Solve Using a Matrix by Elimination x-5y=z-7 , x=5+y-z , x+y+5z=23

$x - 5 y = z - 7$ , $x = 5 + y - z$ , $x + y + 5 z = 23$

Step 1

Subtract $z$ from both sides of the equation.

$x - 5 y - z = - 7$

$x = 5 + y - z$

$x + y + 5 z = 23$

Step 2

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

Subtract $y$ from both sides of the equation.

$x - y = 5 - z$

$x - 5 y - z = - 7$

$x + y + 5 z = 23$

Add $z$ to both sides of the equation.

$x - y + z = 5$

$x - 5 y - z = - 7$

$x + y + 5 z = 23$

$x - y + z = 5$

$x - 5 y - z = - 7$

$x + y + 5 z = 23$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ 1 & - 1 & 1 & 5 \\ 1 & 1 & 5 & 23 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} \\ 1 & 1 & 5 & 23 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (- 5) - 1 & (- 1) \cdot (- 1) + 1 & (- 1) \cdot (- 7) + 5 \\ 1 & 1 & 5 & 23 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ 0 & 4 & 2 & 12 \\ 1 & 1 & 5 & 23 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ 0 & 4 & 2 & 12 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ 0 & 4 & 2 & 12 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (- 5) + 1 & (- 1) \cdot (- 1) + 5 & (- 1) \cdot (- 7) + 23 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ 0 & 4 & 2 & 12 \\ 0 & 6 & 6 & 30 \end{array}]$

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ \frac{1}{4} R_{2} & \frac{1}{4} R_{2} & \frac{1}{4} R_{2} & \frac{1}{4} R_{2} \\ 0 & 6 & 6 & 30 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ (\frac{1}{4}) \cdot (0) & (\frac{1}{4}) \cdot (4) & (\frac{1}{4}) \cdot (2) & (\frac{1}{4}) \cdot (12) \\ 0 & 6 & 6 & 30 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 5 & - 1 & - 7 \\ 0 & 1 & \frac{1}{2} & 3 \\ 0 & 6 & 6 & 30 \end{array}]$

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

$[\begin{array}{c} 5 \cdot R_{2} + R_{1} & 5 \cdot R_{2} + R_{1} & 5 \cdot R_{2} + R_{1} & 5 \cdot R_{2} + R_{1} \\ 0 & 1 & \frac{1}{2} & 3 \\ 0 & 6 & 6 & 30 \end{array}]$

$R_{1} = 5 \cdot R_{2} + R_{1}$

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

$[\begin{array}{c} (5) \cdot (0) + 1 & (5) \cdot (1) - 5 & (5) \cdot (\frac{1}{2}) - 1 & (5) \cdot (3) - 7 \\ 0 & 1 & \frac{1}{2} & 3 \\ 0 & 6 & 6 & 30 \end{array}]$

$R_{1} = 5 \cdot R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 8 \\ 0 & 1 & \frac{1}{2} & 3 \\ 0 & 6 & 6 & 30 \end{array}]$

Perform the row operation $R_{3} = - 6 \cdot R_{2} + R_{3}$ 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 $R_{3} = - 6 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 8 \\ 0 & 1 & \frac{1}{2} & 3 \\ - 6 \cdot R_{2} + R_{3} & - 6 \cdot R_{2} + R_{3} & - 6 \cdot R_{2} + R_{3} & - 6 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 6 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 6 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 8 \\ 0 & 1 & \frac{1}{2} & 3 \\ (- 6) \cdot (0) + 0 & (- 6) \cdot (1) + 6 & (- 6) \cdot (\frac{1}{2}) + 6 & (- 6) \cdot (3) + 30 \end{array}]$

$R_{3} = - 6 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 8 \\ 0 & 1 & \frac{1}{2} & 3 \\ 0 & 0 & 3 & 12 \end{array}]$

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

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

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

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 8 \\ 0 & 1 & \frac{1}{2} & 3 \\ (\frac{1}{3}) \cdot (0) & (\frac{1}{3}) \cdot (0) & (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (12) \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & 8 \\ 0 & 1 & \frac{1}{2} & 3 \\ 0 & 0 & 1 & 4 \end{array}]$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Step 5

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

$x = 2$

$y = 1$

$z = 4$

Step 6

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

$(2, 1, 4)$

Solve Using a Matrix by Elimination x-5y=z-7 , x=5+y-z , x+y+5z=23

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