Solve Using a Matrix by Elimination x+y+z=9 2x-3y+4z=7

$x + y + z = 9$ $2 x - 3 y + 4 z = 7$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 2 & - 3 & 4 & 7 \end{array}]$

Find the reduced row echelon form of the matrix.

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

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

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

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (1) - 3 & (- 2) \cdot (1) + 4 & (- 2) \cdot (9) + 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & - 5 & 2 & - 11 \end{array}]$

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} \end{array}]$

$R_{2} = - \frac{1}{5} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (2) & (- \frac{1}{5}) \cdot (- 11) \end{array}]$

$R_{2} = - \frac{1}{5} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \end{array}]$

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

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

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (- \frac{2}{5}) + 1 & (- 1) \cdot (\frac{11}{5}) + 9 \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{7}{5} & \frac{34}{5} \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \end{array}]$

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

$x + \frac{7 z}{5} = \frac{34}{5}$

$y - \frac{2 z}{5} = \frac{11}{5}$

Subtract $\frac{7 z}{5}$ from both sides of the equation.

$x = \frac{34}{5} - \frac{7 z}{5}$

$y - \frac{2 z}{5} = \frac{11}{5}$

Add $\frac{2 z}{5}$ to both sides of the equation.

$y = \frac{11}{5} + \frac{2 z}{5}$

$x = \frac{34}{5} - \frac{7 z}{5}$

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

$(\frac{34}{5} - \frac{7 z}{5}, \frac{11}{5} + \frac{2 z}{5}, z)$

Solve Using a Matrix by Elimination x+y+z=9 2x-3y+4z=7

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