Solve Using a Matrix by Elimination x+9y=z+23 , x=3+y-z , x+y+z=9

$x + 9 y = z + 23$ , $x = 3 + y - z$ , $x + y + z = 9$

Subtract $z$ from both sides of the equation.

$x + 9 y - z = 23$

$x = 3 + y - z$

$x + y + z = 9$

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

Subtract $y$ from both sides of the equation.

$x - y = 3 - z$

$x + 9 y - z = 23$

$x + y + z = 9$

Add $z$ to both sides of the equation.

$x - y + z = 3$

$x + 9 y - z = 23$

$x + y + z = 9$

$x - y + z = 3$

$x + 9 y - z = 23$

$x + y + z = 9$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ 1 & - 1 & 1 & 3 \\ 1 & 1 & 1 & 9 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} \\ 1 & 1 & 1 & 9 \end{array}]$

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

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

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (9) - 1 & (- 1) \cdot (- 1) + 1 & (- 1) \cdot (23) + 3 \\ 1 & 1 & 1 & 9 \end{array}]$

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

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

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ 0 & - 10 & 2 & - 20 \\ 1 & 1 & 1 & 9 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ 0 & - 10 & 2 & - 20 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ 0 & - 10 & 2 & - 20 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (9) + 1 & (- 1) \cdot (- 1) + 1 & (- 1) \cdot (23) + 9 \end{array}]$

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

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

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ 0 & - 10 & 2 & - 20 \\ 0 & - 8 & 2 & - 14 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{10} R_{2}$ 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 $R_{2} = - \frac{1}{10} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ - \frac{1}{10} R_{2} & - \frac{1}{10} R_{2} & - \frac{1}{10} R_{2} & - \frac{1}{10} R_{2} \\ 0 & - 8 & 2 & - 14 \end{array}]$

$R_{2} = - \frac{1}{10} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{10} R_{2}$ .

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ (- \frac{1}{10}) \cdot (0) & (- \frac{1}{10}) \cdot (- 10) & (- \frac{1}{10}) \cdot (2) & (- \frac{1}{10}) \cdot (- 20) \\ 0 & - 8 & 2 & - 14 \end{array}]$

$R_{2} = - \frac{1}{10} R_{2}$

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

$[\begin{array}{c} 1 & 9 & - 1 & 23 \\ 0 & 1 & - \frac{1}{5} & 2 \\ 0 & - 8 & 2 & - 14 \end{array}]$

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 $0$ .

$[\begin{array}{c} - 9 \cdot R_{2} + R_{1} & - 9 \cdot R_{2} + R_{1} & - 9 \cdot R_{2} + R_{1} & - 9 \cdot R_{2} + R_{1} \\ 0 & 1 & - \frac{1}{5} & 2 \\ 0 & - 8 & 2 & - 14 \end{array}]$

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

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

$[\begin{array}{c} (- 9) \cdot (0) + 1 & (- 9) \cdot (1) + 9 & (- 9) \cdot (- \frac{1}{5}) - 1 & (- 9) \cdot (2) + 23 \\ 0 & 1 & - \frac{1}{5} & 2 \\ 0 & - 8 & 2 & - 14 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{4}{5} & 5 \\ 0 & 1 & - \frac{1}{5} & 2 \\ 0 & - 8 & 2 & - 14 \end{array}]$

Perform the row operation $R_{3} = 8 \cdot R_{2} + R_{3}$ 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 $R_{3} = 8 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{4}{5} & 5 \\ 0 & 1 & - \frac{1}{5} & 2 \\ 8 \cdot R_{2} + R_{3} & 8 \cdot R_{2} + R_{3} & 8 \cdot R_{2} + R_{3} & 8 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = 8 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = 8 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{4}{5} & 5 \\ 0 & 1 & - \frac{1}{5} & 2 \\ (8) \cdot (0) + 0 & (8) \cdot (1) - 8 & (8) \cdot (- \frac{1}{5}) + 2 & (8) \cdot (2) - 14 \end{array}]$

$R_{3} = 8 \cdot R_{2} + R_{3}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Perform the row operation $R_{2} = \frac{1}{5} R_{3} + R_{2}$ 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 $R_{2} = \frac{1}{5} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = \frac{1}{5} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{5} R_{3} + R_{2}$ .

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

$R_{2} = \frac{1}{5} R_{3} + R_{2}$

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

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

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

$x = 1$

$y = 3$

$z = 5$

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

$(1, 3, 5)$

Solve Using a Matrix by Elimination x+9y=z+23 , x=3+y-z , x+y+z=9

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