Find Reduced Row Echelon Form [[7,4,-1,32],[4,1,-3,-10],[1,2,1,22]]

$[\begin{array}{c} 7 & 4 & - 1 & 32 \\ 4 & 1 & - 3 & - 10 \\ 1 & 2 & 1 & 22 \end{array}]$

Perform the row operation $R_{1} = \frac{1}{7} R_{1}$ 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 $R_{1} = \frac{1}{7} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} \\ 4 & 1 & - 3 & - 10 \\ 1 & 2 & 1 & 22 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{7}) \cdot (7) & (\frac{1}{7}) \cdot (4) & (\frac{1}{7}) \cdot (- 1) & (\frac{1}{7}) \cdot (32) \\ 4 & 1 & - 3 & - 10 \\ 1 & 2 & 1 & 22 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ 4 & 1 & - 3 & - 10 \\ 1 & 2 & 1 & 22 \end{array}]$

Perform the row operation $R_{2} = - 4 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 4 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ - 4 \cdot R_{1} + R_{2} & - 4 \cdot R_{1} + R_{2} & - 4 \cdot R_{1} + R_{2} & - 4 \cdot R_{1} + R_{2} \\ 1 & 2 & 1 & 22 \end{array}]$

$R_{2} = - 4 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ (- 4) \cdot (1) + 4 & (- 4) \cdot (\frac{4}{7}) + 1 & (- 4) \cdot (- \frac{1}{7}) - 3 & (- 4) \cdot (\frac{32}{7}) - 10 \\ 1 & 2 & 1 & 22 \end{array}]$

$R_{2} = - 4 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ 0 & - \frac{9}{7} & - \frac{17}{7} & - \frac{198}{7} \\ 1 & 2 & 1 & 22 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ 0 & - \frac{9}{7} & - \frac{17}{7} & - \frac{198}{7} \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ 0 & - \frac{9}{7} & - \frac{17}{7} & - \frac{198}{7} \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (\frac{4}{7}) + 2 & (- 1) \cdot (- \frac{1}{7}) + 1 & (- 1) \cdot (\frac{32}{7}) + 22 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ 0 & - \frac{9}{7} & - \frac{17}{7} & - \frac{198}{7} \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

Perform the row operation $R_{2} = - \frac{7}{9} R_{2}$ 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 $R_{2} = - \frac{7}{9} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ - \frac{7}{9} R_{2} & - \frac{7}{9} R_{2} & - \frac{7}{9} R_{2} & - \frac{7}{9} R_{2} \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

$R_{2} = - \frac{7}{9} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{7}{9} R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ (- \frac{7}{9}) \cdot (0) & (- \frac{7}{9}) \cdot (- \frac{9}{7}) & (- \frac{7}{9}) \cdot (- \frac{17}{7}) & (- \frac{7}{9}) \cdot (- \frac{198}{7}) \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

$R_{2} = - \frac{7}{9} R_{2}$

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & \frac{32}{7} \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

Perform the row operation $R_{1} = - \frac{4}{7} R_{2} + R_{1}$ 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 $R_{1} = - \frac{4}{7} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{4}{7} R_{2} + R_{1} & - \frac{4}{7} R_{2} + R_{1} & - \frac{4}{7} R_{2} + R_{1} & - \frac{4}{7} R_{2} + R_{1} \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

$R_{1} = - \frac{4}{7} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{4}{7} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{4}{7}) \cdot (0) + 1 & (- \frac{4}{7}) \cdot (1) + \frac{4}{7} & (- \frac{4}{7}) \cdot (\frac{17}{9}) - \frac{1}{7} & (- \frac{4}{7}) \cdot (22) + \frac{32}{7} \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

$R_{1} = - \frac{4}{7} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & \frac{10}{7} & \frac{8}{7} & \frac{122}{7} \end{array}]$

Perform the row operation $R_{3} = - \frac{10}{7} R_{2} + R_{3}$ 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 $R_{3} = - \frac{10}{7} R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ - \frac{10}{7} R_{2} + R_{3} & - \frac{10}{7} R_{2} + R_{3} & - \frac{10}{7} R_{2} + R_{3} & - \frac{10}{7} R_{2} + R_{3} \end{array}]$

$R_{3} = - \frac{10}{7} R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{10}{7} R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ (- \frac{10}{7}) \cdot (0) + 0 & (- \frac{10}{7}) \cdot (1) + \frac{10}{7} & (- \frac{10}{7}) \cdot (\frac{17}{9}) + \frac{8}{7} & (- \frac{10}{7}) \cdot (22) + \frac{122}{7} \end{array}]$

$R_{3} = - \frac{10}{7} R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & 0 & - \frac{14}{9} & - 14 \end{array}]$

Perform the row operation $R_{3} = - \frac{9}{14} R_{3}$ 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 $R_{3} = - \frac{9}{14} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ - \frac{9}{14} R_{3} & - \frac{9}{14} R_{3} & - \frac{9}{14} R_{3} & - \frac{9}{14} R_{3} \end{array}]$

$R_{3} = - \frac{9}{14} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{9}{14} R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ (- \frac{9}{14}) \cdot (0) & (- \frac{9}{14}) \cdot (0) & (- \frac{9}{14}) \cdot (- \frac{14}{9}) & (- \frac{9}{14}) \cdot (- 14) \end{array}]$

$R_{3} = - \frac{9}{14} R_{3}$

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

$[\begin{array}{c} 1 & 0 & - \frac{11}{9} & - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & 0 & 1 & 9 \end{array}]$

Perform the row operation $R_{1} = \frac{11}{9} R_{3} + R_{1}$ 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 $R_{1} = \frac{11}{9} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{11}{9} R_{3} + R_{1} & \frac{11}{9} R_{3} + R_{1} & \frac{11}{9} R_{3} + R_{1} & \frac{11}{9} R_{3} + R_{1} \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & 0 & 1 & 9 \end{array}]$

$R_{1} = \frac{11}{9} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{11}{9} R_{3} + R_{1}$ .

$[\begin{array}{c} (\frac{11}{9}) \cdot (0) + 1 & (\frac{11}{9}) \cdot (0) + 0 & (\frac{11}{9}) \cdot (1) - \frac{11}{9} & (\frac{11}{9}) \cdot (9) - 8 \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & 0 & 1 & 9 \end{array}]$

$R_{1} = \frac{11}{9} R_{3} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & 0 & 3 \\ 0 & 1 & \frac{17}{9} & 22 \\ 0 & 0 & 1 & 9 \end{array}]$

Perform the row operation $R_{2} = - \frac{17}{9} R_{3} + R_{2}$ 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 $R_{2} = - \frac{17}{9} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 3 \\ - \frac{17}{9} R_{3} + R_{2} & - \frac{17}{9} R_{3} + R_{2} & - \frac{17}{9} R_{3} + R_{2} & - \frac{17}{9} R_{3} + R_{2} \\ 0 & 0 & 1 & 9 \end{array}]$

$R_{2} = - \frac{17}{9} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{17}{9} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & 3 \\ (- \frac{17}{9}) \cdot (0) + 0 & (- \frac{17}{9}) \cdot (0) + 1 & (- \frac{17}{9}) \cdot (1) + \frac{17}{9} & (- \frac{17}{9}) \cdot (9) + 22 \\ 0 & 0 & 1 & 9 \end{array}]$

$R_{2} = - \frac{17}{9} R_{3} + R_{2}$

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

$[\begin{array}{c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 9 \end{array}]$

Find Reduced Row Echelon Form [[7,4,-1,32],[4,1,-3,-10],[1,2,1,22]]

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