Solve Using a Matrix by Elimination x-2y+4z=-14 , 3x+y-z=-6 , 2x+3y-6z=7

$x - 2 y + 4 z = - 14$ , $3 x + y - z = - 6$ , $2 x + 3 y - 6 z = 7$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ 3 & 1 & - 1 & - 6 \\ 2 & 3 & - 6 & 7 \end{array}]$

Step 2

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 $0$ .

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} \\ 2 & 3 & - 6 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (- 2) + 1 & (- 3) \cdot (4) - 1 & (- 3) \cdot (- 14) - 6 \\ 2 & 3 & - 6 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ 0 & 7 & - 13 & 36 \\ 2 & 3 & - 6 & 7 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ 0 & 7 & - 13 & 36 \\ - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ 0 & 7 & - 13 & 36 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (- 2) + 3 & (- 2) \cdot (4) - 6 & (- 2) \cdot (- 14) + 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ 0 & 7 & - 13 & 36 \\ 0 & 7 & - 14 & 35 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{7} R_{2}$ 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 $R_{2} = \frac{1}{7} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ \frac{1}{7} R_{2} & \frac{1}{7} R_{2} & \frac{1}{7} R_{2} & \frac{1}{7} R_{2} \\ 0 & 7 & - 14 & 35 \end{array}]$

$R_{2} = \frac{1}{7} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{7} R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ (\frac{1}{7}) \cdot (0) & (\frac{1}{7}) \cdot (7) & (\frac{1}{7}) \cdot (- 13) & (\frac{1}{7}) \cdot (36) \\ 0 & 7 & - 14 & 35 \end{array}]$

$R_{2} = \frac{1}{7} R_{2}$

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

$[\begin{array}{c} 1 & - 2 & 4 & - 14 \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 7 & - 14 & 35 \end{array}]$

Perform the row operation $R_{1} = 2 \cdot R_{2} + R_{1}$ 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 $R_{1} = 2 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 2 \cdot R_{2} + R_{1} & 2 \cdot R_{2} + R_{1} & 2 \cdot R_{2} + R_{1} & 2 \cdot R_{2} + R_{1} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 7 & - 14 & 35 \end{array}]$

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

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

$[\begin{array}{c} (2) \cdot (0) + 1 & (2) \cdot (1) - 2 & (2) \cdot (- \frac{13}{7}) + 4 & (2) \cdot (\frac{36}{7}) - 14 \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 7 & - 14 & 35 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 7 & - 14 & 35 \end{array}]$

Perform the row operation $R_{3} = - 7 \cdot R_{2} + R_{3}$ 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 $R_{3} = - 7 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ - 7 \cdot R_{2} + R_{3} & - 7 \cdot R_{2} + R_{3} & - 7 \cdot R_{2} + R_{3} & - 7 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 7 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ (- 7) \cdot (0) + 0 & (- 7) \cdot (1) + 7 & (- 7) \cdot (- \frac{13}{7}) - 14 & (- 7) \cdot (\frac{36}{7}) + 35 \end{array}]$

$R_{3} = - 7 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 0 & - 1 & - 1 \end{array}]$

Perform the row operation $R_{3} = - 1 \cdot R_{3}$ 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 $R_{3} = - 1 \cdot R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ - 1 \cdot R_{3} & - 1 \cdot R_{3} & - 1 \cdot R_{3} & - 1 \cdot R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ (- 1) \cdot (0) & (- 1) \cdot (0) & (- 1) \cdot (- 1) & (- 1) \cdot (- 1) \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 0 & 1 & 1 \end{array}]$

Perform the row operation $R_{1} = - \frac{2}{7} R_{3} + R_{1}$ 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 $R_{1} = - \frac{2}{7} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{2}{7} R_{3} + R_{1} & - \frac{2}{7} R_{3} + R_{1} & - \frac{2}{7} R_{3} + R_{1} & - \frac{2}{7} R_{3} + R_{1} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 0 & 1 & 1 \end{array}]$

$R_{1} = - \frac{2}{7} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{2}{7} R_{3} + R_{1}$ .

$[\begin{array}{c} (- \frac{2}{7}) \cdot (0) + 1 & (- \frac{2}{7}) \cdot (0) + 0 & (- \frac{2}{7}) \cdot (1) + \frac{2}{7} & (- \frac{2}{7}) \cdot (1) - \frac{26}{7} \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 0 & 1 & 1 \end{array}]$

$R_{1} = - \frac{2}{7} R_{3} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & 0 & - 4 \\ 0 & 1 & - \frac{13}{7} & \frac{36}{7} \\ 0 & 0 & 1 & 1 \end{array}]$

Perform the row operation $R_{2} = \frac{13}{7} R_{3} + R_{2}$ 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 $R_{2} = \frac{13}{7} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & - 4 \\ \frac{13}{7} R_{3} + R_{2} & \frac{13}{7} R_{3} + R_{2} & \frac{13}{7} R_{3} + R_{2} & \frac{13}{7} R_{3} + R_{2} \\ 0 & 0 & 1 & 1 \end{array}]$

$R_{2} = \frac{13}{7} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{13}{7} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 4 \\ (\frac{13}{7}) \cdot (0) + 0 & (\frac{13}{7}) \cdot (0) + 1 & (\frac{13}{7}) \cdot (1) - \frac{13}{7} & (\frac{13}{7}) \cdot (1) + \frac{36}{7} \\ 0 & 0 & 1 & 1 \end{array}]$

$R_{2} = \frac{13}{7} R_{3} + R_{2}$

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

$[\begin{array}{c} 1 & 0 & 0 & - 4 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3

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

$x = - 4$

$y = 7$

$z = 1$

Step 4

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

$(- 4, 7, 1)$

Solve Using a Matrix by Elimination x-2y+4z=-14 , 3x+y-z=-6 , 2x+3y-6z=7

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