Solve Using a Matrix by Elimination x+3y=z+15 , x=-3+y-z , x+y+7z=13

$x + 3 y = z + 15$ , $x = - 3 + y - z$ , $x + y + 7 z = 13$

Step 1

Subtract $z$ from both sides of the equation.

$x + 3 y - z = 15$

$x = - 3 + y - z$

$x + y + 7 z = 13$

Step 2

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

Subtract $y$ from both sides of the equation.

$x - y = - 3 - z$

$x + 3 y - z = 15$

$x + y + 7 z = 13$

Add $z$ to both sides of the equation.

$x - y + z = - 3$

$x + 3 y - z = 15$

$x + y + 7 z = 13$

$x - y + z = - 3$

$x + 3 y - z = 15$

$x + y + 7 z = 13$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ 1 & - 1 & 1 & - 3 \\ 1 & 1 & 7 & 13 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} \\ 1 & 1 & 7 & 13 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (3) - 1 & (- 1) \cdot (- 1) + 1 & (- 1) \cdot (15) - 3 \\ 1 & 1 & 7 & 13 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ 0 & - 4 & 2 & - 18 \\ 1 & 1 & 7 & 13 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ 0 & - 4 & 2 & - 18 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ 0 & - 4 & 2 & - 18 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (3) + 1 & (- 1) \cdot (- 1) + 7 & (- 1) \cdot (15) + 13 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ 0 & - 4 & 2 & - 18 \\ 0 & - 2 & 8 & - 2 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ - \frac{1}{4} R_{2} & - \frac{1}{4} R_{2} & - \frac{1}{4} R_{2} & - \frac{1}{4} R_{2} \\ 0 & - 2 & 8 & - 2 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ (- \frac{1}{4}) \cdot (0) & (- \frac{1}{4}) \cdot (- 4) & (- \frac{1}{4}) \cdot (2) & (- \frac{1}{4}) \cdot (- 18) \\ 0 & - 2 & 8 & - 2 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 1 & 15 \\ 0 & 1 & - \frac{1}{2} & \frac{9}{2} \\ 0 & - 2 & 8 & - 2 \end{array}]$

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

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

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

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

$[\begin{array}{c} (- 3) \cdot (0) + 1 & (- 3) \cdot (1) + 3 & (- 3) \cdot (- \frac{1}{2}) - 1 & (- 3) \cdot (\frac{9}{2}) + 15 \\ 0 & 1 & - \frac{1}{2} & \frac{9}{2} \\ 0 & - 2 & 8 & - 2 \end{array}]$

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

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

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

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} & \frac{9}{2} \\ 0 & 0 & 7 & 7 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} & \frac{9}{2} \\ \frac{1}{7} R_{3} & \frac{1}{7} R_{3} & \frac{1}{7} R_{3} & \frac{1}{7} R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} & \frac{9}{2} \\ (\frac{1}{7}) \cdot (0) & (\frac{1}{7}) \cdot (0) & (\frac{1}{7}) \cdot (7) & (\frac{1}{7}) \cdot (7) \end{array}]$

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

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

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

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

$[\begin{array}{c} - \frac{1}{2} R_{3} + R_{1} & - \frac{1}{2} R_{3} + R_{1} & - \frac{1}{2} R_{3} + R_{1} & - \frac{1}{2} R_{3} + R_{1} \\ 0 & 1 & - \frac{1}{2} & \frac{9}{2} \\ 0 & 0 & 1 & 1 \end{array}]$

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

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

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

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

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

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

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 $0$ .

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

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

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

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

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

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

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

Step 5

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

$x = 1$

$y = 5$

$z = 1$

Step 6

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

$(1, 5, 1)$

Solve Using a Matrix by Elimination x+3y=z+15 , x=-3+y-z , x+y+7z=13

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