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

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

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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

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

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

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

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

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

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

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

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 $0$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & - \frac{1}{2} & 0 \\ 0 & - 2 & 1 & 0 \end{array}]$

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

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

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

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 7 \\ 0 & 1 & - \frac{1}{2} & 0 \\ (2) \cdot (0) + 0 & (2) \cdot (1) - 2 & (2) \cdot (- \frac{1}{2}) + 1 & (2) \cdot (0) + 0 \end{array}]$

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

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

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

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

$x + \frac{3 z}{2} = 7$

$y - \frac{z}{2} = 0$

Subtract $\frac{3 z}{2}$ from both sides of the equation.

$x = 7 - \frac{3 z}{2}$

$y - \frac{z}{2} = 0$

Add $\frac{z}{2}$ to both sides of the equation.

$y = \frac{z}{2}$

$x = 7 - \frac{3 z}{2}$

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

$(7 - \frac{3 z}{2}, \frac{z}{2}, z)$

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

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