Solve Using a Matrix by Elimination x_1+6x_2-2x_3=10 3x_1+8x_3=-4 -2x_1+5x_2+3x_3=22

$x_{1} + 6 x_{2} - 2 x_{3} = 10$ $3 x_{1} + 8 x_{3} = - 4$ $- 2 x_{1} + 5 x_{2} + 3 x_{3} = 22$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ 3 & 0 & 8 & - 4 \\ - 2 & 5 & 3 & 22 \end{array}]$

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} \\ - 2 & 5 & 3 & 22 \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (6) + 0 & (- 3) \cdot (- 2) + 8 & (- 3) \cdot (10) - 4 \\ - 2 & 5 & 3 & 22 \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ 0 & - 18 & 14 & - 34 \\ - 2 & 5 & 3 & 22 \end{array}]$

Perform the row operation $R_{3} = 2 \cdot R_{1} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = 2 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ 0 & - 18 & 14 & - 34 \\ 2 \cdot R_{1} + R_{3} & 2 \cdot R_{1} + R_{3} & 2 \cdot R_{1} + R_{3} & 2 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = 2 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = 2 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ 0 & - 18 & 14 & - 34 \\ (2) \cdot (1) - 2 & (2) \cdot (6) + 5 & (2) \cdot (- 2) + 3 & (2) \cdot (10) + 22 \end{array}]$

$R_{3} = 2 \cdot R_{1} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ 0 & - 18 & 14 & - 34 \\ 0 & 17 & - 1 & 42 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{18} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - \frac{1}{18} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ - \frac{1}{18} R_{2} & - \frac{1}{18} R_{2} & - \frac{1}{18} R_{2} & - \frac{1}{18} R_{2} \\ 0 & 17 & - 1 & 42 \end{array}]$

$R_{2} = - \frac{1}{18} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{18} R_{2}$ .

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ (- \frac{1}{18}) \cdot (0) & (- \frac{1}{18}) \cdot (- 18) & (- \frac{1}{18}) \cdot (14) & (- \frac{1}{18}) \cdot (- 34) \\ 0 & 17 & - 1 & 42 \end{array}]$

$R_{2} = - \frac{1}{18} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 6 & - 2 & 10 \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 17 & - 1 & 42 \end{array}]$

Perform the row operation $R_{1} = - 6 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - 6 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - 6 \cdot R_{2} + R_{1} & - 6 \cdot R_{2} + R_{1} & - 6 \cdot R_{2} + R_{1} & - 6 \cdot R_{2} + R_{1} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 17 & - 1 & 42 \end{array}]$

$R_{1} = - 6 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - 6 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (- 6) \cdot (0) + 1 & (- 6) \cdot (1) + 6 & (- 6) \cdot (- \frac{7}{9}) - 2 & (- 6) \cdot (\frac{17}{9}) + 10 \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 17 & - 1 & 42 \end{array}]$

$R_{1} = - 6 \cdot R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 17 & - 1 & 42 \end{array}]$

Perform the row operation $R_{3} = - 17 \cdot R_{2} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 17 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ - 17 \cdot R_{2} + R_{3} & - 17 \cdot R_{2} + R_{3} & - 17 \cdot R_{2} + R_{3} & - 17 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 17 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 17 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ (- 17) \cdot (0) + 0 & (- 17) \cdot (1) + 17 & (- 17) \cdot (- \frac{7}{9}) - 1 & (- 17) \cdot (\frac{17}{9}) + 42 \end{array}]$

$R_{3} = - 17 \cdot R_{2} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 0 & \frac{110}{9} & \frac{89}{9} \end{array}]$

Perform the row operation $R_{3} = \frac{9}{110} R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $1$ .

Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = \frac{9}{110} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ \frac{9}{110} R_{3} & \frac{9}{110} R_{3} & \frac{9}{110} R_{3} & \frac{9}{110} R_{3} \end{array}]$

$R_{3} = \frac{9}{110} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = \frac{9}{110} R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ (\frac{9}{110}) \cdot (0) & (\frac{9}{110}) \cdot (0) & (\frac{9}{110}) \cdot (\frac{110}{9}) & (\frac{9}{110}) \cdot (\frac{89}{9}) \end{array}]$

$R_{3} = \frac{9}{110} R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & \frac{8}{3} & - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

Perform the row operation $R_{1} = - \frac{8}{3} R_{3} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{8}{3} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{8}{3} R_{3} + R_{1} & - \frac{8}{3} R_{3} + R_{1} & - \frac{8}{3} R_{3} + R_{1} & - \frac{8}{3} R_{3} + R_{1} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

$R_{1} = - \frac{8}{3} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{8}{3} R_{3} + R_{1}$ .

$[\begin{array}{c} (- \frac{8}{3}) \cdot (0) + 1 & (- \frac{8}{3}) \cdot (0) + 0 & (- \frac{8}{3}) \cdot (1) + \frac{8}{3} & (- \frac{8}{3}) \cdot (\frac{89}{110}) - \frac{4}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

$R_{1} = - \frac{8}{3} R_{3} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & 0 & - \frac{192}{55} \\ 0 & 1 & - \frac{7}{9} & \frac{17}{9} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

Perform the row operation $R_{2} = \frac{7}{9} R_{3} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{7}{9} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{192}{55} \\ \frac{7}{9} R_{3} + R_{2} & \frac{7}{9} R_{3} + R_{2} & \frac{7}{9} R_{3} + R_{2} & \frac{7}{9} R_{3} + R_{2} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

$R_{2} = \frac{7}{9} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{7}{9} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{192}{55} \\ (\frac{7}{9}) \cdot (0) + 0 & (\frac{7}{9}) \cdot (0) + 1 & (\frac{7}{9}) \cdot (1) - \frac{7}{9} & (\frac{7}{9}) \cdot (\frac{89}{110}) + \frac{17}{9} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

$R_{2} = \frac{7}{9} R_{3} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 0 & 0 & - \frac{192}{55} \\ 0 & 1 & 0 & \frac{277}{110} \\ 0 & 0 & 1 & \frac{89}{110} \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x_{1} = - \frac{192}{55}$

$x_{2} = \frac{277}{110}$

$x_{3} = \frac{89}{110}$

The solution is the set of ordered pairs that makes the system true.

$(- \frac{192}{55}, \frac{277}{110}, \frac{89}{110})$

Solve Using a Matrix by Elimination x_1+6x_2-2x_3=10 3x_1+8x_3=-4 -2x_1+5x_2+3x_3=22

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