\n<\/ol>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\" Figure 2<\/b>[\/caption]\r\n\r\nThe algebra is as follows:\r\n
    [latex]|A|={a}_{1}{b}_{2}{c}_{3}+{b}_{1}{c}_{2}{a}_{3}+{c}_{1}{a}_{2}{b}_{3}-{a}_{3}{b}_{2}{c}_{1}-{b}_{3}{c}_{2}{a}_{1}-{c}_{3}{a}_{2}{b}_{1}[\/latex]<\/div>\r\n
    \r\n

    Example 3: Finding the Determinant of a 3 \u00d7 3 Matrix<\/h3>\r\nFind the determinant of the 3 \u00d7 3 matrix given\r\n
    [latex]A=\\left[\\begin{array}{ccc}0& 2& 1\\\\ 3& -1& 1\\\\ 4& 0& 1\\end{array}\\right][\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nAugment the matrix with the first two columns and then follow the formula. Thus,\r\n
    [latex]\\begin{array}{l}|A|=\\left|\\begin{array}{ccc} {0}& {2}& {1}\\\\ 3& -1& 1\\\\ 4& 0& 1\\end{array}\\right.\\left|\\begin{array}{c}0\\\\ 3\\\\ 4\\end{array}\\begin{array}{c}2\\\\ -1\\\\ 0\\end{array}\\right|\\hfill \\\\ =0\\left(-1\\right)\\left(1\\right)+2\\left(1\\right)\\left(4\\right)+1\\left(3\\right)\\left(0\\right)-4\\left(-1\\right)\\left(1\\right)-0\\left(1\\right)\\left(0\\right)-1\\left(3\\right)\\left(2\\right)\\hfill \\\\ =0+8+0+4 - 0-6\\hfill \\\\ =6\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 2<\/h3>\r\nFind the determinant of the 3 \u00d7 3 matrix.\r\n
    [latex]\\mathrm{det}\\left(A\\right)=\\left|\\begin{array}{ccc}1& -3& 7\\\\ 1& 1& 1\\\\ 1& -2& 3\\end{array}\\right|[\/latex]<\/div>\r\n
    [reveal-answer q=\"188689\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"188689\"][latex]-10[\/latex][\/hidden-answer]<\/div>\r\n
    <\/div>\r\n<\/div>\r\n
    \r\n

    Q & A<\/h3>\r\n

    Can we use the same method to find the determinant of a larger matrix?<\/h3>\r\nNo, this method only works for [latex]2\\text{ }\\times \\text{ }2[\/latex] and [latex]\\text{3}\\text{ }\\times \\text{ }3[\/latex] matrices. For larger matrices it is best to use a graphing utility or computer software.<\/em>\r\n\r\n<\/div>\r\n

    Using Cramer\u2019s Rule to Solve a System of Three Equations in Three Variables<\/h2>\r\nNow that we can find the determinant<\/strong> of a 3 \u00d7 3 matrix, we can apply Cramer\u2019s Rule<\/strong> to solve a system of three equations in three variables<\/strong>. Cramer\u2019s Rule is straightforward, following a pattern consistent with Cramer\u2019s Rule for 2 \u00d7 2 matrices. As the order of the matrix increases to 3 \u00d7 3, however, there are many more calculations required.\r\n\r\nWhen we calculate the determinant to be zero, Cramer\u2019s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system.\r\n\r\nConsider a 3 \u00d7 3 system of equations.\r\n\r\n[caption id=\"attachment_12665\" align=\"aligncenter\" width=\"243\"]\"\"<\/a> Figure 3<\/strong>[\/caption]\r\n\r\n
    [latex]x=\\frac{{D}_{x}}{D},y=\\frac{{D}_{y}}{D},z=\\frac{{D}_{z}}{D},D\\ne 0[\/latex]<\/div>\r\nwhere\r\n\r\n[caption id=\"attachment_12667\" align=\"aligncenter\" width=\"856\"]\"\"<\/a> Figure 4<\/strong>[\/caption]\r\n\r\nIf we are writing the determinant [latex]{D}_{x}[\/latex], we replace the [latex]x[\/latex] column with the constant column. If we are writing the determinant [latex]{D}_{y}[\/latex], we replace the [latex]y[\/latex] column with the constant column. If we are writing the determinant [latex]{D}_{z}[\/latex], we replace the [latex]z[\/latex] column with the constant column. Always check the answer.\r\n
    \r\n

