Solve Using an Augmented Matrix x+y+z=10 , 2x-3y+2z=10 , 4x+y-3z=6

$x + y + z = 10$ , $2 x - 3 y + 2 z = 10$ , $4 x + y - 3 z = 6$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} 1 & 1 & 1 & 10 \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \\ 4 & 1 & - 3 & 6 \end{array}]$

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

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

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

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

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

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

Perform the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ 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 $R_{3} = - 4 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 10 \\ 0 & - 5 & 0 & - 10 \\ - 4 \cdot R_{1} + R_{3} & - 4 \cdot R_{1} + R_{3} & - 4 \cdot R_{1} + R_{3} & - 4 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 4 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ .

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

$R_{3} = - 4 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & 1 & 1 & 10 \\ 0 & - 5 & 0 & - 10 \\ 0 & - 3 & - 7 & - 34 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{5} R_{2}$ 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 $R_{2} = - \frac{1}{5} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 10 \\ - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} \\ 0 & - 3 & - 7 & - 34 \end{array}]$

$R_{2} = - \frac{1}{5} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{5} R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 10 \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 10) \\ 0 & - 3 & - 7 & - 34 \end{array}]$

$R_{2} = - \frac{1}{5} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 1 & 10 \\ 0 & 1 & 0 & 2 \\ 0 & - 3 & - 7 & - 34 \end{array}]$

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 $0$ .

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & 0 & 2 \\ 0 & - 3 & - 7 & - 34 \end{array}]$

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (0) + 1 & (- 1) \cdot (2) + 10 \\ 0 & 1 & 0 & 2 \\ 0 & - 3 & - 7 & - 34 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 1 & 8 \\ 0 & 1 & 0 & 2 \\ 0 & - 3 & - 7 & - 34 \end{array}]$

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

$[\begin{array}{c} 1 & 0 & 1 & 8 \\ 0 & 1 & 0 & 2 \\ 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 1 & 8 \\ 0 & 1 & 0 & 2 \\ (3) \cdot (0) + 0 & (3) \cdot (1) - 3 & (3) \cdot (0) - 7 & (3) \cdot (2) - 34 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 1 & 8 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & - 7 & - 28 \end{array}]$

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

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

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

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

$[\begin{array}{c} 1 & 0 & 1 & 8 \\ 0 & 1 & 0 & 2 \\ (- \frac{1}{7}) \cdot (0) & (- \frac{1}{7}) \cdot (0) & (- \frac{1}{7}) \cdot (- 7) & (- \frac{1}{7}) \cdot (- 28) \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 1 & 8 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 4 \end{array}]$

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 4 \end{array}]$

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

$x = 4$

$y = 2$

$z = 4$

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

$(4, 2, 4)$

Solve Using an Augmented Matrix x+y+z=10 , 2x-3y+2z=10 , 4x+y-3z=6

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