Algebra Examples

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ 6000 & 12000 & 24000 & 420000 \end{array}]$

Perform the row operation $R_{2} = - 6000 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 6000 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ - 6000 \cdot R_{1} + R_{2} & - 6000 \cdot R_{1} + R_{2} & - 6000 \cdot R_{1} + R_{2} & - 6000 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 6000 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 6000 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ (- 6000) \cdot (1) + 6000 & (- 6000) \cdot (1) + 12000 & (- 6000) \cdot (1) + 24000 & (- 6000) \cdot (28) + 420000 \end{array}]$

$R_{2} = - 6000 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ 0 & 6000 & 18000 & 252000 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{6000} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{1}{6000} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ \frac{1}{6000} R_{2} & \frac{1}{6000} R_{2} & \frac{1}{6000} R_{2} & \frac{1}{6000} R_{2} \end{array}]$

$R_{2} = \frac{1}{6000} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{6000} R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ (\frac{1}{6000}) \cdot (0) & (\frac{1}{6000}) \cdot (6000) & (\frac{1}{6000}) \cdot (18000) & (\frac{1}{6000}) \cdot (252000) \end{array}]$

$R_{2} = \frac{1}{6000} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 1 & 28 \\ 0 & 1 & 3 & 42 \end{array}]$

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 $0$ .

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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 $0$ .

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & 3 & 42 \end{array}]$

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (3) + 1 & (- 1) \cdot (42) + 28 \\ 0 & 1 & 3 & 42 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - 2 & - 14 \\ 0 & 1 & 3 & 42 \end{array}]$