    Example 4: Solving a 3 \u00d7 3 System Using Cramer\u2019s Rule<\/h3>\r\nFind the solution to the given 3 \u00d7 3 system using Cramer\u2019s Rule.\r\n
    [latex]\\begin{array}{c}x+y-z=6\\\\ 3x - 2y+z=-5\\\\ x+3y - 2z=14\\end{array}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nUse Cramer\u2019s Rule.\r\n
    [latex]D=\\left|\\begin{array}{ccc}1& 1& -1\\\\ 3& -2& 1\\\\ 1& 3& -2\\end{array}\\right|,{D}_{x}=\\left|\\begin{array}{ccc}6& 1& -1\\\\ -5& -2& 1\\\\ 14& 3& -2\\end{array}\\right|,{D}_{y}=\\left|\\begin{array}{ccc}1& 6& -1\\\\ 3& -5& 1\\\\ 1& 14& -2\\end{array}\\right|,{D}_{z}=\\left|\\begin{array}{ccc}1& 1& 6\\\\ 3& -2& -5\\\\ 1& 3& 14\\end{array}\\right|[\/latex]<\/div>\r\nThen,\r\n
    [latex]\\begin{array}{l}x=\\frac{{D}_{x}}{D}=\\frac{-3}{-3}=1\\hfill \\\\ y=\\frac{{D}_{y}}{D}=\\frac{-9}{-3}=3\\hfill \\\\ z=\\frac{{D}_{z}}{D}=\\frac{6}{-3}=-2\\hfill \\end{array}[\/latex]<\/div>\r\nThe solution is [latex]\\left(1,3,-2\\right)[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 3<\/h3>\r\nUse Cramer\u2019s Rule to solve the 3 \u00d7 3 system of equations.\r\n
    [latex]\\begin{array}{r}\\hfill x - 3y+7z=13\\\\ \\hfill x+y+z=1\\\\ \\hfill x - 2y+3z=4\\end{array}[\/latex]<\/div>\r\n
    <\/div>\r\n
    [reveal-answer q=\"389404\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"389404\"][latex](\u22122, \u200b\\frac{3}{5}\u200b\u200b, \u200b\\frac{12}{5} \u200b\u200b )[\/latex][\/hidden-answer]<\/div>\r\n
    <\/div>\r\n
    <\/div>\r\n<\/div>\r\n
    \r\n

    Example 5: Using Cramer\u2019s Rule to Solve an Inconsistent System<\/h3>\r\nSolve the system of equations using Cramer\u2019s Rule.\r\n
    [latex]\\begin{array}{l}3x - 2y=4\\text{ }\\left(1\\right)\\\\ 6x - 4y=0\\text{ }\\left(2\\right)\\end{array}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nWe begin by finding the determinants [latex]D,{D}_{x},\\text{and }{D}_{y}[\/latex].\r\n
    [latex]D=\\left|\\begin{array}{cc}3& -2\\\\ 6& -4\\end{array}\\right|=3\\left(-4\\right)-6\\left(-2\\right)=0[\/latex]<\/div>\r\nWe know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables.\r\n
      \r\n \t
    1. Multiply equation (1) by [latex]-2[\/latex].<\/li>\r\n \t
    2. Add the result to equation [latex]\\left(2\\right)[\/latex].<\/li>\r\n<\/ol>\r\n
      [latex]\\begin{array}\\text{ }\\hfill\u22126x+4y=\u22128 \\\\ \\hfill6x\u22124y=0 \\\\ \\hfill\\text{_____________} \\\\ \\hfill 0=8\\end{array}[\/latex]<\/div>\r\nWe obtain the equation [latex]0=-8[\/latex], which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 5<\/b>[\/caption]\r\n\r\n<\/div>\r\n
      \r\n

      Example 6: Use Cramer\u2019s Rule to Solve a Dependent System<\/h3>\r\nSolve the system with an infinite number of solutions.\r\n
      [latex]\\begin{array}{rr}\\hfill x - 2y+3z=0& \\hfill \\left(1\\right)\\\\ \\hfill 3x+y - 2z=0& \\hfill \\left(2\\right)\\\\ \\hfill 2x - 4y+6z=0& \\hfill \\left(3\\right)\\end{array}[\/latex]<\/div>\r\n<\/div>\r\n
      \r\n

