Solve Using a Matrix by Elimination x-9y=z-17 , x=7+y-z , x+y-z=3

$x - 9 y = z - 17$ , $x = 7 + y - z$ , $x + y - z = 3$

Step 1

Subtract $z$ from both sides of the equation.

$x - 9 y - z = - 17$

$x = 7 + y - z$

$x + y - z = 3$

Step 2

Move all terms containing variables to the left side of the equation.

Subtract $y$ from both sides of the equation.

$x - y = 7 - z$

$x - 9 y - z = - 17$

$x + y - z = 3$

Add $z$ to both sides of the equation.

$x - y + z = 7$

$x - 9 y - z = - 17$

$x + y - z = 3$

$x - y + z = 7$

$x - 9 y - z = - 17$

$x + y - z = 3$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ 1 & - 1 & 1 & 7 \\ 1 & 1 & - 1 & 3 \end{array}]$

Step 4

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 $0$ .

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

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

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

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

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

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

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ 0 & 8 & 2 & 24 \\ 1 & 1 & - 1 & 3 \end{array}]$

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

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ 0 & 8 & 2 & 24 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ 0 & 8 & 2 & 24 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (- 9) + 1 & (- 1) \cdot (- 1) - 1 & (- 1) \cdot (- 17) + 3 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ 0 & 8 & 2 & 24 \\ 0 & 10 & 0 & 20 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{8} R_{2}$ 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 $R_{2} = \frac{1}{8} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ \frac{1}{8} R_{2} & \frac{1}{8} R_{2} & \frac{1}{8} R_{2} & \frac{1}{8} R_{2} \\ 0 & 10 & 0 & 20 \end{array}]$

$R_{2} = \frac{1}{8} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{8} R_{2}$ .

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ (\frac{1}{8}) \cdot (0) & (\frac{1}{8}) \cdot (8) & (\frac{1}{8}) \cdot (2) & (\frac{1}{8}) \cdot (24) \\ 0 & 10 & 0 & 20 \end{array}]$

$R_{2} = \frac{1}{8} R_{2}$

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

$[\begin{array}{c} 1 & - 9 & - 1 & - 17 \\ 0 & 1 & \frac{1}{4} & 3 \\ 0 & 10 & 0 & 20 \end{array}]$

Perform the row operation $R_{1} = 9 \cdot R_{2} + R_{1}$ 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 $R_{1} = 9 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 9 \cdot R_{2} + R_{1} & 9 \cdot R_{2} + R_{1} & 9 \cdot R_{2} + R_{1} & 9 \cdot R_{2} + R_{1} \\ 0 & 1 & \frac{1}{4} & 3 \\ 0 & 10 & 0 & 20 \end{array}]$

$R_{1} = 9 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = 9 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (9) \cdot (0) + 1 & (9) \cdot (1) - 9 & (9) \cdot (\frac{1}{4}) - 1 & (9) \cdot (3) - 17 \\ 0 & 1 & \frac{1}{4} & 3 \\ 0 & 10 & 0 & 20 \end{array}]$

$R_{1} = 9 \cdot R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & \frac{5}{4} & 10 \\ 0 & 1 & \frac{1}{4} & 3 \\ 0 & 10 & 0 & 20 \end{array}]$

Perform the row operation $R_{3} = - 10 \cdot R_{2} + R_{3}$ 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 $R_{3} = - 10 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{5}{4} & 10 \\ 0 & 1 & \frac{1}{4} & 3 \\ - 10 \cdot R_{2} + R_{3} & - 10 \cdot R_{2} + R_{3} & - 10 \cdot R_{2} + R_{3} & - 10 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 10 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 10 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{5}{4} & 10 \\ 0 & 1 & \frac{1}{4} & 3 \\ (- 10) \cdot (0) + 0 & (- 10) \cdot (1) + 10 & (- 10) \cdot (\frac{1}{4}) + 0 & (- 10) \cdot (3) + 20 \end{array}]$

$R_{3} = - 10 \cdot R_{2} + R_{3}$

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

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

Perform the row operation $R_{3} = - \frac{2}{5} R_{3}$ 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 $R_{3} = - \frac{2}{5} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & \frac{5}{4} & 10 \\ 0 & 1 & \frac{1}{4} & 3 \\ - \frac{2}{5} R_{3} & - \frac{2}{5} R_{3} & - \frac{2}{5} R_{3} & - \frac{2}{5} R_{3} \end{array}]$

$R_{3} = - \frac{2}{5} R_{3}$

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

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

$R_{3} = - \frac{2}{5} R_{3}$

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

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

Perform the row operation $R_{1} = - \frac{5}{4} R_{3} + R_{1}$ 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 $R_{1} = - \frac{5}{4} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{1} = - \frac{5}{4} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{5}{4} R_{3} + R_{1}$ .

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

$R_{1} = - \frac{5}{4} R_{3} + R_{1}$

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

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

Perform the row operation $R_{2} = - \frac{1}{4} R_{3} + R_{2}$ 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 $R_{2} = - \frac{1}{4} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = - \frac{1}{4} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{4} R_{3} + R_{2}$ .

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

$R_{2} = - \frac{1}{4} R_{3} + R_{2}$

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

$[\begin{array}{c} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 4 \end{array}]$

Step 5

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

$x = 5$

$y = 2$

$z = 4$

Step 6

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

$(5, 2, 4)$

Solve Using a Matrix by Elimination x-9y=z-17 , x=7+y-z , x+y-z=3

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