Solve Using an Augmented Matrix 2x+y+z=4 , 6y+5z=-33x , 5y-z=27

$2 x + y + z = 4$ , $6 y + 5 z = - 33 x$ , $5 y - z = 27$

Add $33 x$ to both sides of the equation.

$6 y + 5 z + 33 x = 0$

$2 x + y + z = 4$

$5 y - z = 27$

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & 1 & 1 & 4 \\ 33 & 6 & 5 & 0 \\ 0 & 5 & - 1 & 27 \end{array}]$

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{1} = \frac{1}{2} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{2} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ 33 & 6 & 5 & 0 \\ 0 & 5 & - 1 & 27 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (1) & (\frac{1}{2}) \cdot (1) & (\frac{1}{2}) \cdot (4) \\ 33 & 6 & 5 & 0 \\ 0 & 5 & - 1 & 27 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ 33 & 6 & 5 & 0 \\ 0 & 5 & - 1 & 27 \end{array}]$

Perform the row operation $R_{2} = - 33 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 33 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ - 33 \cdot R_{1} + R_{2} & - 33 \cdot R_{1} + R_{2} & - 33 \cdot R_{1} + R_{2} & - 33 \cdot R_{1} + R_{2} \\ 0 & 5 & - 1 & 27 \end{array}]$

$R_{2} = - 33 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 33 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ (- 33) \cdot (1) + 33 & (- 33) \cdot (\frac{1}{2}) + 6 & (- 33) \cdot (\frac{1}{2}) + 5 & (- 33) \cdot (2) + 0 \\ 0 & 5 & - 1 & 27 \end{array}]$

$R_{2} = - 33 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ 0 & - \frac{21}{2} & - \frac{23}{2} & - 66 \\ 0 & 5 & - 1 & 27 \end{array}]$

Perform the row operation $R_{2} = - \frac{2}{21} R_{2}$ 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 $R_{2} = - \frac{2}{21} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ - \frac{2}{21} R_{2} & - \frac{2}{21} R_{2} & - \frac{2}{21} R_{2} & - \frac{2}{21} R_{2} \\ 0 & 5 & - 1 & 27 \end{array}]$

$R_{2} = - \frac{2}{21} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{2}{21} R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ (- \frac{2}{21}) \cdot (0) & (- \frac{2}{21}) \cdot (- \frac{21}{2}) & (- \frac{2}{21}) \cdot (- \frac{23}{2}) & (- \frac{2}{21}) \cdot (- 66) \\ 0 & 5 & - 1 & 27 \end{array}]$

$R_{2} = - \frac{2}{21} R_{2}$

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

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 2 \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 5 & - 1 & 27 \end{array}]$

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

$[\begin{array}{c} - \frac{1}{2} R_{2} + R_{1} & - \frac{1}{2} R_{2} + R_{1} & - \frac{1}{2} R_{2} + R_{1} & - \frac{1}{2} R_{2} + R_{1} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 5 & - 1 & 27 \end{array}]$

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

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

$[\begin{array}{c} (- \frac{1}{2}) \cdot (0) + 1 & (- \frac{1}{2}) \cdot (1) + \frac{1}{2} & (- \frac{1}{2}) \cdot (\frac{23}{21}) + \frac{1}{2} & (- \frac{1}{2}) \cdot (\frac{44}{7}) + 2 \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 5 & - 1 & 27 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 5 & - 1 & 27 \end{array}]$

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 5 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ (- 5) \cdot (0) + 0 & (- 5) \cdot (1) + 5 & (- 5) \cdot (\frac{23}{21}) - 1 & (- 5) \cdot (\frac{44}{7}) + 27 \end{array}]$

$R_{3} = - 5 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 0 & - \frac{136}{21} & - \frac{31}{7} \end{array}]$

Perform the row operation $R_{3} = - \frac{21}{136} R_{3}$ 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 $R_{3} = - \frac{21}{136} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ - \frac{21}{136} R_{3} & - \frac{21}{136} R_{3} & - \frac{21}{136} R_{3} & - \frac{21}{136} R_{3} \end{array}]$

$R_{3} = - \frac{21}{136} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{21}{136} R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ (- \frac{21}{136}) \cdot (0) & (- \frac{21}{136}) \cdot (0) & (- \frac{21}{136}) \cdot (- \frac{136}{21}) & (- \frac{21}{136}) \cdot (- \frac{31}{7}) \end{array}]$

$R_{3} = - \frac{21}{136} R_{3}$

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{21} & - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

Perform the row operation $R_{1} = \frac{1}{21} R_{3} + R_{1}$ 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 $R_{1} = \frac{1}{21} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{1}{21} R_{3} + R_{1} & \frac{1}{21} R_{3} + R_{1} & \frac{1}{21} R_{3} + R_{1} & \frac{1}{21} R_{3} + R_{1} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

$R_{1} = \frac{1}{21} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{21} R_{3} + R_{1}$ .

$[\begin{array}{c} (\frac{1}{21}) \cdot (0) + 1 & (\frac{1}{21}) \cdot (0) + 0 & (\frac{1}{21}) \cdot (1) - \frac{1}{21} & (\frac{1}{21}) \cdot (\frac{93}{136}) - \frac{8}{7} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

$R_{1} = \frac{1}{21} R_{3} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{151}{136} \\ 0 & 1 & \frac{23}{21} & \frac{44}{7} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

Perform the row operation $R_{2} = - \frac{23}{21} R_{3} + R_{2}$ 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 $R_{2} = - \frac{23}{21} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{151}{136} \\ - \frac{23}{21} R_{3} + R_{2} & - \frac{23}{21} R_{3} + R_{2} & - \frac{23}{21} R_{3} + R_{2} & - \frac{23}{21} R_{3} + R_{2} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

$R_{2} = - \frac{23}{21} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{23}{21} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{151}{136} \\ (- \frac{23}{21}) \cdot (0) + 0 & (- \frac{23}{21}) \cdot (0) + 1 & (- \frac{23}{21}) \cdot (1) + \frac{23}{21} & (- \frac{23}{21}) \cdot (\frac{93}{136}) + \frac{44}{7} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

$R_{2} = - \frac{23}{21} R_{3} + R_{2}$

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{151}{136} \\ 0 & 1 & 0 & \frac{753}{136} \\ 0 & 0 & 1 & \frac{93}{136} \end{array}]$

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

$x = - \frac{151}{136}$

$y = \frac{753}{136}$

$z = \frac{93}{136}$

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

$(- \frac{151}{136}, \frac{753}{136}, \frac{93}{136})$

Solve Using an Augmented Matrix 2x+y+z=4 , 6y+5z=-33x , 5y-z=27

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