      Solution<\/h3>\r\nLet\u2019s find the determinant first. Set up a matrix augmented by the first two columns.\r\n
      [latex]\\left|\\begin{array}{rrr}\\hfill 1& \\hfill -2& \\hfill 3\\\\ \\hfill 3& \\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -4& \\hfill 6\\end{array}\\text{ }\\right.\\left|\\text{ }\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 3& \\hfill 1\\\\ \\hfill 2& \\hfill -4\\end{array}\\right|[\/latex]<\/div>\r\nThen,\r\n
      [latex]1\\left(1\\right)\\left(6\\right)+\\left(-2\\right)\\left(-2\\right)\\left(2\\right)+3\\left(3\\right)\\left(-4\\right)-2\\left(1\\right)\\left(3\\right)-\\left(-4\\right)\\left(-2\\right)\\left(1\\right)-6\\left(3\\right)\\left(-2\\right)=0[\/latex]<\/div>\r\nAs the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out.\r\n
        \r\n \t
      1. Multiply equation (1) by [latex]-2[\/latex] and add the result to equation (3):\r\n
        [latex]\\frac{\\begin{array}{r}\\hfill -2x+4y - 6x=0\\\\ \\hfill 2x - 4y+6z=0\\end{array}}{0=0}[\/latex]<\/div><\/li>\r\n \t
      2. Obtaining an answer of [latex]0=0[\/latex], a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line.<\/li>\r\n<\/ol>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Two Figure 6<\/b>[\/caption]\r\n\r\n<\/div>","rendered":"

        Evaluating the Determinant of a 3 \u00d7 3 Matrix<\/h2>\n

        Finding the determinant of a 2\u00d72 matrix is straightforward, but finding the determinant of a 3\u00d73 matrix is more complicated. One method is to augment the 3\u00d73 matrix with a repetition of the first two columns, giving a 3\u00d75 matrix. Then we calculate the sum of the products of entries down<\/em> each of the three diagonals (upper left to lower right), and subtract the products of entries up<\/em> each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example.<\/p>\n

        Find the determinant<\/strong> of the 3\u00d73 matrix.<\/p>\n

        [latex]A=\\left[\\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\\\ {a}_{2}& {b}_{2}& {c}_{2}\\\\ {a}_{3}& {b}_{3}& {c}_{3}\\end{array}\\right][\/latex]<\/div>\n
          \n
        1. Augment [latex]A[\/latex] with the first two columns.\n
          [latex]\\mathrm{det}\\left(A\\right)=\\left|\\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\\\ {a}_{2}& {b}_{2}& {c}_{2}\\\\ {a}_{3}& {b}_{3}& {c}_{3}\\end{array}\\right|\\left.\\begin{array}{c}{a}_{1}\\\\ {a}_{2}\\\\ {a}_{3}\\end{array}\\begin{array}{c}{b}_{1}\\\\ {b}_{2}\\\\ {b}_{3}\\end{array}\\right|[\/latex]<\/div>\n<\/li>\n
        2. From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal.<\/li>\n
        3. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal.<\/li>\n<\/ol>\n
          \"\"<\/p>\n

          Figure 2<\/b><\/p>\n<\/div>\n

          The algebra is as follows:<\/p>\n

          [latex]|A|={a}_{1}{b}_{2}{c}_{3}+{b}_{1}{c}_{2}{a}_{3}+{c}_{1}{a}_{2}{b}_{3}-{a}_{3}{b}_{2}{c}_{1}-{b}_{3}{c}_{2}{a}_{1}-{c}_{3}{a}_{2}{b}_{1}[\/latex]<\/div>\n
          \n

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

          Find the determinant of the 3 \u00d7 3 matrix given<\/p>\n

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

          Solution<\/h3>\n

          Augment the matrix with the first two columns and then follow the formula. Thus,<\/p>\n

          [latex]\\begin{array}{l}|A|=\\left|\\begin{array}{ccc} {0}& {2}& {1}\\\\ 3& -1& 1\\\\ 4& 0& 1\\end{array}\\right.\\left|\\begin{array}{c}0\\\\ 3\\\\ 4\\end{array}\\begin{array}{c}2\\\\ -1\\\\ 0\\end{array}\\right|\\hfill \\\\ =0\\left(-1\\right)\\left(1\\right)+2\\left(1\\right)\\left(4\\right)+1\\left(3\\right)\\left(0\\right)-4\\left(-1\\right)\\left(1\\right)-0\\left(1\\right)\\left(0\\right)-1\\left(3\\right)\\left(2\\right)\\hfill \\\\ =0+8+0+4 - 0-6\\hfill \\\\ =6\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
          \n

