Find Reduced Row Echelon Form [[3,-5,-16,-26],[1,-2,-6,-11]]

$[\begin{array}{c} 3 & - 5 & - 16 & - 26 \\ 1 & - 2 & - 6 & - 11 \end{array}]$

Step 1

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

$[\begin{array}{c} \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} \\ 1 & - 2 & - 6 & - 11 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (- 5) & (\frac{1}{3}) \cdot (- 16) & (\frac{1}{3}) \cdot (- 26) \\ 1 & - 2 & - 6 & - 11 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ 1 & - 2 & - 6 & - 11 \end{array}]$

Step 2

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 $0$ .

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (- \frac{5}{3}) - 2 & (- 1) \cdot (- \frac{16}{3}) - 6 & (- 1) \cdot (- \frac{26}{3}) - 11 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ 0 & - \frac{1}{3} & - \frac{2}{3} & - \frac{7}{3} \end{array}]$

Step 3

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

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ - 3 \cdot R_{2} & - 3 \cdot R_{2} & - 3 \cdot R_{2} & - 3 \cdot R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ (- 3) \cdot (0) & (- 3) \cdot (- \frac{1}{3}) & (- 3) \cdot (- \frac{2}{3}) & (- 3) \cdot (- \frac{7}{3}) \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{3} & - \frac{16}{3} & - \frac{26}{3} \\ 0 & 1 & 2 & 7 \end{array}]$

Step 4

Perform the row operation $R_{1} = \frac{5}{3} R_{2} + R_{1}$ 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 $R_{1} = \frac{5}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{5}{3} R_{2} + R_{1} & \frac{5}{3} R_{2} + R_{1} & \frac{5}{3} R_{2} + R_{1} & \frac{5}{3} R_{2} + R_{1} \\ 0 & 1 & 2 & 7 \end{array}]$

$R_{1} = \frac{5}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{5}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{5}{3}) \cdot (0) + 1 & (\frac{5}{3}) \cdot (1) - \frac{5}{3} & (\frac{5}{3}) \cdot (2) - \frac{16}{3} & (\frac{5}{3}) \cdot (7) - \frac{26}{3} \\ 0 & 1 & 2 & 7 \end{array}]$

$R_{1} = \frac{5}{3} R_{2} + R_{1}$

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

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

Find Reduced Row Echelon Form [[3,-5,-16,-26],[1,-2,-6,-11]]

Related Questions

Support LCMGCF.com by disabling your adblocker.