Solve Using an Augmented Matrix 5x-4y-6z=-3 , x-3y+z=-1 , -3x-6y+7z=1

$5 x - 4 y - 6 z = - 3$ , $x - 3 y + z = - 1$ , $- 3 x - 6 y + 7 z = 1$

Write the system of equations in matrix form.

$[\begin{array}{c} 5 & - 4 & - 6 & - 3 \\ 1 & - 3 & 1 & - 1 \\ - 3 & - 6 & 7 & 1 \end{array}]$

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{1} = \frac{1}{5} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{5} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} \\ 1 & - 3 & 1 & - 1 \\ - 3 & - 6 & 7 & 1 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{5}) \cdot (5) & (\frac{1}{5}) \cdot (- 4) & (\frac{1}{5}) \cdot (- 6) & (\frac{1}{5}) \cdot (- 3) \\ 1 & - 3 & 1 & - 1 \\ - 3 & - 6 & 7 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ 1 & - 3 & 1 & - 1 \\ - 3 & - 6 & 7 & 1 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} \\ - 3 & - 6 & 7 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (- \frac{4}{5}) - 3 & (- 1) \cdot (- \frac{6}{5}) + 1 & (- 1) \cdot (- \frac{3}{5}) - 1 \\ - 3 & - 6 & 7 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ 0 & - \frac{11}{5} & \frac{11}{5} & - \frac{2}{5} \\ - 3 & - 6 & 7 & 1 \end{array}]$

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ 0 & - \frac{11}{5} & \frac{11}{5} & - \frac{2}{5} \\ 3 \cdot R_{1} + R_{3} & 3 \cdot R_{1} + R_{3} & 3 \cdot R_{1} + R_{3} & 3 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ 0 & - \frac{11}{5} & \frac{11}{5} & - \frac{2}{5} \\ (3) \cdot (1) - 3 & (3) \cdot (- \frac{4}{5}) - 6 & (3) \cdot (- \frac{6}{5}) + 7 & (3) \cdot (- \frac{3}{5}) + 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ 0 & - \frac{11}{5} & \frac{11}{5} & - \frac{2}{5} \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

Perform the row operation $R_{2} = - \frac{5}{11} R_{2}$ 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 $R_{2} = - \frac{5}{11} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ - \frac{5}{11} R_{2} & - \frac{5}{11} R_{2} & - \frac{5}{11} R_{2} & - \frac{5}{11} R_{2} \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

$R_{2} = - \frac{5}{11} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{5}{11} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ (- \frac{5}{11}) \cdot (0) & (- \frac{5}{11}) \cdot (- \frac{11}{5}) & (- \frac{5}{11}) \cdot (\frac{11}{5}) & (- \frac{5}{11}) \cdot (- \frac{2}{5}) \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

$R_{2} = - \frac{5}{11} R_{2}$

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

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{6}{5} & - \frac{3}{5} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

Perform the row operation $R_{1} = \frac{4}{5} R_{2} + R_{1}$ 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 $R_{1} = \frac{4}{5} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{4}{5} R_{2} + R_{1} & \frac{4}{5} R_{2} + R_{1} & \frac{4}{5} R_{2} + R_{1} & \frac{4}{5} R_{2} + R_{1} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

$R_{1} = \frac{4}{5} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{4}{5} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{4}{5}) \cdot (0) + 1 & (\frac{4}{5}) \cdot (1) - \frac{4}{5} & (\frac{4}{5}) \cdot (- 1) - \frac{6}{5} & (\frac{4}{5}) \cdot (\frac{2}{11}) - \frac{3}{5} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

$R_{1} = \frac{4}{5} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & - \frac{42}{5} & \frac{17}{5} & - \frac{4}{5} \end{array}]$

Perform the row operation $R_{3} = \frac{42}{5} R_{2} + R_{3}$ 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 $R_{3} = \frac{42}{5} R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ \frac{42}{5} R_{2} + R_{3} & \frac{42}{5} R_{2} + R_{3} & \frac{42}{5} R_{2} + R_{3} & \frac{42}{5} R_{2} + R_{3} \end{array}]$

$R_{3} = \frac{42}{5} R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = \frac{42}{5} R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ (\frac{42}{5}) \cdot (0) + 0 & (\frac{42}{5}) \cdot (1) - \frac{42}{5} & (\frac{42}{5}) \cdot (- 1) + \frac{17}{5} & (\frac{42}{5}) \cdot (\frac{2}{11}) - \frac{4}{5} \end{array}]$

$R_{3} = \frac{42}{5} R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & 0 & - 5 & \frac{8}{11} \end{array}]$

Perform the row operation $R_{3} = - \frac{1}{5} R_{3}$ 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 $R_{3} = - \frac{1}{5} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ - \frac{1}{5} R_{3} & - \frac{1}{5} R_{3} & - \frac{1}{5} R_{3} & - \frac{1}{5} R_{3} \end{array}]$

$R_{3} = - \frac{1}{5} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{1}{5} R_{3}$ .

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (\frac{8}{11}) \end{array}]$

$R_{3} = - \frac{1}{5} R_{3}$

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

$[\begin{array}{c} 1 & 0 & - 2 & - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

$[\begin{array}{c} 2 \cdot R_{3} + R_{1} & 2 \cdot R_{3} + R_{1} & 2 \cdot R_{3} + R_{1} & 2 \cdot R_{3} + R_{1} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

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

$[\begin{array}{c} (2) \cdot (0) + 1 & (2) \cdot (0) + 0 & (2) \cdot (1) - 2 & (2) \cdot (- \frac{8}{55}) - \frac{5}{11} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{41}{55} \\ 0 & 1 & - 1 & \frac{2}{11} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{41}{55} \\ R_{3} + R_{2} & R_{3} + R_{2} & R_{3} + R_{2} & R_{3} + R_{2} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{41}{55} \\ 0 + 0 & 0 + 1 & 1 - 1 & - \frac{8}{55} + \frac{2}{11} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{41}{55} \\ 0 & 1 & 0 & \frac{2}{55} \\ 0 & 0 & 1 & - \frac{8}{55} \end{array}]$

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

$x = - \frac{41}{55}$

$y = \frac{2}{55}$

$z = - \frac{8}{55}$

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

$(- \frac{41}{55}, \frac{2}{55}, - \frac{8}{55})$

Solve Using an Augmented Matrix 5x-4y-6z=-3 , x-3y+z=-1 , -3x-6y+7z=1

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