          Try It 2<\/h3>\n

          Find the determinant of the 3 \u00d7 3 matrix.<\/p>\n

          [latex]\\mathrm{det}\\left(A\\right)=\\left|\\begin{array}{ccc}1& -3& 7\\\\ 1& 1& 1\\\\ 1& -2& 3\\end{array}\\right|[\/latex]<\/div>\n
          \n
          Show Answer<\/span><\/p>\n
          [latex]-10[\/latex]<\/div>\n<\/div>\n<\/div>\n
          <\/div>\n<\/div>\n
          \n

          Q & A<\/h3>\n

          Can we use the same method to find the determinant of a larger matrix?<\/h3>\n

          No, this method only works for [latex]2\\text{ }\\times \\text{ }2[\/latex] and [latex]\\text{3}\\text{ }\\times \\text{ }3[\/latex] matrices. For larger matrices it is best to use a graphing utility or computer software.<\/em><\/p>\n<\/div>\n

          Using Cramer\u2019s Rule to Solve a System of Three Equations in Three Variables<\/h2>\n

          Now that we can find the determinant<\/strong> of a 3 \u00d7 3 matrix, we can apply Cramer\u2019s Rule<\/strong> to solve a system of three equations in three variables<\/strong>. Cramer\u2019s Rule is straightforward, following a pattern consistent with Cramer\u2019s Rule for 2 \u00d7 2 matrices. As the order of the matrix increases to 3 \u00d7 3, however, there are many more calculations required.<\/p>\n

          When we calculate the determinant to be zero, Cramer\u2019s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system.<\/p>\n

          Consider a 3 \u00d7 3 system of equations.<\/p>\n

          \"\"<\/a><\/p>\n

          Figure 3<\/strong><\/p>\n<\/div>\n

          [latex]x=\\frac{{D}_{x}}{D},y=\\frac{{D}_{y}}{D},z=\\frac{{D}_{z}}{D},D\\ne 0[\/latex]<\/div>\n

          where<\/p>\n

          \"\"<\/a><\/p>\n

          Figure 4<\/strong><\/p>\n<\/div>\n

          If we are writing the determinant [latex]{D}_{x}[\/latex], we replace the [latex]x[\/latex] column with the constant column. If we are writing the determinant [latex]{D}_{y}[\/latex], we replace the [latex]y[\/latex] column with the constant column. If we are writing the determinant [latex]{D}_{z}[\/latex], we replace the [latex]z[\/latex] column with the constant column. Always check the answer.<\/p>\n

          \n

          Example 4: Solving a 3 \u00d7 3 System Using Cramer\u2019s Rule<\/h3>\n

          Find the solution to the given 3 \u00d7 3 system using Cramer\u2019s Rule.<\/p>\n

          [latex]\\begin{array}{c}x+y-z=6\\\\ 3x - 2y+z=-5\\\\ x+3y - 2z=14\\end{array}[\/latex]<\/div>\n<\/div>\n
          \n

          Solution<\/h3>\n

          Use Cramer\u2019s Rule.<\/p>\n

          [latex]D=\\left|\\begin{array}{ccc}1& 1& -1\\\\ 3& -2& 1\\\\ 1& 3& -2\\end{array}\\right|,{D}_{x}=\\left|\\begin{array}{ccc}6& 1& -1\\\\ -5& -2& 1\\\\ 14& 3& -2\\end{array}\\right|,{D}_{y}=\\left|\\begin{array}{ccc}1& 6& -1\\\\ 3& -5& 1\\\\ 1& 14& -2\\end{array}\\right|,{D}_{z}=\\left|\\begin{array}{ccc}1& 1& 6\\\\ 3& -2& -5\\\\ 1& 3& 14\\end{array}\\right|[\/latex]<\/div>\n

          Then,<\/p>\n

          [latex]\\begin{array}{l}x=\\frac{{D}_{x}}{D}=\\frac{-3}{-3}=1\\hfill \\\\ y=\\frac{{D}_{y}}{D}=\\frac{-9}{-3}=3\\hfill \\\\ z=\\frac{{D}_{z}}{D}=\\frac{6}{-3}=-2\\hfill \\end{array}[\/latex]<\/div>\n

          The solution is [latex]\\left(1,3,-2\\right)[\/latex].<\/p>\n<\/div>\n

          \n

          Try It 3<\/h3>\n

          Use Cramer\u2019s Rule to solve the 3 \u00d7 3 system of equations.<\/p>\n

          [latex]\\begin{array}{r}\\hfill x - 3y+7z=13\\\\ \\hfill x+y+z=1\\\\ \\hfill x - 2y+3z=4\\end{array}[\/latex]<\/div>\n
          <\/div>\n
          \n
          Show Answer<\/span><\/p>\n
          [latex](\u22122, \u200b\\frac{3}{5}\u200b\u200b, \u200b\\frac{12}{5} \u200b\u200b )[\/latex]<\/div>\n<\/div>\n<\/div>\n
          <\/div>\n
          <\/div>\n<\/div>\n
          \n

          Example 5: Using Cramer\u2019s Rule to Solve an Inconsistent System<\/h3>\n

          Solve the system of equations using Cramer\u2019s Rule.<\/p>\n

          [latex]\\begin{array}{l}3x - 2y=4\\text{ }\\left(1\\right)\\\\ 6x - 4y=0\\text{ }\\left(2\\right)\\end{array}[\/latex]<\/div>\n<\/div>\n
          \n

          Solution<\/h3>\n

          We begin by finding the determinants [latex]D,{D}_{x},\\text{and }{D}_{y}[\/latex].<\/p>\n

          [latex]D=\\left|\\begin{array}{cc}3& -2\\\\ 6& -4\\end{array}\\right|=3\\left(-4\\right)-6\\left(-2\\right)=0[\/latex]<\/div>\n

          We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables.<\/p>\n

            \n
          1. Multiply equation (1) by [latex]-2[\/latex].<\/li>\n
          2. Add the result to equation [latex]\\left(2\\right)[\/latex].<\/li>\n<\/ol>\n
            [latex]\\begin{array}\\text{ }\\hfill\u22126x+4y=\u22128 \\\\ \\hfill6x\u22124y=0 \\\\ \\hfill\\text{_____________} \\\\ \\hfill 0=8\\end{array}[\/latex]<\/div>\n

            We obtain the equation [latex]0=-8[\/latex], which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines.<\/p>\n

            \"Graph<\/p>\n

            Figure 5<\/b><\/p>\n<\/div>\n<\/div>\n

            \n

            Example 6: Use Cramer\u2019s Rule to Solve a Dependent System<\/h3>\n

            Solve the system with an infinite number of solutions.<\/p>\n

            [latex]\\begin{array}{rr}\\hfill x - 2y+3z=0& \\hfill \\left(1\\right)\\\\ \\hfill 3x+y - 2z=0& \\hfill \\left(2\\right)\\\\ \\hfill 2x - 4y+6z=0& \\hfill \\left(3\\right)\\end{array}[\/latex]<\/div>\n<\/div>\n
            \n

            Solution<\/h3>\n

            Let\u2019s find the determinant first. Set up a matrix augmented by the first two columns.<\/p>\n

            [latex]\\left|\\begin{array}{rrr}\\hfill 1& \\hfill -2& \\hfill 3\\\\ \\hfill 3& \\hfill 1& \\hfill -2\\\\ \\hfill 2& \\hfill -4& \\hfill 6\\end{array}\\text{ }\\right.\\left|\\text{ }\\begin{array}{rr}\\hfill 1& \\hfill -2\\\\ \\hfill 3& \\hfill 1\\\\ \\hfill 2& \\hfill -4\\end{array}\\right|[\/latex]<\/div>\n

            Then,<\/p>\n

            [latex]1\\left(1\\right)\\left(6\\right)+\\left(-2\\right)\\left(-2\\right)\\left(2\\right)+3\\left(3\\right)\\left(-4\\right)-2\\left(1\\right)\\left(3\\right)-\\left(-4\\right)\\left(-2\\right)\\left(1\\right)-6\\left(3\\right)\\left(-2\\right)=0[\/latex]<\/div>\n

            As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out.<\/p>\n

              \n
            1. Multiply equation (1) by [latex]-2[\/latex] and add the result to equation (3):\n
              [latex]\\frac{\\begin{array}{r}\\hfill -2x+4y - 6x=0\\\\ \\hfill 2x - 4y+6z=0\\end{array}}{0=0}[\/latex]<\/div>\n<\/li>\n
            2. Obtaining an answer of [latex]0=0[\/latex], a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line.<\/li>\n<\/ol>\n
              \"Two<\/p>\n

              Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n\n\t\t\t

              \n\t\t\t

              Candela Citations<\/h3>\n\t\t\t\t\t
              \n\t\t\t\t\t\t
              \n\t\t\t\t\t\t\t
              CC licensed content, Specific attribution<\/div>